11 121 132 143 154 165 176 187 198 209 22012 144 156 168 180
192 204 216 228 24013 169 182 195 208 221 234 247 26014 196 210 224
238 252 266 28015 225 240 255 270 285 30016 256 272 288 304 32017
289 306 323 34018 324 342 36019 361 38020 400For quickest math-1.
Many diagrams are given in almost every example which would be the
reminder of theprocess when ucompletely learn the techniques.2. Try
to understand the colors, every color is explaining the steps and
other important things.Edited byRAKES PRASADFirst we learn how to
find square, square root and cube roots with multiplication
techniques.Rule no 1. You are to remember squares of 1 to 32. They
are given as followsNo Square No Sq No Sq No Sqno sq1 1 9 81 17 289
25 625 33 10892 4 10 100 18 324 26 676 34 11563 9 11 121 19 361 27
729 35 12254 16 12 144 20 400 28 784 36 12965 25 13 169 21 441 29
841 38 14446 36 14 196 22 484 30 900 41 16817 49 15 225 23 529 31
9618 64 16 256 24 576 32 1024N.B. The square of 38 contains 444 and
the sq of an odd prime no 41 contains two perfectsquares as 42= 16
and 92= 81.Rule No 2. Squares of numbers ending in 5 :we are
showing by givinganexample.For the number 25, the last digit is 5
and the previous digit is 2. to multiply the previous digit 2by
onemore , that is, by 3. It becomes the L.H.S of the result, that
is, 2 X 3 = 6. The R.H.S of the resultis
52, that is, 25. Thus 252= 2 X 3 / 25 = 625. In the same way,
1052= 10 X 11/25 = 11025; 1352= 13 X14/25 = 18225; see the figure
below.Now we are extending the formula asRule no 3: If sum of the
last two digits give 10 then you can use the formula butL.H.S
shouldbe same. Check the examplesEx 1 : 47 X 43 .See the end digits
sum 7 + 3 = 10 ; then 47 x 43 = ( 4+ 1 ) x 4 / 7 x 3= 20 / 21 =
2021.Example 2: 62 x 682 + 8 = 10, L.H.S. portion remains the same
i.e.,, 6.Next of 6 gives 7 and 62 x 68 = ( 6 x 7 ) / ( 2 x 8 ) = 42
/ 16 =4216.Example 3: 127 x 123127 x 123 = 12 x 13 / 7 x 3 = 156 /
21 = 15621. (see the fig )Example 4. 39523952= 395 x 395 = 39 x 40
/ 5 x 5 = 1560 / 25 = 156025.You can rem the formula as
1252=(122+12)25=(144+12)25=15625. Bothare same things but it can
use in bigger case easily.Rule 4: Also we can extend the formula
where the sum of last two digitsbeing 5 asEx 1. 82 x 83=(82+8/2) /3
x 2=(64+4) / 06=6806 i.e.( n2+n/2) n is even but note thatthe R.H.S
term should be of two digits as 06 .( in this ex.)
Ex 2.181 x 184=(182+9) / 04=(324+9) / 04=33304.But we know what
are u thinking of? Yes, if the number be odd then whathappen?And ur
ans is as follows:Ex 3: 91 x 94= (92+9/2) / 04=(81+4) / 54.
Surprised? We just pass as 5 in the nextrow since belongs to
hundreds place so 1/2 x 100=50. Just pass 5 in suchcases.See the
fig.Ex 4 . 51 x 54= (25+2) / 54 =2754 Ex 5. 171 x 174= (289+8)
/54=29754Rule no 5 . Now we are discussing the general method of
finding squares ofanynumber. See the chart first.See 52=25And 102=
100 , so on.Now u can understand the tricks.base range Trick
alternative50 26-74 25()( read as 25 plusminus)100 76-126 100()2()
Given no()150 126-174 225()3()200 176-224 400()4()250 226-274
625()5()300 276-224 900()6()350 326-374 1225()7()400 376-424
1600()8()450 426-474 2025()9()500 476-524 2500()10() .. .1000
976-1024 100,00()20()1500 1476-1524 225,00,00()30()2000 1976-2024
4,000,000()40()For base 50: Now we are explaining one by one
through some examples.Ex 1. 432
step 1. Find base no as 50-43=07Step 2. Sq the no as
072=49.write it on the right side.Step 3. Subtract the no from 25
as 25-7=18.write it on extreme left.Step 4. Now this is Ur ans.
1849Ex 2. 572step 1. 57-50=7Step2. 72=49Step3. 25+7=32step4. Ans is
3249.Ex 3. 692step1. 69-60=19 , 192=361Step2. carry over 3 and
write 61 in right side.Step3 . 25+19=44, now add 44 with 3 (carry
number) and writeit with extreme left.i.e. (25+19+3)Step 4. ur ans
is 4761Ex4 . 38 2 step1. 50-38=12 , 122=144 ,carry 1 and write 44
at right side .Step2. now (25-12)+1=14 i.e. Carry number always
adds (itnever subtract). So ur ans is 1444.So, 1. At first see the
given no is it more or less than Ur base. If it is morethen
addition rule and if it be less, the n subtraction rule will
beapplied. Thats why () sign is given2. And always add the carried
no.3. range is chosen making () 24 from base .4. Trick is done by
squaring the base excluding a 0 (zero) such for base 50 :5 2
=25.for base 350 , 35 2 = 12255. for() divide the base without zero
by 5 as for base 100: 10/5=2,so trickis100 ()2() and for base 350:
35/5=7 ,and the trick is 1225()7() andso on.
Now we are giving more examples to clear the fact, a regular
practice can makeumaster of the art.ok, seeThe following
examples.For base 100Ex 5. 892Step1. 100-89=11 ,112=121 ,carry
1Step2. 100-(2 x 11 )+1=79, so the ans is 7921Also u can use the
alternative formula here,given no () such asStep 1. 100-89=11,sq is
121,carry 1Step2. 89-11+1=79 ,ass is 7921(I think this process is
better for numbers whose base is near 100)Ex 6. 117Step1.
117-17=17, 172=289 ,carry 2Step2. 117+17=134, 134+2=136 ,ans is
13689For base 150Ex 6. 1391. 150-139=11 , sq is 121 ,carry 12.
225-311=225-33=192, 192+1=193 ,ans is 19, 321Ex 7. 164 1.
164-150=14, 14 =196, carry 12. 225+3(14)=225+42=267 ,267+1=268 ,ans
is 26,896Ex 8. 512 1. 12 =144 ,carry 1, 2500+10(12)+1=2621, ans is
262144Try yourself to various numbers to learn the technique
quickly. now we learntofind the square root .before starting note
thatRule no 6:unit digit of perfectsquare1 4569unit digit of square
root 1,9(1+9)=102,82+8=1055+5=104,64+6=10
3,73+7=10Notes : 1 . any no end with 2,3,7,8 cant be a perfect
square.2. any no ends with odd no zeros cant be a perfect
square.Now check the given boxesrange sq range sq range sq range1=1
1-3 9=81 81-99 17=289 289-323 25=625 625-6752=4 4-8 10=100 100-120
18=324 324-360 26=676 676-7283=9 9-15 11=121 121-143 19=361 361-399
27=729 729-7834=16 16-24 12=144 144-168 20=400 400-440 28=784
784-8405=25 25-35 13=169 169-195 21=441 441-483 29=841 841-8996=36
36-48 14=196 196-224 22=484 484-528 30=900 900-9607=49 49-63 15=225
225-255 23=529 529-575 31=961 961-10238=64 64-80 16=256 256-288
24=576 576-624 32=1024 1024-1088Rule: 1. separate the two numbers
of extreme right.2. find the range of the square of the rest, write
it on left side.3. Multiply the range with its preceding number and
see that the separatednumber is more or less than the multiplied
number.4. if more, then choose the large no of unit digit as the
R.H.S.Ex1. (2601)1. 26 / 012. 26 falls on the range of 5 ,write 5
in the left.3. now 56=30 ,26 is smaller than 30.4. so choose 1 as
the right side no. ans is 51Ex2. (6241)1. 62 / 412. 62 falls in
range of 7 ,write 7 in left.3. 78=56 i.e.,62 is more than 56,so we
choose 9 as the rightsideEx3. 27041. 27 / 042. 27 falls in range of
5, so 56=30 ,but 27 is less than 30, soWe choose 2 between 2 &
8 (allocated for 4)3.so the ans is 52But it is easy if the number
ends with 5 such asEx 4. (99225)1. 992 /252. Here u can insert 5 as
right side because there is only 5 allocatedfor 5 and no choice
need here.
3. clearly 992 falls in the range of 31 and the ans is
315Exercise 1. 34225 2. 105625 3. (0.00126025) 4. 22095.
___________________2916 6. 2116 7. 15129 8. 161299. 55696 10. 66564
11. 8.8804 12. 0.0010112413. 0.125316Rule no 7: let us know how to
find cube root quickly. See the box first.unit digit ofperfect
cube1 2 3 4 5 6 7 8 9unit digit ofcube root1 82+8=1073+7=104 5 6
37+3=1028+2=109n.b. 1, 4 , 5, 6, 9 has no change as they were in
the sq formula and 2,3,7,8has therecompliment with 10, i.e.,
2+8=10, 3+7=10 etc.Ex1. (46656)1/31. 46 / 6562. write 6 for 6 on
right side3. see 46 falls in the range of 3 34.so the ans is 36cube
range cube range cube range1 =1 1-7 6 =216 216-342 11 =1331
1331-17272 =8 8-26 7 =343 343-511 12 =1728 1728-21963 =27 27-63 8
=512 512-728 13 =2197 2197-27434 =64 64-124 9 =729 729-999 14 =2744
2744-33745 =125 125-215 10 =1000 1000-1330 15 =3375
3375-409521=926125=15625Ex2 . cube of 0.00200483831. 0.0020048 /
3832. write 7 for 3 , and see 2048 falls in the range of 12
3. so the ns be 0 .123n.b. (se 3 places to shift 1 decemal
place. )Rule no 8: Let us know to find the cube of any two digits
number.Ex 1. (18) 3Step1. Find the ratio of the numbers such as 1:8
in this example.Step 2. Write cube of first digit and then write
three successive terms inhorizontal line which are multiplied by
the ratio. in this ex asStep 3. Make double of second and third
terms and write them just belowtheir own positionsStep 4. Add
successive terms (carry over if more than one digit) and u willget
the required result .Lets see the methods 1. 18 , that is 1:84 24
512. 1 3 =1 / 1x8=8 / 8 x 8=64 / 64 x 8 =5123. 8 x 2=16 64 x
2=1284.
------------------------------------------------------------( 4 +1)
( 24 +8+16) ( 51 +64+128) 51 25 = 4 8 = 24 35. So the ans is
5832.Ex 2. (33)3Rakes Prasad , 2008It contains some vedic methds
for multiplication of two,threeandmore digits,finding h.c.f. ofr
two polynomial equations.Edited byRakes Prasad, 2008Rule no (1) If
R.H.S. contains less number of digits than the number ofzeros in
the base, the remaining digits are filled up by giving zero or
zeroeson theleft side of the R.Note: If the number of digits are
more than the number of zeroes in hebase, the excess digit or
digits are to be added to L.H.S of theanswer.Case 1: Let N1 and N2
be two numbers near to a given base inpowers of 10, and D1 and D2
are their respectivedeviations(difference)from the base. Then N1 X
N2 can be represented asEx. 1: Find 97 X 94. Here base is 100.
.Ex. 2: 98 X 97 Base is 100.ans: 95/06=9506Ex. 3: 75X95. Base
is 100. .Ex. 4: 986 X 989. Base is 1000Ex. 5: 994X988. Base is 1000
.Ex. 6: 750X995Case 2:Ex. 7: 13X12. Base is 10.Ex. 8: 18X14. Base
is 10Ex. 9: 104X102. Base is 100.104 04102 02Ans is 106 / 4x2 =
10608 ( rule - f ) .Ex. 10: 1275X1004. Base is 1000.1275 2751004
004...............................................1279 /275x4 =1279
/ 1 / 100 (rule f) =1280100Case ( iii ): One number is more and the
other is less than the base.Ex.11: 13X7. Base is 10Ex. 12: 108 X
94. Base is 100.Ex. 13: 998 X 1025. Base is 1000.Find the following
products by the formula.1) 7 X 4 2) 93 X 85 3) 875 X 9944) 1234 X
1002 5) 1003 X 997 6) 11112 X 99987) 1234 X 1002 8) 118 X 105Rules
no 2,3,4 of chapter no 1 can also be extended asEg. 1: consider 292
x 208. Here 92 + 08 = 100, L.H.S portionis same i.e. 2292 x 208 = (
2 x 3 ) / 92 x 8 / =736 ( for 100 raise theL.H.S. product by 0 ) =
60736.Eg. 2: 848 X 852Here 48 + 52 = 100, L.H.S portion is 8 and
its next number is 9.848 x 852 = 8 x 9 / 48 x 52720 = 2496=
722496.[Since L.H.S product is to be multiplied by 10 and 2 to be
carriedover as the base is 100].Eg. 3: 693 x 607693 x 607 = 6 x 7 /
93 x 7 = 420 / 651 = 420651.Find the following products .
