woRKouT 45.6.7.

Solution - Problem #1The guestion reolly osks for the number of positive integer divisors of 24. We could list the

possibi l i t ies: t ,2,3,4,6,8, t2and24. Remember,weknowshedoesnot.hovejusto-neguest,so. ,there ore 7 possible volues for m. For numbers larger than 24, though, listing the foctors could becumbersome. ft would be nice to hove o method for counting the divisors of on integer, withouthoving to list oll of them. Eoch divisor is the producf of some of f he_prime foctors of ?4, so b^eginby prime foctoring 24 as 23'3. Then, each divisor is of the form 2o'3b, where o is either O,1,2, or3 ' (4possibi l i t ies)ondbisei ther0or l (2possibi l i t ies). This impl iesthotthereare4'2=Sfactorsof 24. The foctors ore shown in the table below:

Gonnection to... UPG codes (Problem #4)Universol Product Codes (UPCs) ore constructed so thot the first digit represents the type of

product, the next five digits identify the monufocturer, thefive ofter thot lobel the specificiroduct and the finol digit is o "check digit." A computer sconner con then moke the computotiondescribed in this problem. If the result isn't divisible by 10, the computer knows thot it hos sconnedthe numbers incorrectly. Test theUPC code on some producfs oround your home or school.

This type of check digit is olso designed to detect tronspositions, or the switching of twoodjocent digits. Try swopping two digits to see if the number still sotisfies the checking criterio.

lnvestigation & Exploration (Problem #8)This problem does not focus on whole numbers of pounds. fnsteod it focuses on the

remoinders. This type of orithmetic is colled "modulor" orithmetic; in this problem, we ore usingmod 10. The eguation we're trying to solve, then, is 7n: 3 (mod 10). In regular olg'ebro, we'd solveon eguation like this by multiplyingbV G/7). fn mod 10 orithmetic, though, thct would. give us onanswer of 3/7, which isn't reolly whot we meon. Insfeod, notice thot 3'7 is 21, which hos o unitsdigit of 1, or o remoinder of Lwhen divided by 10. This meons thot 21: 1 in mod 10. Thus,mult iplying both sides of the eguation by 3 yields ?tn= n: 3'3 : 9, so n is 9.

Answers1.7 (C,E, G)

(c)(F, M)

8.99. 6.310. 36

(P, T, C)(P, C, G)(e, c, G)

(c)

2.3.4.

22319

6614.0510

(P, T)(F, M)

(P, F, C)

90 MATHCOUNTS 200142 MATHCOUNTS

2.

3.

4.

5.

6.

7.

9.

10.

f, '. grtWorkout 3Problem 1. Substituting 1 for P and 9.8 for g, we must solve the equation I = 2zSquaring both sides, we Cet L = 2* = 0.25 meters.

Problem 2, If 61% live in Asia, then39o/o live in the rest of the world. .39x.14x.49--.026754x 2.7o/".

Problem 3. The factors 4 through 50 in the numerator will cancel the factors 4 through 50 in the denominator, teauing1;ffg = # .probtem4.f fh"+#x160minutes=633,600fet . 633,600feetxf f i -=120miles. Thecartraveledl20miles.problem i,-fU{weglculetgj.yEch isj!9 rglrinerimeter. s = &# = 7.5 . Using this value and the three side lengths in the formula, weget t =,ll.s(t.s- +)( z.s - s)(z.s - 6) =,1 98.437 s o 9.9 square inches'

Problem 6. This perimeter of20 cm is one diameter plus one halfofthe circumference. Hence 20 = n + ! rO . Solving for D, we getO = -4- *7.78 cm and the radius is half that or about 3.89 cm. The area this half circle is thus j:z x 3.892 o 23.8 square centimeters.

I r inproblemT. Firstofal l ,weknowthati twi l l takelessthanonehourforl ighttogol2l,000,000milessinceittravels6T0,000,000milesinaful l

rT{es for Z. With some rearranging, we have 17 Jn

alL =-; .

t25=150,eachof

) on the second

havel the 280

fs= t8s) bet'veen

0.002 x 140 = 0.28 .

the left has a baselrean triple.)we could use

pwer large

p 2002"d evenI all the leadingThere are 45ie t hundreds

ints. The threetscore. So the

I in the centers['s

ficious applesBenson wants for

[=5.r-25.

I,Enot greaterle. When wewery x=44,

hour. Div id ingl2lby610,weget0.1806hours,or0.1806x60=10.836minutes,or10.836x60p650seconds.problemS. Tangentlines-sreperpendiculartoradii. Thismeans thatAlul and tN areparallelandthelengthofthe same as the length of lC. This length can be found-.14i-th the Pythagorean theorem:.7V = Jzs= - rz =,t625 - a,s = rlsi6 =24. Therefore rVl[ isalso 24inches.Problem 9. Problem solved in the answer key to Workout 3.problem 10. The proper factors of284 are 1,2,4,71 and 142. These factors have a sum of220. The properfactors of220 arel,2,4,5, 10, 11, 20,22,44,55 and 110' These factors have a sumof284.

