ALGORITMA ALTERNATIF

(I) LONG METHOD (WORDS)

25 = 2 TENS 5 ONES + 56 = 5 TENS 6 ONES

7 TENS 11 ONES = 8 TENS 1 ONES

ALGORITHM FOR ADDITION

(II) LONG METHOD (NUMBERS)25 = 20 + 5

+ 56 = 50 + 6 70 + 11 = (70 + ( 10 + 1) = (70 + 10) + 1 = 80 + 1

= 81

ALGORITHM FOR ADDITION

(III) PARTIAL SUM TENS ONES 2 5 2 5+ 5 6 OR 5 6 1 1 1 1 7 0 7 0 8 1 8 1

ALGORITHM FOR ADDITION

(IV) CONVENTIONAL ALGORITHM

2 5 + 5 6

8 1

ALGORITHM FOR ADDITION

I EXPANDED NOTATION METHOD (words) 3 TENS and 5 ONES = 2 TENS and 15 ONES - 1 TENS and 7 ONES

2 TENS and 15 ONES- 1 TENS and 7 ONES 1 TENS and 8 ONES

ALGORITHM FOR SUBTRACTION

II EQUAL ADDITION METHOD (alternate method) • A subtraction problem that requires “trading“

can be worked by "adding" the same value to both the minuend and the subtrahend to avoid “trading". This method only works under subtraction.

ALGORITHM FOR SUBTRACTION

II EQUAL ADDITION METHOD (alternate method)

(35 + 3) = 38 35- (17 + 3) = - 20 - 17

18 18

ALGORITHM FOR SUBTRACTION

III WRITING TWO-DIGIT NUMBER AS SOMETHING PLUS A "TEEN" NUMBER

Subtract 29 from 57.57 = 50 + 7 = 40 + 1729 = (-)20 + 957 - 29 = 20 + 8 = 28Notice that only 57 was put into something and

"teen" form.

ALGORITHM FOR SUBTRACTION

IV USING BIGGER NUMBERS.You simply have one or more extra zeroes at the end

Subtract 129 from 321321 = 300 + 20 + 1 = 300 + 10 + 11 = 200 + 110 + 11 321 = 200 + 110 + 11 (-) 129 = (-) 100 + 20 + 9

321 – 129 = 100 + 90 + 2 = 192

ALGORITHM FOR SUBTRACTION

I LONG MULTIPLICATION

23958233 × 5830 00000000 (= 23,958,233 × 0) 71874699 (= 23,958,233 × 30) 191665864 (= 23,958,233 × 800)+119791165 (= 23,958,233 × 5,000) 139676498390 (= 139,676,498,390)

ALGORITHM FOR MULTIPLICATION

II LATTICE OR SIEVE MULTIPLICATION345 X 2

LATTICE FORM

ALGORITHM FOR MULTIPLICATION

I SUBTRACTION PROCEDURE15 ÷ 3 = 5 3 1 5

-3 …………1 1 2 -3…………..2 9 -3…………..3 6 -3……………4 3 3…………….5 0

ALGORITHM FOR DIVISION