Question

A number is successively divided by 8, 6 and 5 leaving 1, 5 and 4 as remainder respectively

The sum of remainders when order of divisors

be reversed is​

Answer 1

Answer:

So, the answer is, 6, 4 and 2 would be the remainders when division sequence is reversed. Hope this helps. N = 3[5(8q+7)+4]+1 = 3[40q+39]+1 = 120q+118. Therefore, the remainders when divided by 8, 5, and 3 are 6, 4, and 2 respectively.

Answer 2

Solution :-

Starting from the end we get,

→ 5 * 1 + 4 = 9

→ 6 * 9 + 5 = 59

→ 8 * 59 + 1 = 473

So, the given number is equal to 473 .

Now, dividing 473 successively by taking divisors in reversed order as 5, 6 and 8,

→ 473 ÷ 5 = 94 quotient , 3 remainder

→ 94 ÷ 6 = 15 quotient , 4 remainder

→ 15 ÷ 8 = 1 quotient , 7 remainder

therefore,

→ Sum of remainder = 3 + 4 + 7 = 14 (Ans.)

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