
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 

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 

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 

4.  In case of additional remainders, reevaluate 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 = k^{2}a, d = k^{3}a When any number, x is divided by these successive numbers: x/b = x/(ka) x/c = x/(k^{2}a) x/d = x/(k^{3}a) 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 k^{2}a a k^{2} k^{2} a x x x 1 1 R And,  =  =  .  =  (Q ) d k^{3}a a k^{3} k^{3} a 
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