1. 318 x 312 2. 425 x 475 3. 796 x 7444. 902 x 998 5. 397 x 393
6. 551 x 549(2) One less than the previous1) The use of this sutra
in case of multiplication by 9,99,999.. is asfollowsa) The left
hand side digit (digits) is ( are) obtained by deduction 1fromthe
left side digit (digits) .b) The right hand side digit is the
complement or difference betweenthemultiplier and the left hand
side digit (digits)c) The two numbers give the answerExample 1: 8 x
9Step ( a ) gives 8 1 = 7 ( L.H.S. Digit )Step ( b ) gives 9 7 = 2
( R.H.S. Digit )Step ( c ) gives the answer 72Example 2: 15 x
99Step ( a ) : 15 1 = 14 Step ( b ) : 99 14 = 85 ( or 100 15 )Step
( c ) : 15 x 99 = 1485Example 3: 24 x 99 Answer :Example 5: 878 x
9999 Answer :find out the products64 x 99 723 x 999 3251 x 999943 x
999 256 x 9999 1857 x 99999(3) Multiplication of two 2 digit
numbers.Ex.1: Find the product 14 X 12The symbols are operated from
right to left .Step i) : Step ii) :Step iii) :Ex.4: 32 X 24Step (i)
: 2 X 4 = 8Step (ii) : 3 X 4 = 12; 2 X 2 = 4; 12 + 4 = 16.Here 6 is
to be retained. 1 is to be carried out to left side.Step (iii) : 3
X 2 = 6. Now the carried over digit 1 of 16 is to beadded. i.e., 6
+ 1 = 7. Thus 32 X 24 = 768Note that the carried over digit from
the result (3X4) + (2X2)= 12+4 = 16 i.e., 1 is placed under the
previous digit 3 X 2 = 6 andadded.After sufficient practice, you
feel no necessity of writing in this wayandsimply operate or
perform mentally.(4) Consider the multiplication of two 3 digit
numbers.
Ex 1. 124 X 132 =16368.Proceeding from right to lefti) 4 X 2 =
8. First digit = 8ii) (2 X 2) + (3 X 4) = 4 + 12 = 16. The digit 6
is retained and 1 is carriedover to left side. Second digit =
6.iii) (1 X 2) + (2 X 3) + (1 X 4) = 2 + 6 + 4 =12. The carried
over 1 ofabovestep is added i.e., 12 + 1 = 13. Now 3 is retained
and 1 is carried overto left side. Thus third digit = 3.iv) ( 1X 3
) + ( 2 X 1 ) = 3 + 2 = 5. the carried over 1 of above step isadded
i.e., 5 + 1 = 6 . It is retained. Thus fourth digit = 6v) ( 1 X 1 )
= 1. As there is no carried over number from the previousstep it is
retained. Thus fifth digit = 1(5) Cubing of Numbers:Example : Find
the cube of the number 106. We proceed as follows:i) For 106, Base
is 100. The surplus is 6. Here we add double of thesurplus i.e.
106+12 = 118. (Recall in squaring, we directly addthe surplus) This
makes the left-hand -most part of the answer.i.e. answer proceeds
like 118 / - - - - -ii) Put down the new surplus i.e. 118-100=18
multipliedby the initial surplus i.e. 6=108.Since base is 100, we
write 108 in carried over form 108 i.e. .As this is middle portion
of the answer, the answer proceedslike 118 / 108 /....iii) Write
down the cube of initial surplus i.e. 63 = 216 as the
lastportioni.e. right hand side last portion of the answer. Since
base is 100,write 216 as 216 as 2 is to be carried over. Answer is
118 / 108 / 216Now proceeding from right to left and adjusting the
carried over,we get the answer 119 / 10 / 16 = 1191016.Eg.(1): 1023
= (102 + 4) / 6 X 2 / 23 = 106/ 12 / 08 = 1061208.Observe initial
surplus = 2, next surplus =6 and base = 100.Eg.(2): 943Observe that
the nearest base = 100.i) Deficit = -6. Twice of it -6 X 2 = -12
add it to the number = 94 -12=82.ii) New deficit is -18. Product of
new deficit x initial deficit = -18 x -6 =108iii) deficit3 = (-6)3
= -216.__ Hence the answer is 82 / 108 / -216Since 100 is base 1
and -2 are the carried over. Adjusting the carriedover
in order, we get the answer ( 82 + 1 ) / ( 08 03 ) / ( 100 16
)= 83 / 05 / 84 = 830584 . 16 becomes 84 after taking1 from
middlemost portion i.e. 100. (100-16=84). _ Now 08 - 01 = 07
remains in themiddle portion, and 2 or 2 carried to it makes the
middleas 07 - 02 = 05. Thus we get the above result.Eg.(3): 9983
Base = 1000; initial deficit = - 2.9983 = (998 2 x 2) / (- 6 x 2) /
(- 2)3 = 994 / 012 / -008= 994 / 011 / 1000 - 008 = 994 / 011 / 992
= 994011992.Find the cubes of the following numbers using yavadunam
sutra.1. 105 2. 114 3. 1003 4. 10007 5. 926. 96 7. 993 8. 9991 9.
1000008 10. 999992.(6) Highest common factor:Example 1: Find the
H.C.F. of x2 + 5x + 4 and x2 + 7x + 6.1. Factorization method:x2 +
5x + 4 = (x + 4) (x + 1)x2 + 7x + 6 = (x + 6) (x + 1)H.C.F. is ( x
+ 1 ).2. Continuous division process.x2 + 5x + 4 ) x2 + 7x + 6 (
1x2 + 5x + 4___________2x + 2 ) x2 + 5x + 4 ( xx2 + x__________4x +
4 ) 2x + 2 ( 2x + 2______0Thus 4x + 4 i.e., ( x + 1 ) is H.C.F.OUR
PROCESSi.e.,, (x + 1) is H.C.FExample 2: Find H.C.F. of 2x2 x 3 and
2x2 + x 6Example 3: x3 7x 6 and x3 + 8x2 + 17x + 10.Example 4: x3 +
6x2 + 5x 12 and x3 + 8x2 + 19x + 12.(or)Example 5: 2x3 + x2 9 and
x4 + 2x2 + 9Add: (2x3 + x2 9) + (x4 + 2x2 + 9) = x4 + 2x3 + 3x2. x2
gives x2 + 2x + 3 ------ (i)Subtract after multiplying the first by
x and the second by 2.Thus (2x4 + x3 9x) - (2x4 + 4x2 + 18) = x3 -
4x2 9x 18 ------ ( ii )Multiply (i) by x and subtract from (ii)x3
4x2 9x 18 (x3 + 2x2 + 3x) = - 6x2 12x 18
- 6 gives x2 + 2x + 3.Thus ( x2 + 2x + 3 ) is the H.C.F. of the
given expressions.Find the H.C.F. in each of the following cases
using Vedic sutras:1 x2 + 2x 8, x2 6x + 82 x3 3x2 4x + 12, x3 7x2 +
16x - 123 x3 + 6x2 + 11x + 6, x3 x2 - 10x - 84 6x4 11x3 + 16x2 22x
+ 8, 6x4 11x3 8x2 + 22x 8.Edited byRakes Prasad , 2008Contains: (1)
Use the formula ALL FROM 9 AND THE LASTFROM 10 to perform instant
subtractions.(2) Multiplying numbers just over 100.(3) The easy way
to add and subtract fractions.(4) Multiplying a number by 11.(5)
Method for diving(6) SOME BASIC RELATIONS AND REVIEW OF SQUAREAND
CUBE FORMULA:(7) SQUARE ROOTS AND MULTIPLICATION FORMULA :(8) TWO
INTERESTING FACTS:(9) ROMAN NUMERALS(10) NUMER REPRESENTATION:(11)
DECIMAL REPRENTATION AND PREFIXS:(12) POWER OF INTIGERS:(13) SOME
PROOFS BUT WITHOUT WORDS:(14) THE LARGEST PRIME NUMBERS DISCOVERED
SOFAR:Edited byRakes Prasad, 2008(1) Use the formula ALL FROM 9 AND
THE LAST FROM 10 toperforminstant subtractions. For example 1000 -
357 = 643 We simply take each figure in 357 from 9 and thelast
figurefrom 10.So the answer is 1000 - 357 = 643And thats all there
is to it!This always works for subtractions from numbers consisting
of a 1 followed bynoughts: 100;1000; 10,000 etc.
Similarly 10,000 - 1049 = 8951For 1000 - 83, in which we have
more zeros than figures in the numbers beingsubtracted, wesimply
suppose 83 is 083. So 1000 - 83 becomes 1000 - 083 = 917Try some
yourself:1) 1000 - 777 2) 1000 - 283 3) 1000 - 505 4) 10,000 - 2345
5) 10000 - 9876 6)10,000 -1101(2) Multiplying numbers just over
100. 103 x 104 = 10712The answer is in two parts: 107 and 12, 107
is just 103 + 4 (or 104 + 3), and 12 isjust 3 x 4. Similarly 107 x
106 = 11342 107 + 6 = 113 and 7 x 6 = 42Again, just for mental
arithmetic Try a few:1) 102 x 107 = 2) 106 x 103 = 1) 104 x 104 =
4) 109 x 108 =(3) The easy way to add and subtract fractions.Use
VERTICALLY AND CROSSWISE to write the answer straight down!Multiply
crosswise and add to get the top of the answer: 2 x 5 = 10 and 1 x
3 = 3.Then 10 + 3 =13. The bottom of the fraction is just 3 x 5 =
15..You multiply the bottom number together.Subtracting is just as
easy: multiply crosswise as before, but the subtract:Try a few:(4)
Multiplying a number by 11.To multiply any 2-figure number by 11 we
just put the total ofthe two figures between the 2 figures. 26 x 11
= 286Notice that the outerfigures in 286 are the26 being
multiplied.And the middle figure is just 2 and 6 added up. So 72 x
11 = 792Multiply by 11:1) 43 = ) 81 = 3) 15 = 4) 44 = 5) 11 = 77 x
11 = 847This involves a carry figure because 7 + 7 = 14 we get 77 x
11 = 7147 = 847.Multiply by 11:1) 88 = 2) 84 = 3) 48 = 4) 73 = 5)
56 = 234 x 11 = 2574
We put the 2 and the 4 at the ends. We add the first pair 2 + 3
= 5. and we add thelast pair: 3 +4 = 7.Multiply by 11:1) 151 = 2)
527 = 3) 333 = 4) 714 = 5) 909 =(5) Method for diving by 9.23 / 9 =
2 remainder 5The first figure of 23 is 2, and this is the answer.
The remainder is just 2 and 3added up! 43 / 9 = 4 remainder 7The
first figure 4 is the answer and 4 + 3 = 7 is the remainder - could
it be easier?Divide by 9:1) 61 = remainder 2) 33 = remainder 3) 44
= remainder 4) 53 = remainder 134 / 9 = 14 remainder 8The answer
consists of 1,4 and 8.1 is just the first figure of 134.4 is the
total of thefirst twofigures 1+ 3 = 4,and 8 is the total of all
three figures 1+ 3 + 4 = 8.Divide by 9:6) 232 = 7) 151 = 8) 303 =
9) 212 = remainder remainder remainder remainder 842 / 9 = 812
remainder 14 = 92 remainder 14Actually a remainder of 9 or more is
not usually permitted because we are trying tofind howmany 9s there
are in 842.Since the remainder, 14 has one more 9 with 5 left over
the final answer will be 93remainder5Divide these by 9:1) 771 2)
942 3) 565 4) 555 5) 777 6) 2382(6) SOME BASIC RELATIONS AND REVIEW
OF SQUARE AND CUBEFORMULA:Relations :If a = b , b = c then a= cIf a
> b , b > c then a> cIf a < b , b < c then a <
cIf a b< b c then a < cIf ab > bc then b > cIf a > b
, c> d then a + c> b+ dIf a > b , c< d then a -c> b
dIf a < b , c< d then a -c< b -dIf a > b & a ,b
both positive then 1/a < 1/bx = n then x = n or -n 2 2(a + b) =
a + 2ab + b 2 2 2
(a b) = a 2ab + b 2 2 2(a + b) = a + 3a b + 3ab + b 3 3 2 2 3(a
- b) = a -3a b + 3ab -b 3 3 2 2 3(a + b+ c) = a + b + c + 2ab +
2bc+ 2ac 2 2 2 2Factorizationa b = ( a + b )(a b) 2 2a + b = (a +
b) ( a -ab + b ) 3 3 2 2a -b = (a -b ) (a + ab+ b ) 3 3 2
2Identities(a+ b) + (a b) = 2(a + b ) 2 2 2 2(a+ b) -(ab) = 4ab 2
2In dicesamxan=a(m+n)am/an=a(m-n)a=1 0a =1/a -mma = ma 1/m(a *b ) =
amx b m m(a /b ) = a mm/bm(a ) = a m n m*nLogarithmsa = n then log
a n= x xlog a (mn )= log am + log anlog a (m/n) = log a m -log
anlog a (m) = n log am nlog b n= log an/log abSurd( a ) = a n nn a
* b = ab n nn a / b = a /b n nm a = a = a n mn n mm a = a = a p p
mp p mAngle MeasurementTotal of Interior Angles in DegreesTriangle
180Rectangle 360Square 360Pentagon 540Circle 360Some FactsIn a
triangle, interior opposite angle is always less than exterior
angle.
