Steps | 103 ÷ 64 | ||
1. | Find the Ratio of the composition of the Divisor. | In this case, 64 = 2 × 4 × 8 where, 2:4 = 4:8 = 8:64 = 1:2 |
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2. | Find the Quotient and Remainder by one of the smaller proportions. | In this case, Let us consider 4, then 103/4 = 3 25 - 4 Quotient = 25 Remainder = 3 |
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3. | The Remainder being the same, find the desired Quotient by dividing the quotient with the proportional Ratio. | In this case, So, the desired Quotient = 25/16 = 1 And, Remainder = 9 |
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4. | In case of additional remainders, re-evaluate the final remainder by adding both of them. | In this case, the desired Remainder = 3 9 3 + 36 -- + -- = ------ 64 16 64 That gives us: 39 1 -- 64, which is the answer! |
27 = 3 × 9, where 3:9 = 9:27 = 1:3 Then, 1063/9 = 118, Remainder = 1 Again, 118/3 = 39, Remainder = 1 Thus, 1063 ÷ 27 = 1 1 10 39 (-- + --) = 39 -- 3 27 27 |
343 = 7 × 49, where 7:49 = 49:343 = 1:7 Then, 70804/7 = 10114, Remainder = 6 Again, 10114/49 = 206, Remainder = 20 Thus, 70804 ÷ 343 = 6 20 146 206 (--- + --) = 206 --- 343 49 343 |
For any successive numbers, with ratio k: a:b:c:d Then, b = ka, c = k2a, d = k3a When any number, x is divided by these successive numbers: x/b = x/(ka) x/c = x/(k2a) x/d = x/(k3a) And so on. It should be noted that not only does this series have a constant Remainder, the Quotients are also in (reducing) proportions in the same ratio. For example, If x/a gives quotient Q, and remainder R x R - = Q - a a x x x 1 1 R And, - = -- = - . - = - (Q -) b ka a k k a x x x 1 1 R And, - = --- = - . -- = -- (Q -) c k2a a k2 k2 a x x x 1 1 R And, - = --- = - . -- = -- (Q -) d k3a a k3 k3 a |
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