Workout 4Problem 1, Problem solved in the answer key to Workout 4.problem 2. The sum ofthe first 40 natural numbers is af = 329 , so the set of40 consecutive integers we're looking for must start in thenegative numbers. Ifthe list starts at -19, it would have to end at 20, having a sum of20. Ifwe slide up 1 integer, and take the sum oftheintegers- l8through2l ,wegetasumof60. Sl id ingupl integermore,wegetthesumoftheintegers-17through22,whichhasasumofl00.probleml. f fk4<106,thenJfo.JtOu +k2<103+&2<1000.312=961 but322=1024,sothelargest integerfrcanbeis3l .

Problem 4. Tripling the sum of the odd digits and adding the even digits, we get:n=(O+8+e+l+0+6) x3+(7+0+4+l+0) =3:"3+12=99+12=111. Tomakethisvaluediv is ib lebyl0, the126digi txmustbe9.problem5. Takingthecuberootof 10,626wi l lgetusintheneighborhoodofthethreeconsecut iveintegers.3/10526 x21.98. Checkingthe

Problem 8. We are looking for the smallest positive multiple of 0.7 that leaves a 3 in the tenths place. That would be the9th multiple of 0.7 which is 6.3, so the fewest number of apples Dot could have in the bag is 9 apples .

Problem 9. The horse that is 8 feet from the center of the merry-go-round will travel 1 6a feet in the 9 seconds of one rotation which isfiSr\ t g feer per second. The ho$e that is 17 feet from the center will travel 34r feet in 9 seconds which is par) / 9 feet per second. The' 34', 16'-=18o =2o= 6.3feetDersecondfasterthanthefirsthorse'secono norse goes

l--l- 9probleml0. Theratioofthecircumferenceofwheell totheinnercircumferenceofwheelBisl2tr.2n,or6:1. ThismeansthatwheelBwil lrevolve 6 times for each revolution of wheel l. Similarly, wheel C will revolve 6 times for each revolution of wheel B. Combining the two,wheel C will make 36 revolutions for each revolution of wheel ,4.

Workout 5problem 1. Thereare 0.05x4000x1=200 eaningsonthepeoplewearingoneearr ing. Thereare 0.5x0.95x 4O00x2= 3800 earr ings onthe people with fwo. There are 0.10 x 0.5 x 0.95 x 4000 x 3 = 570 earrings on the people with three. That's 200 + 3800 + 570 = 4570 earrings.

problem 2. All 100 ofthe numbers in the 300's contain the digit 3 at least once. In each ofthe other 8 hundreds, there are l0 numbers in the30'sandgothernumbersthatcontainthedigi t3. Inal l , that 's100+8x19=252numbers.problem 3. Using the Pythagorean theorem, we can find the.length of the other leg: x2 +122 =(Zx)z >144= 3x2 = x2 = 48 , so*=JqS =4J3 cent imeters. Theareaofthetr iangleisthus fx4{3 x12=24^t3 squarecent imeters.

problem 4. The larger digits should go in the left-most places to create the maximum value. Trial and error will lead eventually to the

product 21x 22x23,wefindthati tdoesequal 10,626. T\e sumoftheseintegersis66.problem 6. The rate of $0.25 per { mile is the same as $2.00 per mile. The fare for the cab ride will be1.60+2x13.25--1.60+26.50 = $28.10. Since the two passengers will share the fare equally, they each owe $14.05.problem 7. By looking at a $aph, it's obvipuslhat the only midpoint worth considering is the midpoint of side ,4B.point (the star) has coordinates (6,8!and is {62 +82 =,tX+e+ = JtOO = 10 units fromthe origin, closer thanB (16 units), A (12 units) or C (10^12 x 14'1 units).

following solution: 9632 + (84 x 75) = 15,93r.

Problem 5. To find the area of the triangle we need to find a base and a height in terms the radius of circle G. Sincetnanele CEH is a 30-60-90, we know that segment EC = r, segment GH = r/2 and segrpent Efl = (rrl3 )tZ. Therefore,eC= rJt. Sofindingtheareaoftheequilateraltr iangle,wehave !"rJi" |r=ffr2. Thesquarel.BDFhasa F

diagonal of 2r, so using the formula for the area of a rhombus, its area is (l/2)(2r)(2r) = 2t'. The ratio of the area of the

triangle to the area of the square is -4= +2r2 8

MATHCOUNTS 2OO1-02Solutions113