Sum of 2 interior opposite angles of a triangle is always equal
to exterior angle.Triangle can have at one most obtuse angle.Angle
made by altitude of a triangle with side on which it is drawn is
equal to 90 degrees.in parallelogram opposite angles are
equal.SquaresNumber Square Number Square Number Square1 1 11 121 21
4412 4 12 144 22 4843 9 13 169 23 5294 16 14 196 24 5765 25 15 225
25 6256 36 16 256 26 6767 49 17 289 27 7298 64 18 324 28 7849 81 19
361 29 84110 100 20 400 30 900Cubes & Other PowersNumber ( X)
X3 X4 X511112 8 16 323 27 81 2434 64 256 10245 125 625 31256 216
1296 77767 343 2401 168078 512 4096 327689 729 6561 5904910 1000
10000 100000(7) SQUARE ROOTS AND MULTIPLICATION FORMULA :No Roots
No Roots No Roots No Roots1 1 5 2.23 9 3 36 62 1.41 6 2.44 10 3.16
49 73 1.73 7 2.64 16 4 64 84 2 8 2.82 25 5 81 91 to 10 :1 2 3 4 5 6
7 8 9 101 1 2 3 4 5 6 7 8 9 102 2 4 6 8 10 12 14 16 18 203 3 6 9 12
15 18 21 24 27 304 4 8 12 16 20 24 28 32 36 405 5 10 15 20 25 30 35
40 45 506 6 12 18 24 30 36 42 48 54 607 7 14 21 28 35 42 49 56 63
708 8 16 24 32 40 48 56 64 72 809 9 18 27 36 45 54 63 72 81 9010 10
20 30 40 50 60 70 80 90 10011to 20
11 12 13 14 15 16 17 18 19 201 11 12 13 14 15 16 17 18 19 202
22 24 26 28 30 32 34 36 38 403 33 36 39 42 45 48 51 54 57 604 44 48
52 56 60 64 68 72 76 805 55 60 65 70 75 80 85 90 95 1006 66 72 78
84 90 96 102 108 114 1207 77 84 91 98 105 112 119 126 133 1408 88
96 104 112 120 128 136 144 152 1609 99 108 117 126 135 144 153 162
171 18010 110 120 130 140 150 160 170 180 190 200(8) TWO
INTERESTING FACTS:(9) ROMAN NUMERALS(10) NUMER REPRESENTATION:(11)
DECIMAL REPRENTATION AND PREFIXS:(12) POWER OF INTIGERS:(13) SOME
PROOFS BUT WITHOUT WORDS:(14) THE LARGEST PRIME NUMBERS DISCOVERED
SO FAR:THE ENDEdited byRakes Prasad ,2008Contains: 1. Some
different techniques of squaringending with 1,2,3,4,6,7,8,9,etc .2.
adding different number sequence3. finding the difference of
squares.4. dividing a 6-digit/3-digit number
by13,7,37037,11,41,15873 etc5. dividing numbers by 75,125,625,12
2/3 etc..6. know why divisibility rules works?7. finding percentage
quickly.8. various types of multiplication by99,72,84,9. squaring
of some specialnumbersRakes Prasad2008.Squaring a 2-digit number
ending in 1
.Take a 2-digit number ending in 1. Subtract 1 from thenumber.
Squarethe difference. Add the difference twice to its square. Add
1.Example:If the number is 41, subtract 1: 41 - 1 = 40. 40 x 40 =
1600(square thedifference). 1600 + 40 + 40 = 1680 (add the
difference twice toits square).1680 + 1 = 1681 (add 1). So 41 x 41
= 1681.See the pattern?For 71 x 71, subtract 1: 71 - 1 = 70.70 x 70
= 4900 (square the difference).4900 + 70 + 70 = 5040 (add the
difference twice to itssquare). . So 71 x 71= 5041.Squaring a
2-digit number ending in 2I Take a 2-digit number ending in 2. f
The last digit will be _ __ 4. tMultiply the first digit by 4: the
2nd number will be h the nextto the lastdigit: _ _ X 4. e n Square
the first digit and add the numbercarried from uthe previous step:
X X _ _. m b e Example: r is 52, the last digit is _ _ _ 4. 4 x 5 =
20 (four times the firstdigit): _ _ 0 4. 5x 5 = 25 (square the
first digit), 25 + 2 = 27 (add carry): 2 7 0 4.For 82 x82, the last
digit is _ _ _ 4. 4 x 8 = 32 (four times the firstdigit): _ _ 2 4.
8 x8 = 64 (square the first digit), 64 + 3 = 67 (add carry): 6 7 2
4.Squaring a 2-digit number ending in 3 Take a 2-digitnumberending
in 3. 2 The last digit will be _ _ _ 9. 3 Multiply the firstdigit
by 6:
the 2nd number will be the next to the last digit: _ _ X 9.
4Square the firstdigit and add the number carried from the previous
step: X X_ _.Example:1 If the number is 43, the last digit is _ _ _
9. 2 6 x 4 = 24 (sixtimes thefirst digit): _ _ 4 9. 3 4 x 4 = 16
(square the first digit), 16 + 2 =18 (addcarry): 1 8 4 9.4 So 43 x
43 = 1849.See the pattern?1 For 83 x 83, the last digit is _ _ _ 9.
2 6 x 8 = 48 (six timesthe first digit):_ _ 8 9.38 x 8 = 6 4 ( s q
u a r e t he first digit), 64 + 4 = 68 (add carry):6 8 8 9. 4So 83
x 83 = 6889.Squaring a 2-digit number ending in 41 Take a 2-digit
number ending in 4. 2 Square the 4; the lastdigit is 6: _ __ 6
(keep carry, 1.)3 Multiply the first digit by 8 and add the carry
(1); the 2ndnumber will bethe next to the last digit: _ _ X 6 (keep
carry). 4 Square thefirst digit andadd the carry: X X _ _.Example:1
If the number is 34, 4 x 4 = 16 (keep carry, 1); the last digitis _
_ _ 6. 2 8x 3 = 24 (multiply the first digit by 8), 24 + 1 = 25
(add thecarry): the nextdigit is 5: _ _ 5 6. (Keep carry, 2.) 3
Square the first digit andadd the
carry, 2: 1 1 5 6. 4 So 34 x 34 = 1156.See the pattern?1 For 84
x 84, 4 x 4 = 16 (keep carry, 1);the last digit is _ _ _ 6.2 8 x 8
= 64 (multiply the first digit by 8),64 + 1 = 65 (add the
carry):the next digit is 5: _ _ 5 6. (Keep carry, 6.)Square the
first digit and add the carry, 6: 7 0 5 6.So 84 x 84 = 7056.See
previous chapters1 Choose a 2-digit number ending in 6. 2 Square
the seconddigit (keepthe carry): the last digit of the answer is
always 6: _ _ _ 63 Multiply the first digit by 2 and add the carry
(keep thecarry): _ _ X _4 Multiply the first digit by the next
consecutive number andadd thecarry: the product is the first two
digits: XX _ _.Example:1 If the number is 46, square the second
digit : 6 x 6 = 36; thelast digit ofthe answer is 6 (keep carry 3):
_ _ _ 6 2 Multiply the first digit(4) by 2 andadd the carry (keep
the carry): 2 x 4 = 8, 8 + 3 = 11; the nextdigit of theanswer is 1:
_ _ 1 63 Multiply the first digit (4) by the next number (5) and
addthe carry: 4 x5 = 20, 20 + 1 = 21 (the first two digits): 2 1 _
_4 So 46 x 46 = 2116.See the pattern?1 For 76 x 76, square 6 and
keep the carry (3): 6 x 6 = 36; thelast digit ofthe answer is 6: _
_ _ 6 2 Multiply the first digit (7) by 2 andadd the carry:
2 x 7 = 14, 14 + 3 = 17; the next digit of the answer is 7
(keepcarry 1): _ _7 6 3 Multiply the first digit (7) by the next
number (8) and addthe carry:7 x 8 = 56, 56 + 1 = 57 (the first two
digits: 5 7 _ _ 4 So 76 x 76= 5776.1 Choose a 2-digit number ending
in 7. 2 The last digit of theanswer isalways 9: _ _ _ 9 3 Multiply
the first digit by 4 and add 4 (keepthe carry):__X_4 Multiply the
first digit by the next consecutive number andadd thecarry: the
product is the first two digits: XX _ _.Example:1 If the number is
47: 2 The last digit of the answer is 9: _ _ _9 3 Multiplythe first
digit (4) by 4 and add 4 (keep the carry): 4 x 4 = 16,16 + 4 =
20;the next digit of the answer is 0 (keep carry 2): _ _ 0 94
Multiply the first digit (4) by the next number (5) and addthe
carry (2): 4x 5 = 20, 20 + 2 = 22 (the first two digits): 2 2 _ _5
So 47 x 47 = 2209.See the pattern?1 For 67 x 67 2 The last digit of
the answer is 9: _ _ _ 9 3Multiply the firstdigit (6) by 4 and add
4 (keep the carry): 4 x 6 = 24, 24 + 4 =28; the nextdigit of the
answer is 0 (keep carry 2): _ _ 8 9 4 Multiply thefirst digit (6)by
the next number (7) and add5 the carry (2): 6 x 7 = 42, 42 + 2 = 44
(the first two digits): 4 6So 67 x 67= 4489.
1 Choose a 2-digit number ending in 8. 2 The last digit of
theanswer isalways 4: _ _ _ 4 3 Multiply the first digit by 6 and
add 6 (keepthe carry):__X_4 Multiply the first digit by the next
consecutive number andadd thecarry: the product is the first two
digits: XX _ _.Example:1 If the number is 78: 2 The last digit of
the answer is 4: _ _ _4 3 Multiplythe first digit (7) by 6 and add
6 (keep the carry): 7 x 6 = 42,42 + 6 = 48;the next digit of the
answer is 8 (keep carry 4): _ _ 8 4 4Multiply the firstdigit (7) by
the next number (8) and add the carry (4): 7 x 8 =56, 56 + 4 =60
(the first two digits): 6 0 _ _ 5 So 78 x 78 = 6084.See the
pattern?1 For 38 x 38The last digit of the answer is 4: _ _ _
4Multiply the first digit (3) by 6 and add 6 (keep thecarry): 3 x 6
= 18, 18 + 6 = 24; the next digit of theanswer is 4 (keep carry 2):
_ _ 4 4Multiply the first digit (3) by the next number (4)and add
the carry (2):3 x 4 = 12, 12 + 2 = 14 (the first two digits): 1 4 _
_So 38 x 38 = 1444Squaring a 2-digit number ending in 91 Choose a
2-digit number ending in 9. 2 The last digit of theansweris
always1. Multiply the first digit by 8 and add 8 (keep the carry):
_ _ X_3
Multiply the first digit by the next consecutive number andadd
thecarry: the product is the first two digits: XX _ _.Example:1 If
the number is 39: 2 The last digit of the answer is 1: _ _
_13Multiply the first digit (3) by 8 and add 8 (keep the carry): 8
x3=24, 24 + 8 = 32; the next digit of the answer is 2 (keep
carry3): _ _2 1 4 Multiply the first digit (3) by the next number
(4) and addthecarry (3): 3 x 4 = 12, 12 + 3 = 15 (the first two
digits): 1 5 _ _5 So 39 x 39 = 1521. rn?1 For 79 x 792 The last
digit of the answer is 1: _ _ _ 1Multiply the first digit (7) by 8
and add 8 (keep the carry): 8 x7 = 56,56 + 8 = 64; the next digit
of the answer is 4 (keep carry 6): __41Multiply the first digit (7)
by the next number (8) and add thecarry (6): 7 x 8 = 56, 56 + 6 =
62 (the first two digits): 6 2 _ _So 79 x 79 = 6241...Choose two
2-digit numbers less than 20 (no limits forexperts). Add allthe
numbers between them:1 Add the numbers; 2 Subtract the numbers and
add 1; 3Multiply halfthe sum by this difference + 1, OR Multiply
the sum by halfthe difference+ 1.Example:
1 If the two numbers selected are 6 and 19: 2 Add thenumbers: 6
+ 19 =25. 3 Subtract the numbers: 19 - 6 = 13. Add 1: 13 + 1 = 14.
4Multiply 25by half of 14: 25 x 7 = 175. 5 So the sum of the
numbers from6 through 19is 175.Therefore
6+7+8+9+10+11+12+13+14+15+16+17+18+19=175See the pattern?1 If the
two numbers selected are 4 and 18: 2 Add thenumbers: 4 + 18 =22. 3
Subtract the numbers: 18 - 4 = 14. Add 1: 14 + 1 = 15.Multiply half
of 22 by 15: 11 x 15 = 165 (10 x 15 + 15). So thesum of thenumbers
from 4 through 18 is 165.1 Choose a 2-digit odd number. Add all the
odd numbersstarting withone through this 2-digit number: 2 Add one
to the 2-digitnumber. 3Divide this sum by 2 (take half of it). 4
Square this number.This is thesum of all odd numbers from 1 through
the 2-digit numberchosen.Example:1 If the 2-digit odd number
selected is 35: 2 35+1 = 36 (add 1).3 36/2 = 18(divide by 2) or 1/2
x 36 = 18 (multiply by 1/2). 4 18 x 18 = 324(square 18):18 x 18 =
(20 - 2)(18) = (20 x 18) - (2 x 18) = 360 - 36 = 360 -30 -6 = 324.
5So the sum of all the odd numbers from one through 35 is324.See
the pattern?1 If the 2-digit odd number selected is 79: 2 79+1 = 80
(add 1).3 80/2 = 40
(divide by 2) or 1/2 x 80 = 40 (multiply by 1/2). 4 40 x 40 =
1600(square40).5. So the sum of all the odd numbers from one
through 79 is1600.1 Have a friend choose a a single digit number.
(Norestrictions forexperts.) 2 Ask your friend to jot down a series
of doubles(where thenext term is always double the preceding one),
and tell youthe last term.3 Ask your friend to add up all these
terms. 4 You will givethe answerbefore he or she can finish: The
sum of all the terms of thisseries will betwo times the last term
minus the first term.Example:if the number selected is 9: 1 The
series jotted down is: 9, 18,36, 72, 144.2 Two times the last term
(144) minus the first (9): 2 x 144 =288; 288 - 9 =279.3 So the sum
of the doubles from 9 through 144 is 279.See the pattern? Heres one
for the experts:1 The number selected is 32: 2 The series jotted
down is: 64,128, 256,512. 3 Two times the last term (512) minus the
first (64): 2 x512 = 1024;1024 - 32 = 1024 - 30 - 2 = 994 - 2 =
992.4 So the sum of the doubles from 32 through 512 is 992.Remember
to subtract in steps from left to right. Withpractice you will
beexpert in summing series.restrictions for experts.) Ask your
friend to jot down a seriesof
quadruples (where the next term is always four times
thepreceding one),and tell you only the last term. Ask your friend
to add up allthese terms.You will give the answer before he or she
can finish: Thesum of all theterms of this series will be four
times the last term minus thefirst term,divided by 3. Example:If
the number selected is 5:1 The series jotted down is: 5, 20, 80,
320, 1280. 2 Four timesthe last term(1280) minus the first (5):
4000 + 800 + 320 - 5 = 5120 - 5 =5115 Divide by3: 5115/3 = 17053 So
the susum of the quadruples from 5 through 1280 is1705.See the
pattern? Heres one for the experts:1 The number selected is 32: 2
The series jotted down is: 32,128, 512,2048. 3 Four times the last
term (2048) minus the first (32):8000 + 160 +32 - 32 = 8,160Divide
by 3: 8160/3 = 2720.4. So the sum of the quadruples from 32 through
2048 is2720.Practice multiplying from left to right and dividing by
3. Withpractice youwill be anexpert quad adder.Add a sequence from
one to a selected 2-digit number1 Choose a 2-digit number. 2
Multiply the 2-digit number byhalf the nextnumber, orMultiply half
the 2-digit number by the nextnumber.
Example:1 If the 2-digit even number selected is 51: 2 The
nextnumber is 52.Multiply 51 times half of 52. 3 51 x 26: (50 x 20)
+ (50 x 6) + 1 x26) = 1000+ 300 + 26 = 13264 So the sum of all
numbers from 1 through 51 is 1326.See the pattern?1 If the 2-digit
even number selected is 34: 2 The nextnumber is 35.Multiply half of
34 x 35. 3 17 x 35: (10 x 35) + (7 x 30) + (7 x 5)= 350 + 210+ 35 =
560 + 35 = 5954 So the sum of all numbers from 1 through 34 is
595.With some multiplication practice you will be able to findthese
sums ofsequential numbers easily and faster than someone using
acalculator!1 Choose a 1-digit number. 2 Square it.Example:1 If the
1-digit number selected is 7: 2 To add 1 + 2 + 3 + 4 + 5+6+7+6+5 +
4 + 3 + 2 + 1 3 Square 7: 49 4 So the sum of all numbersfrom
1through 7 and back is 49.See the pattern?1 If the 1-digit number
selected is 9: 2 To add 1 + 2 + 3 + 4 + 5+6+7+8+9 + 8 + 7 + 6 + 5 +
4 + 3 + 2 + 1 3 Square 9: 81 4 So the sum ofall numbersfrom 1
through 9 and back is 81.Add sequences of numbers in the 10s1
Choose a 2-digit number in the 10s. To add all the 10sfrom 10
up
through this number and down from it: 2 Square the 2nd digitof
thenumber (keep the carry) _ _ X 3 The number of terms is 2 xthe
2nd digit +1. 4 The first digits = the number of terms (+ carry).
nbsp; X X_Example:1 If the 2-digit number in the 10s selected is
16: (10 + 11 +12 ... 16 + 15 +14 ... 10) 2 Square the 2nd digit of
the number: 6 x 6 = 36(keep carry 3) _ _6 3 No. of terms = 2 x 2nd
digit + 1: 2 x 6 + 1 = 134 No. of terms (+ carry): 13 + 3 = 16 1 6
_ 5 So the sum of thesequence is166.See the pattern?1. If the
2-digit number in the 10s selected is 18:(10 + 11 + 12 + ... 18 +
17 + 16 + 15 ... 10)1 Square the 2nd digit of the number: 8 x 8 =
64 (keep carry6) _ _ 4 2 No.of terms = 2 x 2nd digit + 1: 2 x 8 + 1
= 17 3 No. of terms (+carry): 17 + 6 =23 2 3 _ 4 So the sum of the
sequence is 234.1 Choose a 2-digit number in the 20s. To add all
the 20sfrom 20 upthrough this number and down from it: 2 Square the
2nd digitof thenumber (keep the carry) _ _ X 3 The number of terms
is 2 xthe 2nd digit +1. 4 Multiply the number of terms by 2 (+
carry). X X _Example:1 If the 2-digit number in the 20s selected is
23: (20 + 21 + 22+ 23 + 22 +21 + 20) 2 Square the 2nd digit of the
number: 3 x 3 = 9 _ _ 9 3No. of
terms = 2 x 2nd digit + 1: 2 x 3 + 1 = 7 4 2 x no. of terms: 2
x 7= 14 1 4 _ 5So the sum of the sequence is 149.See the pattern?1
If the 2-digit number in the 20s selected is 28: (20 + 21 +22 ... +
28 + 27+ ... 22 + 21 + 20) 2 Square the 2nd digit of the number: 8
x 8= 64 (keepcarry 6) _ _ 4 3 No. of terms = 2 x 2nd digit + 1: 2 x
8 + 1 = 174 2 x no. ofterms (+ carry): 2 x 17 + 6 = 40 4 0 _ 5 So
the sum of thesequence is 404.2 S 1 Choose a 2-digit number in the
30s. To add all the 30sqfrom 30 upthrough this number and down from
it: uare the 2nd digit of the number (keep the carry) _ _ X 3
Thenumber ofterms is 2 x the 2nd digit + 1. 4 Multiply the number
of termsby 3 (+carry) X X _Example:1 If the 2-digit number in the
30s selected is 34: (30 + 31 + 32+ 33 + 34 +33 + 32 + 31 + 30) 2
Square the 2nd digit of the number: 4 x 4= 16 (keepcarry 1) _ _ 6 3
No. of terms = 2 x 2nd digit + 1: 2 x 4 + 1 = 94 3 x no. of terms:
3 x 9 + 1 = 28 2 8 _ 5 So the sum of thesequence is286.See the
pattern?1 If the 2-digit number in the 30s selected is 38: (30 + 31
+ 232 + ... + 38 +37 + ... 32 + 31 + 30) 3 Square the 2nd digit of
the number: 8 x8 = 64 (keepcarry 6) _ _ 4
No. of terms = 2 x 2nd digit + 1: 2 x 8 + 1 = 173 x no. of
terms: 3 x 17 + 6 = 51 + 6 = 57 5 7 _ So the sum ofthe sequenceis
574.1 Choose a 2-digit number in the 40s. To add all the 40sfrom 40
up through this number and down from it:2 Square the 2nd digit of
the number(keep the carry) _ _X3 The number of terms is 2 x the 2nd
digit + 1.4 Multiply the number of terms by 4 (+ carry) X X _e
2-digit number in the 40s selected is 43: (40 + 41 + 42 + 43+ 42 +
41 +40) 1 Square the 2nd digit of the number: 3 x 3 = 9 _ _ 9 2
No.of terms = 2x 2nd digit + 1: 2 x 3 + 1 = 73 4 x no. of terms: 4
x 7 = 28 2 8 _ 4 So the sum of thesequence is 289.See the pattern?1
If the 2-digit number in the 40s selected is 48: (40 + 41 + 42+ ...
+ 48 +47 + ... 42 + 41 + 40) 2 Square the 2nd digit of the number:
8 x8 = 64 (keepcarry 6) _ _ 4 3 No. of terms = 2 x 2nd digit + 1: 2
x 8 + 1 = 174 4 x no. ofterms: 4 x 17 + 6 = 40 + 28 + 6 = 68 + 6 =
74 7 4 _5 So the sum of the sequence is 744.1 Choose a 2-digit
number in the 50s. To add all the 50sfrom 50 upthrough this number
and down from it: 2 Square the 2nd digitof thenumber (keep the
carry) _ _ X 3 The number of terms is 2 xthe 2nd digit +1. 4
Multiply the number of terms by 5. X X _Example:
1 If the 2-digit number in the 50s selected is 53: (50 + 51 +
52+ 53 + 52 +51 + 50) 2 Square the 2nd digit of the number: 3 x 3 =
9 _ _ 9 3No. ofterms = 2 x 2nd digit + 1: 2 x 3 + 1 = 7 4 5 x no.
of terms: 5 x 7= 35 3 5 _ 5So the sum of the sequence is 359.See
the pattern?1 If the 2-digit number in the 50s selected is 57: (50
+ 51 + 52 ... + 57 + 56 + ... 52 + 51 + 50) 2 Square the 2nd digit
of thenumber: 7 x 7= 49 (keep carry 4) _ _ 93 No. of terms = 2 x
2nd digit + 1: 2 x 7 + 1 = 155 x no. of terms (+ carry): 5 x 15 + 4
= 75 + 4 = 79 7 9 _So the sum of the sequence is 799.Add sequences
of numbers in the 60s1 Choose a 2-digit number in the 60s. To add
all the 60sfrom 60 up throughthis number and down from it: 2 Square
the 2nd digit of thenumber (keepthe carry) _ _ X 3 The number of
terms is 2 x the 2nd digit + 1.4 Multiply thenumber of terms by 6
(+ carry) X X _Example:1 If the 2-digit number in the 60s selected
is 63: (60 + 61 + 62+ 63 + 62 + 61 +60) 2 Square the 2nd digit of
the number: 3 x 3 = 9 _ _ 9 3 No.of terms = 2 x2nd digit + 1: 2 x 3
+ 1 = 745 6 x no. of terms: 6 x 7 = 42 4 2 _ So the sum of
thesequence is 429.See the pattern?1 If the 2-digit number in the
60s selected is 68: (60 + 61 +62 + ... + 68 + 67 + ... 62 + 61 +
60) 2 Square the 2nd digit of
the number: 8 x 8 = 64 (keep carry 6) _ _ 4 3 No. of terms =2 x
2nd digit + 1: 2 x 8 + 1 = 17 4 6 x no. of terms: 6 x 17 +6 = 42 +
1 = 102 + 6 = 108 1 0 8 _5oSthe sum of the sequence is 1084.1
Choose a 2-digit number in the 70s. To add all the 70sfrom 70
upthrough this number and down from it: 2 Square the 2nd digitof
thenumber (keep the carry) _ _ X3 The number of terms is 2 x the
2nd digit + 1. 4 Multiply thenumber ofterms by 7 (+ carry) X X
_Example:1 If the 2-digit number in the 70s selected is 73: (70 +
71 + 72+ 73 + 72 +71 + 70) 2 Square the 2nd digit of the number: 3
x 3 = 9 _ _ 9 3No. ofterms = 2 x 2nd digit + 1: 2 x 3 + 1 = 74 7 x
no. of terms: 7 x 7 = 49 4 9 _ 5 So the sum of thesequence is
499.See the pattern?1 If the 2-digit number in the 70s selected is
78: (70 + 71 + 72+ ... + 78 +77 + ... 72 + 71 + 70) 2 Square the
2nd digit of the number: 8 x8 = 64 (keepcarry 6) _ _ 4 3 No. of
terms = 2 x 2nd digit + 1: 2 x 8 + 1 = 174 7 x no. ofterms: 7 x 17
+ 6 = 49 + 6 = 119 + 6 = 125 1 2 5 _5 So the sum of the sequence is
1254.1 Choose a 2-digit number in the 80s. To add all the80sfrom 80
up through this number and down from it:2 Square the 2nd digit of
the number(keep the carry) _ _
umber of terms is 2 x the 2nd digit + 1 4 Multiply the numberof
terms by8 (add the carry) X X _Example:1 If the 2-digit number in
the 80s selected is 83: (80 + 81 + 82+ 83 + 82 +81 + 80) 2 Square
the 2nd digit of the number: 3 x 3 = 9 _ _ 9 3No. ofterms = 2 x 2nd
digit + 1: 2 x 3 + 1 = 7 4 8 x no. of terms: 8 x 7= 56 5 6 _ 5So
the sum of the sequence is 569.See the pattern?1 If the 2-digit
number in the 80s selected is 88: (80 + 81 +82 ... + 88 + 87+ ...
83 + 82 + 81 + 80) 2 Square the 2nd digit of the number: 8x 8 =
64(keep carry 6) _ _ 4 3 No. of terms = 2 x 2nd digit + 1: 2 x 8 +
1= 178 x no. of terms (+ carry): 8 x 17 + 6 = 80 + 56 + 6 = 136 + 6
=42 4 2 _So the sum of the sequence is 1424.1 Have a friend choose
and write down a single-digit number.(Twodigits for experts.) 2 Ask
your friend to name and note a thirdnumberby adding the first two.
3 Name a fourth by adding thesecond and third.Continue in this way,
announcing each number, through tennumbers. 4Ask your friend to add
up the ten numbers. You will give theanswerbefore he or she can
finish:The sum of all the terms of this series will be the
seventhnumbermultiplied by 11.
Example:1 If the numbers selected are 7 and 4:e series jotted
down is: 4, 7, 11, 18, 29, 47, 76, 123, 199, 322. 3Theseventh
number is 76. 11 x 76 = 836 (use the shortcut for 11:7 is the
firstdigit, 6 is the third digit; the middle digit will be 7 + 6,
andcarry the 1:836).4 So the sum of the ten numbers is 836.Here are
some of the calculations that you can do mentallyafterpracticing
the exercises. See how many you can do. Checkyour answerswith a
calculator. Your mental math powers should beimpressive!15 x 15 89
x 89 394 x 101 32 x 38 51 x 59101 x101 x 94 93 x 93 228 x 101 25 x
25147 56 x 56 94 x 96 101 x 448 83 x 87 687 x 101 64 x 66 52 x
52101 x 65435 x 35 11 x 19 101 x61 x 69 89 x 101 61 x 61 34 x 36206
101 x 48 52 x 52 45 x 45 456 x 101 54 x 54 41 x 41 101 x 4132 x 32
83 x101 29 x 29 55 x 55 63 x 67 479 x 101 82 x 82 882 x 101 14 x
16319 x 10173 x 77 65 x 65 101 x 859 13 x 17 101 x 149 41 x 49 82 x
101 75x 75 101 x71 x 79 69 x 101 22 x 22 993 x 101144 51 x 51 85 x
85 101 x 101 129 x 101 69 x 69 31 x 39 738 x101 21 x 2971 x 71 77 x
101 95 x 95 265 x 101 74 x 76 53 x 53 101 x 409 51x 51 98 x101 94 x
94 42 x 42 53 x 57 339 x
62 x 68 101 x 88 19 x 19 238 x 10110143 x 47 96 x 96 21 x 21
101 x 279 81 x 89648 x62 x 62 101 x 57 24 x 26 101 x 668101 99 x 99
23 x 27 22 x 28 515 x 101 97 x 97 79 x 79 101 x 826245 x 10131 x 31
37 x 33 101 x82 x 88 39 x 39 126 x 101 91 x 9921868 x 101 51 x 101
598 x 101 98 x 98 72 x 7291 x 91 54 x 56 101 x 29 57 x 57 771 x 101
81 x 81 559 x 101 52x 58 59 x 5984 x 86 46 x 44 101 x 25 12 x 18
101 x 697 92 x 92 349 x93 x 97 101 x 188 42 x 48 49 x 4910158 x 58
92 x 92 101 x 78 37 x 1011 Select two consecutive 2-digit numbers.
2 Add the two 2-digit numbers!Examples:1 24 + 25 = 49. (Try it on a
calculator and see, or if yourereally sharp, doit mentally: 24 x 24
= 576, 25 x 25 = 625, 625 - 576 = 49.) 2 If 63and 64 areselected,
then 63 + 64 = 127. (For larger number addition, doit in steps:63 +
64 = 63 + 60 + 4 = 123 + 4 = 127.)1 Select two consecutive 2-digit
numbers, one not more than10 largerthan the other (experts need not
use this limitation). 2Subtract thesmaller number from the larger.
3 Add the two numbers. 4Multiply thefirst answer by the
second.Examples:
1 If 71 and 64 are selected: 2 71 - 64 = 7. 3 71 + 64 =Add left
to right: 71 + 64 = 71 + 60 + 4 = 131 + 4 = 135)4 Multiply these
results: 7 x 135 = 945 (Multiply left to right: 7x 135 = 7
x(100+30+5) = 700 + 210 + 35 = 910 + 35 = 945)5 So the difference
of the squares of 71 and 64 is 945.See the pattern?1 If 27 and 36
are selected: 3 2 36 - 27 = 9.36 + 27 = 63 (Think: 27 + 30 + 6 = 57
+ 6 = 63)Multiply these results: 9 x 63 = 567 (Think: 9 x (60+3) =
540 +27 = 567)So the difference of the squares of 27 and 36 is
567.Why do the divisibility rules work?From: [email protected] (Cindy
Smith) To: Dr. Math Subject: Factoring TricksIn reading a popular
math book, I came across severalarithmeticfactoring tricks.
Essentially, if the last digit of a number iszero,then the entire
number is divisible by 10. If the last number iseven, then the
entire number is divisible by 2. If the last twodigits are
divisible by 4, then the whole number is. If the lastthree digits
divide by 8, then the whole number does. If thelastfour digits
divide by 16, then the whole number does, etc. Ifthelast digit is
5, then the whole number divides by 5.Now for the tricky ones.If
you add the digits in a number and the sum is divisible by3,then
the whole number is. Similarly, fif you add the digits ina number
and the sum is divisible by 9, then the wholenumber is. For
example, take the number 1233: 1 + 2 + 3 + 3 =9.
Therefore, the whole number is divisible by 9 and thequotient
is137. The number is also divisible by 3 and the quotient is411.It
works for extremely large numbers too (I checked on mycalculator).
Now heres a really tricky trick. You add up thealternate digits of
a number and then add up the other setof alternate digits. If the
sums of the alternate digits equaleach other then the whole number
is divisible by 11. Also,if the difference of the alternate digits
is 11 or a multiple of11, then the whole number is divisible by 11.
For example,123,456,322. 1 + 3 + 5 + 3 + 2 = 14 and 2 + 4 + 6 + 2 =
14.Therefore, the whole number is divisible by 11 and thequotientis
11,223,302. Also, if a number is divisible by both 3 and 2,thenthe
whole number is divisible by 6.The only single digit number for
which there is no trick listedis 7.I find these rules interesting
and useful, especially whenfactoringlarge numbers in algebraic
expressions. However, Im notsurewhy all these rules work. Can you
explain to me why thesemathtricks work? If I could understand why
they work, I think itwouldimprove my math skills. Thanks in advance
for your help.Cindy Smith [email protected]: Dr. Math
To:[email protected] (Cindy Smith) Subject: Re: Factoring TricksThank
you for this long and very well-written question. I willtry towrite
as clearly as you as I answer. All these digital tests for
divisibility are based on the fact that our system of
numeralsiswritten using the base of 10.The digits in a string of
digits making up a numeral areactually thecoefficients of a
polynomial with 10 substituted for thevariable. Forexample, 1233 =
1*10^3 + 2*10^2 + 3*10^1 + 3*10^0 which isgottenfrom the polynomial
1*x^3 + 2*x^2 + 3*x^1 + 3*x^0 = x^3 +2*x^2 +3*x + 3 by substituting
10 for x. We can explain each of thesetricksin terms of that fact:
10 - numerals ending in 0 representnumbersdivisible by 10:Since the
last digit is zero, and all other terms in thepolynomial formare
divisible by 10, the number is divisible by 10. Similarly,
ifthenumber is divisible by 10, since all the terms except the
lastoneare automatically divisible by 10 no matter what
thecoefficients ordigits are, the number will be divisible by 10
only if the lastdigit is.Since all the digits are smaller than 10,
the last digit has to be0 tobe a multiple of 10.2 - numerals ending
in an even digitrepresentnumbers divisible by 2: f o Same argument
as above about alltheterms except the last one being divisible by
2. The last digitis
divisible by 2 (even) if and only if the whole number is. R 4 -
numerals endingwitha two-digit multiple of 4 represent numbers
divisible by 4:o fSimilar to the above, but since 10 is not a
multiple of 4, but10^2 is,we have to look at the last two digits
instead of just the lastdigit.a n 8 - numerals ending with a
three-digit multiple of 8representnumbers divisible by 8: u Similar
to 4, but now 10^2 is not amultipleof 8, but 10^3 is, so we have to
look at the last three mdigits.era16 - You figure this one!l5 - You
figure this one, too!ook3 - numerals whose sum of digits is
divisible by 3 representnumberss divisible by 3: This one is
different, because 3 does notdivide anypower of 10 evenly. That
means we will have to consider theeffect ofall the l digits. Here
we use this fact: 10^k - 1 = (10 - 1)*(10^(k-1) = ... +10^2 + i 10
+ 1) This is a fancy way of saying 9999...999 =9*1111...111. We use
k this to rewrite our powers of 10 as 10^k = 9*a[k] +1. Now
thepolynomial e this: d[k]*10^k + d[k-1]*10^(k-1) + ... + d[1]*10
+d[0] = d[k]*(9*a[k] + 1) + ... + d[1]*(9*a[1] + 1) + d[0] =
9*(d[k]*a[k] + ... +d[1]*a[1])
+ d[k] + ... + d[1] + d[0] Now notice that 3 divides the
firstpart, so thewhole number is divisible by 3 if and only if the
sum of the digits is.9 - numerals whose sum of digits is divisible
by 9 representnumbersdivisible by 9: Use the same equation as the
previous case.You figurethe rest!11 - numerals whose alternating
sum of digits is divisible by11 representnumbers divisible by
11:Here the phrase "alternating sum" means we alternate thesigns
frompositive to negative to positive to negative, and so on. Weuse
this fact:10 to an odd power plus 1 is divisible by 11, and 10 to
aneven powerminus 1 is divisible by 11. The first part is a fancy
way ofwriting 10000...0001 = 11*9090...9091 (where there are an
even number of0s on theleft-hand side). The second part is a fancy
way of writing99999...9999= 11*9090...0909 (where there are an even
number of 9s inthe lefthand side). We write 10^(2*k) = 11*b[2*k] +
1 and 10^(2*k+1) =11*b[2*k+1] - 1. Here k is any nonnegative
integer. We substitutethatinto the polynomial form,
so:d[2*k]*10^(2*k) + d[2*k-1]*10^(2*k-1) + ... + d[1]*10 + d[0]=
d[2*k]*(11*b[2*k] + 1) + d[2*k-1]*(11*b[2*k-1] - 1) + ...
+ d[1]*(11*a[1] - 1) + d[0] = 11*(d[2*k]*b[2*k] + ... +
d[1]*a[1])+ d[2*k]- d[2*k-1] + ... - d[1] + d[0]The first part is
divisible by 11 no matter what the digits are,so thewhole number is
divisible by 11 if and only if the last part,which isthe
alternating sum of the digits, isdivisible by 11. If you prefer,
you can writed[2*k] - d[2*k-1] + ... - d[1] + d[0] = (d[0] + d[2] +
... + d[2*k]) -(d[1] +d[3] + ... + d[2*k-1]), so that you add up
every other digit,starting fromthe units digit, and then add up the
remaining digits, andsubtract thetwo sums. This will compute the
same result as thealternating sum ofthe digits. 7 - There is a
trick for 7 which is not as well knownas theothers. It makes use of
the fact that 10^(6*k) - 1 is divisible by7, and10^(6*k - 3) + 1 is
divisible by 7. It goes like this: Mark off thedigits ingroups of
threes, just as you do when you put commas inlarge numbers.Starting
from the right, compute the alternating sum of thegroups
asthree-digit numbers. If the result is negative, ignore the
sign.If theresult is greater than 1000, do the same thing to the
resultingnumberuntil you have a result between 0 and 1000
inclusive. That 3-digit
number is divisible by 7 if and only if the original number
istoo.Example:123471023473 = 123,471,023,473, so make the sum473 -
23 + 471 - 123 = 450 + 348 = 798.798 = 7*114, so 798 is divisible
by 7, and 123471023472 is,too.An extra trick is to replace every
digit of 7 by a 0, every 8 by a1, andevery 9 by a 2, before,
during, or after the sum, and the factremains.The sum could also
have been computed as473 - 23 + 471 - 123 --> 403 - 23 + 401 -
123 = 380 + 278 --> 310+ 201 =511 = 7*73. You can figure out why
this "casting out 7s" partworks.There is another way of testing for
7 which uses the fact that7 divides2*10 + 1 = 21. Start with the
numeral for the number you wantto test.Chop off the last digit,
double it, and subtract that from therest of thenumber. Continue
this until you get stuck. The result is 7, 0,or -7, ifand only if
the original number is a multiple of 7.Example:123471023473 -->
12347102347 - 2*3 = 12347102341 -->1234710234- 2*1 = 1234710232
--> 123471023 - 2*2 = 123471019 -->12347101 - 2*9= 12347083
--> 1234708 - 2*3 = 1234702--> 123470 - 2*2 = 123466 -->
12346 - 2*6 = 12334 --> 1233 - 2*4= 1225--> 122 - 2*5 = 112
--> 11 - 2*2 = 7.
13 - The same trick that works for 7 works for 13; that is,
13divides10^(6*k) - 1 and 10^(6*k - 3) + 1, so the alternating sum
ofthree-digitgroups works here, too. 17 - This is harder. You would
havetouse alternating sums of 8-digit groups!1 Select a 3-digit
number. 2 Repeat these digits to make a 6-digitnumber. 3 Divide
these 6 digits by 7, then by 13. 4 Theanswer is 11 times the first
three digits!Example:1 If the 3-digit number selected is 234: 2 The
6-digit numberis 234234.3. Divide by 7, then by 13: multiply by
11-to multiply 234 by 11, work right to left: last digit on right =
___4next digit to left = 3 + 4 = 7: _ _ 7 _ next digit to left = 2
+ 3 =5: _ 5 __ last digit on left = 2 _ _ _3 So 234234 divided by
7, then 13 is 2574.See the pattern?1 If the 3-digit number selected
is 461: 2 The 6-digit numberis 461461.3. Divide by 7, then by 13:
multiply by 11-to multiply 461 by 11, work right to left: last
digit on right = ___1nextdigit to left = 4 + 6 = 10: _ 0 _ _ last
digit on left = 4 + 1(carry) = 5: 5 _ _ _3 So 461461 divided by 7,
then 13 is 5071.Practice multiplying by 11 - this process works
formultiplying anynumber by 11.
1 Select a 3-digit number. 2 Repeat these digits to make a
6-digitnumber. 3 Divide these 6 digits by 13, then by 11. 4
Theansweris 7 times the first three digits!Example:1 If the 3-digit
number selected is 231: 2 The 6-digit numberis 231231.3 Divide by
13, then by 11:7 x 231 = 1400 + 210 + 7 = 1617.4 So 231231 divided
by 13, then 11 is 1617.See the pattern?1 If the 3-digit number
selected is 412: 2 The 6-digit numberis 412412.Remembertomultiply
left to right and add in increments. Then you willbe able to give
these answers quickly and accurately.1 Select a 6-digit number
repeating number. 2 Repeat thesedigits tomake a 6-digit number. 3
Multiply a single digit by 3, then by5.Example:1 If the 6-digit
repeating number selected is 333333: 2Multiply 3 x 3: 9 3Multiply 9
x 5: 45 4 So 333333 divided by 37037 andmultiplied by 5 is 45.See
the pattern?You can expand this exercise by using a different
number inthe final step.Example: multiply by 4:1 If the 6-digit
repeating number selected is 555555: 2Multiply 3 x 5: 15. 3
Multiply 15 x 4: 60 4 So 555555 divided by 37037 andmultiplied
by 4 is 60.By changing the last step you can generate many
extensionsof this3 dsivide by 13, then by 11:7 x 412 = 2800 + 70 +
14 = 2884.4 So 412412 divided by 13, then 11 is 2884.exercise. Be
inventive and create some impressivecalculations.Dividing a
repeating 6-digit number by 7, 11, 13; subtract 101Select a 3-digit
number.Repeat these digits to make a 6-digit number.Divide this
6-digit by 7, then 11, then 13.Subtract 101.The answer is the
original number minus 101!Example:If the 3-digit number selected is
289:The 6-digit number is 289289.Divide by 7, then by 11, then by
13:the answer is 289.Subtract 101: 289 - 101 = 188So 289289 divided
by 7, then 11, then 13 minus 101 is 188.See the pattern?If the
3-digit number selected is 983:The 6-digit number is 983983.Divide
by 7, then by 11, then by 13:the answer is 983.1 Subtract 101: 983
- 101 = 882So 983983 divided by 7, then 11, then 13 minus 101 is
882.1 Select a repeating 3-digit number . 2 The answer is 3
timesoneof the digits plus 41!Example:1 Select 999. 2 Multiply one
digit by 3: 9 x 3 = 27. 3 Add 41: 27+ 41
= 68. 4 So (999 / 37) + 41 = 68.Change the last step to add
other numbers, and thusproduce manynew exercises.1 Select a
repeating 6-digit number . 2 The answer is 7 timesthe firstdigit of
the number!Example:1 777777 / 15873 = 7 x 7 = 49. 2 555555 / 15873
= 7 x 5 = 35. 3999999 /15873 = 7 x 9 = 63. Not very demanding
mental math, butgood for aquick challenge or two.Dividing mixed
numbers by 21 Select a mixed number (a whole number and a
fraction). 2. If the whole number is even, divide by 2 - this is
thewhole numberof the answer. The numerator of the fraction stays
the same; multiplythedenominator by 2. 3. If the whole number is
odd, subtract 1 and divide by 2 -thisis the whole number of the
answer. Add the numerator and the denominator of the fraction
-thiswill be the new numerator of the fraction; Multiply the
denominator by 2.Even whole number example:If the first number
selected is 8 3/4: Divide the whole number(8) by
2: 8/2 = 4 (whole number) Use the same numerator: 3
Multiplythedenominator by 2: 4 x 2 = 8 (denominator) So 8 3/4
divided by2 = 4 3/8.Odd whole number example:1 If the first number
selected is 13 2/5: 2 3 Subtract 1 from thewholenumber and divide
by 2: 13 - 1 = 12, 12/2 = 6.Add the numerator and the denominator:
2 + 5 = 7. This is thenumerator of the fraction. Multiply the
denominator by 2: 5 x2 = 10. So 13 2/5 divided by 2 = 6
7/10.file:///1 Select a 2-digit number. 2 Multiply it by 8 (or by 2
threetimes).3 Move the decimal point 2 places to the left.Example:1
The 2-digit number chosen to multiply by 12 1/2 is 78. 2Multiplyby
2 three times: 2 x 78 = 156 2 x 156 = 312 2 x 312 = 6243 Move the
decimal point 2 places to the left: 6.24 4 So 78dividedby 12 1/2 =
6.24.See the pattern?1 If the 2-digit number chosen to multiply by
12 1/2 is 91: 2Double three times: 182, 364, 728. 3 Move the
decimal point2 places to the left: 7.28 4 So 91 divided by 12 1/2 =
7.28.5ividing a 2-digit number by 151 Select a 2-digit number. 2
Multiply it by 2. 3 Divide theresult by 3.4 Move the decimal point
1 place to the left.Example:1 The 2-digit number chosen to multiply
by 15 is 68. 2Multiply by 2:
2 x 68 = 120 + 16 = 136 3 Divide the result by 3: 136/3 = 45
1/34Move the decimal point 1 place to the left: 4.5 1/3 5 So
68divided by 15 = 4.51/3.See the pattern?1 The 2-digit number
chosen to multiply by 15 is 96. 2Multiplyby 2: 2 x 96 = 180 + 12 =
192 3 Divide the result by 3: 192/3 =644 Move the decimal point 1
place to the left: 6.4 5 So 96/15 =6.4.With this method you will be
able to divide numbers by 15with two quick1 Select a 2-digit
number. (Choose larger numbers when youfeel sureabout the method.)
2 Multiply by 4 (or by 2 twice). 3 Move thedecimall point two
places to the left.Example:1 The 2-digit number chosen to divide by
25 is 38. 2 Multiplyby 4: 4 x 38= 4 x 30 = 120 + 32 = 152. 3 Move
the decimal point 2 placesto the left:1.52 4 So 38 divided by 25 =
1.52.See the pattern?1 The 3-digit number chosen to divide by 25 is
641. 2 Multiplyby 2 twice:2 x 64 = 1282. 2 x 1282 = 2400 + 164 =
2564.3 Move the decimal point 2 places to the left: 25.64. 4 So
641divided by25 = 25.64.1 Select a 2-digit number (progress to
larger ones). 2 Multiplyit by 3. 3Move the decimal point 2 places
to the left.
Example:1 The 2-digit number chosen to multiply by 33 1/3 is
46. 2Multiply by 3:3 x 46 = 3(40 + 6) = 120 + 18 = 138 3 Move the
decimal point 2places tothe left: 1.38 4 So 46 divided by 33 1/3 =
1.38. (If you divide by33.3using a calculator, you will not get the
exact answer.)See the pattern?1 If the 3-digit number chosen to
multiply by 33 1/3 is 650: 2Multiplyby 3: 3 x (600 + 50) = 1800 +
150 = 1950 3 Move the decimalpoint 2places to the left: 19.50 4 So
650 divided by 33 1/3 = 19.5.Practice multiplying left to right and
this procedure willbecome aneasy one - and you will get exact
answers, too.Dividing a 2- or 3-digit number by 351 Select a
2-digit number. (Choose larger numbers when youfeel sureabout the
method.) 2 Multiply by 2. 3 Divide the resultingnumber by 7.4 Move
the decimal point 1 place to the left.Example:If the number chosen
to divide by 35 is 61: Multiply by 2: 2 x61 = 122.Divide by 7:
122/7 = 17 3/7 Move the decimal point 1 place tothe left:1.7 3/7 So
61 divided by 35 = 1.7 3/7.See the pattern?1 If the number chosen
to divide by 35 is 44: 2 Multiply by 2:2 x 44 =88 3 Divide by 7:
88/7 = 12 4/7 4 Move the decimal point 1place to theleft: 1.2 4/7 5
So 44 divided by 35 = 1.2 4/7.
Division done by calculator will give repeating decimals(unless
theoriginal number is a multiple of 7), truncated by the limits
ofthe display.The exact answer must be expressed as a mixed
number.1 Select a 2-digit number. 2 Multiply by 8. 3 Divide
theproduct by 3. 4Example:1 The 2-digit number chosen to divide by
37 1/2 is 32. 2Multiply by 8:8 x 32 = 240 + 16 = 256 3 Divide by 3:
256/3 = 85 1/3 4 Movethe decimalpoint two places to the left: .85
1/3 5 So 32 divided by 37 1/2 =.85 1/3.See the pattern?1 The
2-digit number chosen to divide by 37 1/2 is 51. 2Multiply by 8:8 x
51 = 408 3 Divide by 3: 408/3 = 136 4 Move the decimalpoint
twoplaces to the left: 1.36 5 So 51 divided by 37 1/2 =
1.36.Dividing a 2-digit number by 451 Select a 2-digit number. 2
Divide by 5. 3 Divide the resultingnumberby 9.Example:1 f the
number chosen to divide by 45 is 32: 2 Divide by 5:32/5 = 6.4 3
Divide the result by 9: 6.4/9 = .71 1/9 4 So 32dividedby 45 = .71
1/9.See the pattern?1 If the number chosen to divide by 45 is 61: 2
Divide by 5:61/5 =12.2 3 Divide the result by 9: 12.2/9 = 1.35 5/9
4 So 61 dividedby 451.35 5/9.
1 Select a 2-digit number. (Choose larger numbers when
youfeelsure about the method.) 2 Multiply by 4 (or by 2 twice).
3Move thedecimal point two places to the left. 4 Divide by 3
(expressremainderas a fraction).Example:1 The 2-digit number chosen
to divide by 75 is 82. 2 Multiplyby 4:4 x 82 = 328. 3 Move the
decimal point 2 places to the left:3.28 4Divide by 3: 3.28/3 = 1.09
1/3 5 So 82 divided by 75 = 1.09 1/3.See the pattern?1 The 3-digit
number chosen to divide by 75 is 631. 2 Multiplyby 4(multiply left
to right): 4 x 631 = 2400 + 120 + 4 = 2520 + 4 =2524.3 Move the
decimal point 2 places to the left: 25.24. 4 Divideby 3:25.24/3 =
8.41 1/3 5 So 631 divided by 75 = 8.41 1/3.Select a number.2
Multiply it by 2 3 Divide the result by 3.Example:1 The number
chosen to divide by 1 1/2 is 72. 2 Multiply by 2:2 x 72 =144 3
Divide by 3: 144 / 3 = 48 4 So 72 divided by 1 1/2 = 48.See the
pattern?1 The number chosen to divide by 1 1/2 is 83. 2 Multiply by
2:2 x 83 =166 3 Divide by 3: 166 / 3 = 55 1/3 4 So 83 divided by 1
1/2 =55 1/3.1 Select a 2-digit number. 2 Multiply by 3. 3 Divide by
4.Example:
1 The 2-digit number chosen to divide by 1 1/3 is 47. 2Multiply
by 3:3 x 47 = 120 + 21 = 141 3 Divide by 4: 141/4 = 35 1/4 4 So
47dividedby 1 1/3 = 35 1/4.See the pattern?1 The 2-digit number
chosen to divide by 1 1/3 is 82. 2Multiply by 3:3 x 82 = 246 3
Divide by 4: 246/4 = 61 1/2 4 So 82 divided by1 1/3 = 61 1/2.With
this pattern you will be able to give answers quickly,
butmostimportantly, your answers will be exact. If a calculator
userdividesby 1.3, the answer will NOT be correct.1 Select a
2-digit number. 2 Multiply by 8. 3 Move the decimalpointone place
to the left.Example:1 The 2-digit number chosen to divide by 1 1/4
is 32. 2Multiply by 8:8 x 32 = 240 + 16 = 256 3 Move the decimal
point one place tothe left: 25.6 4 So 32 divided by 1 1/4 =
25.6.See the pattern?1 The 2-digit number chosen to divide by 1 1/4
is 64. 2Multiply by 8:8 x 64 = 480 + 32 = 512 3 Move the decimal
point one place tothe left:51.2 4 So 64 divided by 1 1/4 =
51.2.Multiply from left to right for ease and accuracy. You
willsoon bedoing this division by a mixed number quickly.1 Select a
number. 2 Multiply it by 5 3 Divide the result by 6.Example:
1 The number chosen to divide by 1 1/5 is 24. 2 Multiply by 5:5
x 24= 120 3 Divide by 6: 120/6 = 20 4 So 24 divided by 1 1/5 =
20.See the pattern?1 The number chosen to divide by 1 1/5 is 76. 2
Multiply by 5:5 x 76= 350 + 30 = 380 3 Divide by 6: 380/6 = 63 2/6
4 So 76 dividedby 1 1/5= 63 1/3.Dividing a 2-digit number by 1 2/31
Select a 2-digit number. 2 Multiply by 6 (or by 2 and 3). 3Move the
decimal point one place to the left.Example:1 The 2-digit number
chosen to divide by 1 2/3 is 78. 2Multiplyby 3: 3 x 78 = 210 + 24 =
234 3 Multiply by 2: 2 x 234 = 468 4Movethe decimal point one place
to the left: 46.8 5 So 78 dividedby1 2/3 = 46.8.See the pattern?1
The 2-digit number chosen to divide by 1 2/3 is 32. 2Multiply by
3:3 x 32 = 96 3 Multiply by 2: 2 x 96 = 180 + 12 = 192 4 Move
thedecimalpoint one place to the left: 19.2 5 So 32 divided by 1
2/3 =19.2.Practice multiplying from left to right and you will
becomeadept atmentally dividing a number by 1 2/3.1 Select a
2-digit number. 2 Multiply by 8. 3 Move the decimalpoint3 places to
the left.Example:
1 The 2-digit number chosen to divide by 125 is 72. 2
Multiplyby 8:8 x 72 = 560 + 16 = 576. 3 Move the decimal point 3
places tothe left: .576 4 So 72 divided by 125 = .576.See the
pattern?1 The 2-digit number chosen to divide by 125 is 42. 2
Multiplyby 8:8 x 42 = 320 + 16 = 336.3 ove the decimal point 3
places to the left: .336 4 So 42divided by125 = .336.1 Select a
2-digit number. 2 Multiply by 4. 3 Divide theproduct by 7.Example:1
The 2-digit number chosen to divide by 1 3/4 is 34. 2Multiply by
4:4 x 34 = 136 3 Divide the product by 7: 137/6 = 19 3/7 (If youuse
acalculator, you will get a long, inexact decimal number.)4 So 34
divided by 1 3/4 = 19 3/7.See the pattern?1 The 2-digit number
chosen to divide by 1 3/4 is 56. 2Multiply by 4:4 x 56 = 224 3
Divide the product by 7: 224/7 = 32 4 So 56dividedby 1 3/4 =
32.Notice that numbers divisible by 7 will produce
whole-numberquotients. For numbers not divisible by 7, your
calculator willgiveyou long decimal results that are not exact.1
Select a 2-digit number. 2 Multiply it by 7. 3 Move thedecimalpoint
1 place to the left.
Example:1 The number chosen to multiply by 1 3/7 is 36. 2
Multiply by7:7 x 36 = 210 + 42 = 252 3 Move the decimal point 1
place totheleft: 25.2 4 So 36 divided by 1 3/7 = 25.2.See the
pattern?1 The number chosen to multiply by 1 3/7 is 51. 2 Multiply
by7:7 x 51 = 357 3 Move the decimal point 1 place to the left:
35.74So 36 divided by 1 3/7 = 35.7.1 Select a 2- or 3-digit number.
2 Multiply by 4 (or by 2 twice).3 Move the decimal point one place
to the left.Example:1 The 2-digit number chosen to divide by 2 1/2
is 86. 2Multiplyby 4: 4 x 80 + 4 x 6 = 320 + 24 = 344 3 Move the
decimal pointone place to the left: 34.4 4 So 86 divided by 2 1/2 =
34.4.See the pattern?1 The 3-digit number chosen to divide by 2 1/2
is 624. 2Multiplyby 2: 2 x 624 = 1248 3 Multiply by 2: 2 x 1248 =
2400 + 96 =24964 Move the decimal point one place to the left:
249.6 5 So 624divided by 2 1/2 = 249.6.Multiply by 4 when this is
easy; otherwise use two steps andmultiply by 2 twice.Dividing a
2-digit number by 2 1/31 Select a 2-digit number. 2 Multiply by 3.
3 Divide the resultby 7.Example:1 he 2-digit number chosen to
divide by 2 1/3 is 42. 2 Multiplyby 3:
3 x 42 = 126 3 Divide by 7: 126/7 = 18 4 So 42 divided by 2
1/3= 18.See the pattern?1 The 2-digit number chosen to divide by 2
1/3 is 73. 2Multiply by 3:3 x 73 = 219 3 Divide by 7: 219/7 = 31
2/7 4 So 73 divided by 21/3 = 31 2/7.If the number chosen is
divisible by 7, the quotient will be awholenumber. If the number is
not divisible by 7, a calculator userwillget a long, inexact
decimal, while your answer will be exact.1 Select a 2-digit number.
2 Multiply by 3. 3 Divide by 8.Example:1 The 2-digit number chosen
to divide by 2 2/3 is 32. 2Multiply by 3:3 x 32 = 96 3 Divide by 8:
96/8 = 12 4 So 32 divided by 2 2/3 =12.See the pattern?1 The
2-digit number chosen to divide by 2 2/3 is 61. 2Multiply by 3:3 x
61 = 183 3 Divide by 8: 183/8 = 22 7/8 4 So 61 divided by 22/3 = 22
7/8.Using this method, your answers will be exact. Those
usingcalculators will only get approximations.1 Select a 2-digit
number. 2 Mltmultiiply the number by 2. 3Divide theproduct by
7.Example:1 The 2-digit number chosen to divide by 3 1/2 is 42.
2Multiply by 2: 2 x42 = 84 3 Divide by 7: 84/7 = 12 4 So 42 divided
by 3 1/2 = 12.See the pattern?1 The 2-digit number chosen to divide
by 3 1/2 is 61. 2Multiply by 2:
2 x 61 = 122 3 Divide by 7: 122/7 = 17 3/7 4 So 61 divided by
31/2 = 17 3/7.If the number chosen is divisible by 7, the answer
will be awhole number.For numbers not divisible by 7, a calculator
will get arepeating decimal,but your fractional answer will be
exact.1 Select a 2- or 3-digit number. 2 Multiply by 3. 3 Move
thedecimal pointone place to the left.Example:1 The 2-digit number
chosen to divide by 3 1/3 is 72. 2Multiply by 3:72 x 3 = 216 3 Move
the decimal point one place to the left:21.6 4 So72 divided by 3
1/3 = 21.6.See the pattern?1 The 2-digit number chosen to divide by
3 1/3 is 48. 2Multiply by 3:48 x 3 = 120 + 24 = 144 3 Move the
decimal point one place tothe left:14.4 4 So 48 divided by 3 1/3 =
14.4.After practicing, choose larger numbers. Insist on
exactanswers (youwont get an exact answer if you divide by 3.3
using acalculator).Multiply from left to right in steps and impress
your friendswithyour mental powers.Dividing a 2-digit number by
3751 Select a 2-digit number. 2 Multiply by 8. 3 Divide theproduct
by3 (express remainder as a fraction). 4 Move the decimal
pointthree places to the left.Example:
1 The number chosen to divide by 375 is 32. 2 Multiply by 8:8 x
32 = 240 + 16 = 256 3 Divide by 3: 256/3 = 85.3 1/3 4 Movethe
decimal point 3 places to the left: .0853 1/3 5 So 32divided by 375
= .0853 1/3.See the pattern?1 The number chosen to divide by 375 is
61. 2 Multiply by 8:8 x 61 = 480 + 8 = 488 3 Divide by 3: 488/3 =
162 2/3 4 Movethe decimal point 3 places to the left: .162 2/3 5 So
61 dividedby 375 = .162 2/3.72 3 Divide by 9:72/9 = 8 4 So
36divided by 4 1/2 =8.For numbers notdivisible by 9, yourcalculator
will getarepeating decimal,but your fractionalanswer will
beexact.Dividing a 2-digit number by6251 Select a 2-digit number. 2
Multiply by 8. 3 Divide theproductby 5. 4 Move the decimal point 3
places to the left.Example:1 The 2-digit number chosen to divide by
625 is 65. 2 Multiplyby8: 8 x 65 = 480 + 40 = 520 3 Divide by 5:
520/5 = 104 4 Movethedecimal point 3 places to the left: .104 5 So
65 divided by 625= .104.See the pattern?
1 The 2-digit number chosen to divide by 625 is 32. 2
Multiplyby8: 8 x 32 = 240 + 16 = 256 3 Divide by 5: 256/5 = 51.2 4
Movethedecimal point 3 places to the left: .0512 5 So 32 divided
by625 =.0512.Dividing a 2-digit number by 7 1/21 Select a 2-digit
number. 2 Multiply it by 4 (or by 2 twice). 3Divideby 3. 4 Move the
decimal point 1 place to the left.2M u lt i p lExample: y b 1 The
2-digit number chosen to multiply by 71/2 is 42. y4: 42 x 4 = 168 3
Divide by 3: 168/3 = 56 4 Move the decimalpoint1 place to the left:
5.6 5 So 42 divided by 7 1/2 = 5.6.See the pattern?1 The 2-digit
number chosen to multiply by 7 1/2 is 93. 2Multiply byDividing a
2-digit number by 4 1/221 Select a 2-digit number.2 Multiply the
number by 2.3 Divide the product by 9.Example:1 The 2-digit number
chosen to divide by 4 1/2 is 62.23Multiply by 2: 2 x 62 = 124
Divide by 9: 124/9 = 13 7/94 So 62 divideby 4 1/2 = 13 7/9.See the
pattern?1 The 2-digit number chosen to divide by 4 1/2 is 36.4: 93
x 4 = 360 + 12 = 372 3 Divide by 3: 372/3 = 124 4 Movethe
decimal point 1 place to the left: 12.4 5 So 93 divided by 7
1/2= 12.4.Practice and you will soon be cranking out these
quotientswithspeed and accuracy.Dividing a 2- or 3-digit number by
16 2/31 Select a 2-digit number. (Choose larger numbers when
youfeelsure about the method.) 2 Multiply by 6 (or by 3 and then
2). 3Move the decimal point two places to the left.Example:1 The
2-digit number chosen to divide by 16 2/3 is 72. 2Multiplyby 3: 3 x
72 = 216 3 Multiply by 2: 2 x 216 = 432 4 Move thedecimalpoint 2
places to the left: 4.32 5 So 72 divided by 16 2/3 =4.32.See the
pattern?1 The 2-digit number chosen to divide by 16 2/3 is 212.
2Multiply by3: 3 x 212 = 636 3 Multiply by 2: 2 x 636 = 1200 + 72 =
1272 4Move thedecimal point 2 places to the left: 12.72 5 So 212
divided by16 2/3 = 12.72.Practice multiplying by 3, then by 2, and
you will be able todo theseproblems quickly.Divisibility RulesWhy
do these rules work? - Dr. RobDivisibilidad por 13 y por nmeros
primos (13,17,19...)-en espaol, de la lista SNARKFrom the Archives
of the Math Forums Internet project AskDr. Mathourthanks to Ethan
Dr. Math Magness, Steven Dr. MathSinnott,
nd, for the explanation of why these rules work, Robert L.Ward
(Dr. Rob).Dividing by 3Add up the digits: if the sum is divisible
by three, then thenumber isas well. Examples: 1 111111: the digits
add to 6 so the wholenumberis divisible by three. 2 87687687. The
digits add up to 57, and5 plusseven is 12, so the original number
is divisible by three.Why does the divisibility by 3 rule
work?From: "Dr. Math" To: [email protected] (Kevin
Gallagher)Subject:Re: Divisibility of a number by 3As Kevin
Gallagher wrote to Dr. MathOn 5/11/96 at 21:35:40 (Eastern
Time),>Im looking for a SIMPLE way to explain to several
verybright 2nd>graders why the divisibility by 3 rule works,
i.e. add up allthe >digits;if the sum is evenly divisible by 3,
then the number is as well.>Thanks!>Kevin GallagherThe only
way that I can think of to explain this would be asfollows:Look at
a 2 digit number: 10a+b=9a+(a+b). We know that 9a isdivisible by 3,
so 10a+b will be divisible by 3 if and only if a+bis.Similarly,
100a+10b+c=99a+9b+(a+b+c), and 99a+9b isdivisibleby 3, so the total
will be iff a+b+c is.This explanation also works to prove the
divisibility by 9 test.Itclearly originates from modular arithmetic
ideas, and Im notsure
if its simple enough, but its the only explanation I can
thinkof.Doctor Darren, The Math ForumCheck out our web site
-http://forum.swarthmore.edu/dr.math/Dividing by 4ook at the last
two digits. If they are divisible by 4, thenumber isas well.
Examples:1 100 is divisible by 4. 2 1732782989264864826421834612
isdivisible by four also, because 12 is divisible by four.Dividing
by 5If the last digit is a five or a zero, then the number is
divisibleby 5.Dividing by 6Check 3 and 2. If the number is
divisible by both 3 and 2, it isdivisible by 6 as well. Robert
Rusher writes in:Another easy way to tell if a [multi-digit] number
isdivisible by six . . .is to look at its [ones digit]: if it is
even, and the sum ofthe [digits] isa multiple of 3, then the number
is divisible by 6.Dividing by 7To find out if a number is divisible
by seven, take the lastdigit,double it, and subtract it from the
rest of the number.Example:If you had 203, you would double the
last digit to get six, andubtract that from 20 to get 14. If you
get an answer divisibleby 7(including zero), then the original
number is divisible byseven.If you dont know the new numbers
divisibility, you canapplythe rule again.Dividing by 8
Check the last three digits. Since 1000 is divisible by 8, if
thelast three digits of a number are divisible by 8, then so is
thewhole number. Example: 33333888is divisible by 8; 33333886isnt.
How can you tell whether the last three digits aredivisible by 8?
Phillip McReynolds answers:If the first digit is even, the number
is divisible by 8 if thelast two digits are. If the first digit is
odd, subtract 4 fromthe last two digits; the number will be
divisible by 8 if the resulting last two digits are. So, to
continue the last example,33333888 is divisible by 8 because the
digit in the hundredsplace is an even number, and the last two
digits are 88, whichis divisible by 8. 33333886 is not divisible by
8 because thedigit in the hundreds place is an even number, but the
lasttwo digits are 86, which is not divisible by 8.Dividing by 9Add
the digits. If they are divisible by nine, then the numberis as
well. This holds for any power of three.Dividing by 10If the number
ends in 0, it is divisible by 10.Dividing by 11Lets look at 352,
which is divisible by 11;the answer is 32. 3+2 is 5; another way to
say thisis that 35 -2 is 33. Now look at 3531, which is
alsodivisible by 11. It is not a coincidence that 353-1 is352 and
11 x 321 is 3531. Here is a generalizationof this system. Lets look
at the number 94186565.First we want to find whether it is
divisible by 11, but on theway we are going to save the numbers
that we use: in everystep we will subtract the last digit from the
other digits, thensave the subtracted amount in order. Start
with9418656 - 5 = 9418651 SAVE 5 Then 941865 - 1 = 941864SAVE
1Then94186 - 4 = 94182 SAVE 4 Then 9418 - 2 = 9416 SAVE 2 Then941-
6 = 935
SAVE 6 Then 93 - 5 = 88 SAVE 5 Then 8 - 8 = 0 SAVE 8Now write
the numbers we saved in reverse order, and wehave8562415, which
multiplied by 11 is 94186565.Heres an even easier method,
contributed by ChisForen:Takeany number, such as 365167484.Add the
1,3,5,7,..,digits.....3 + 5 + 6 + 4 + 4 = 22Add the
2,4,6,8,..,digits.....6 + 1 + 7 + 8 =22If the difference, including
0, is divisible by 11, then so isthenumber. 22 - 22 = 0 so
365167484 is evenly divisible by11.See alsoDivisibility by 11 in
the Dr. Math archives.Dividing by 12Check for divisibility by 3 and
4.Dividing by 13Heres a straightforward method supplied by Scott
Fellows:D e lete the last digit from the given number. Then
subtractninetimes the deleted digit from the remaining number. If
what isleftis divisible by 13, then so is the original number.And
heres a more complex method that can be extended tootherformulas:1
= 1 (mod 13)10 = -3 (mod 13) (i.e., 10 - -3 is divisible by13)100=
-4 (mod 13) (i.e., 100 - -4 is divisible by 13)1000 = -1
(mod13)(i.e., 1000 - -1 is divisible by 13)10000 = 3 (mod 13)100000
= 4(mod 13)1000000 = 1 (mod 13)Call the ones digit a, the tens
digit b, the hundreds digit c, .....and you get:a - 3*b - 4*c - d +
3*e + 4*f + g - .....
If this number is divisible by 13, then so is the
originalnumber.You can keep using this technique to get other
formulas fordivisibility for prime numbers. For composite numbers
justcheck for divisibility by divisors.Finding 2 1/2 percent of a
number1 Choose a number (start with 2 digits and advance to 3
withpractice).2 Divide by 4 (or divide twice by 2). 3 Move the
decimal pointone placeto the left.Example:1 If the number selected
is 86: 2 Divide 86 by 4: 86/4 = 21.5 3Move thedecimal point one
place to the left.: 2.15 4 So 2 1/2% of 86 =2.15.See the pattern?1
If the number selected is 648: 2 Divide 648 by 2 twice: 648/2=
324,324/2 = 162 3 Move the decimal point one place to the
left.:16.2 4 So2 1/2% of 648 = 16.2.Practice dividing by 4, or by 2
twice, and you will be able tofind theseanswers faster than with a
calculator.Finding 5 percent of a number1 Choose a large number (or
sum of money). 2 Move thedecimalpoint one place to the left. 3
Divide by 2 (take half of it).Example:1 If the amount of money
selected is $850: 2 Move thedecimal pointone place to the left.: 85
3 Divide by 2: 85/2 = 42.50 4 So 5%of $850 = $42.50.See the
pattern?
1 If the amount of money selected is $4500: 2 Move thedecimal
pointone place to the left.: 450 3 Divide by 2: 450/2 = 225 4 So
5%of $4500 =$225.Finding 15 percent of a number1 Choose a 2-digit
number. 2 Multiply the number by 3. 3Divide by 2. 4 Move the
decimal point one place to the left.Example:1 If the number
selected is 43: 2 Multiply by 3: 3 x 43 = 129 3Divideby 2: 129/2 =
64.5 4 Move the decimal point one place to theleft: 6.455 So 15% of
43 = 6.45.See the pattern?1 If the number selected is 72: 2
Multiply by 3: 3 x 72 = 216 3Divide by2: 216/2 = 108 4 Move the
decimal point one place to the left:10.85S o15%of72=10.8.Finding 45
percent of a number1 Choose a 2-digit number. 2 Multiply the number
by 9. 3Divide by 2.4 Move the decimal point one place to the
left.Example:1 If the number selected is 36: 2 Multiply by 9: 9 x
36 = 270 +54 = 324 3Divide by 2: 324/2 = 163 4 Move the decimal
point one placeto the left: 16.2 5 So 45% of 36 = 16.2.See the
pattern?
If the number selected is 52: Multiply by 9: 9 x 52 = 450 + 18
=468 Divideby 2: 468/2 = 234 Move the decimal point one place to
theleft: 23.4 So45% of 52 = 23.4.Finding 55 percent of a numbern 1
Choose a 2-digit number. e 2 Multiply the number by 11.(Add
digitsfrom right to left - see examples). x 3 Divide by 2. t 4 Move
thedecimalpoint one place to the left. di Example: git 1 If the
number selected is 81: t 2 Multiply by 11: 11 x 81 =891 o
rightdigit is 1 l eft is 1 + 8 = 9 last digit to left is 83 Divide
by 2: 891/2 = 445.5 4 Move the decimal point oneplace to theleft:
44.55 5 So 55% of 81 = 44.55.See the pattern?1 If the number
selected is 59:2. Multiply by 11: 11 x 59 = 649 right digit is 9
next digit to leftis 9 + 5= 14 (use the 4 and carry 1)last digit to
left is 5 + 1 = 62 Divide by 2: 649/2 = 324.53 Move the decimal
point one place to the left: 32.454 So 55% of 59 = 32.45.Why do the
divisibility "rules" work?The Power of Modulo ArithmeticModulo
arithmetic is a powerful tool that can be used to
testfordivisibility by any number. The great disadvantage of
usingmodulo arithmetic to test for divisibility is the fact that it
isusuallya slow method. Under some circumstances, however, the
application of modulo arithmetic leads to divisibility
rulesthat canbe used. These rules are important to mathematics
becausetheysave us lots of time and effort.In fact, many of the
divisibility rules that are commonly used(rules for 3, 9, 11) have
their roots in modulo arithmetic. Thispage will show you how some
of these common divisibilityrules are connected to modulo
arithmetic.Applying the Rules of Modulo ArithmeticThe rules of
modulo arithmetic state that the number N isdivisibleby some number
(P) if the above expression is also divisibleby Pafter the base
(10) is replaced by the remainder of 10 dividedby P.In compact
notation, this remainder is denoted by (10 modP). Thus, N is
divisible by P if the following expression is alsodivisible by
P:[Dn * (10 mod P)^(n-1)] + ... + [D2 * (10 mod P)^1] + [D1 *
(10mod P)^0].Origin of Divisibility Rules for 3, 9, and 11D We can
now see how the divisibility rules for P = 3, 9, and11 arerooted in
modulo arithmetic. nConsider the case where P = 3. .Because 10 mod
3 is equal to 1, any number is divisible by 3if thefollowing
expression is. also divisible by 3: .-[Dn * (1)^n-1] + ... + [D2 *
(1)^1] + [D1* D2(1)^0]. Because 1 raised to any power is equal to
1, the above
+expression can be simplified as: Dn + ... + D2 + D1 +
D0,Dwhich is equal to the sum of digits in the number.
Thusmoduloarithmetic allows us to 1state that any number is
divisible by 3 if the sum of its digits isBecause 10 mod 9 is also
1, any number is divisible by 9 isthesum of its digits are
alsodivisible by 9. Thus, the divisibility rules for 3 and 9
comedirectlyfrom modulo narithmetic.n)Finally, consider the case
where P = 11. aBecause 10 mod 11 is equal to -1, any number is
divisible by11if the n following expression is also divisible by
11: dD[Dn * (-1)^n-1] + ... + [D2 * (-1)^1] + [D1* (-1)^0].nBecause
-1 raised to any even powe