A number when successively divided by 3, 5 and 8 leaves remainders

A number when successively divided by 3,5 and 8 leaves remainders 1,4 and 7 respectively. Find the respective remainders when the order of the divisors is reversed.

A. 8,5,3
B. 4,2,1
C. 3,2,1
D. 6,4,2
E. None of above

My approach would be

Three Divisors: 3, 5, 8
Three Remain: 1, 4, 7

Start from the bottom of the last column i.e. from the third remainder:

Go up diagonally and multiply by the second divisor: 5*7 = 35

Go down and add the second remainder: 35 + 4 = 39

Go up diagonally and multiply by the first divisor: 39* 3 = 117

Go down and add the first remainder: 117 + 1 = 118

Divide 118 by 8, 5, 3
118/8 gives quotient = 14 and remainder = 6
14/5 gives quotient = 2 and remainder = 4
2/3 gives quotient = 0 and remainder = 2

Ans is D i.e. 6,4,2

thanks for the question. What's the source?

Hello, I don't get the red part. Why isn't the number written as 15K + remainder? That's the way we wrote it in the first portion (from which we got 120k+118)
Thanks

Since 15/5=3 and 14/5=2 and gives 4 as remainder. Hence it is 3k+2 remainder 4

Posted from my mobile device

it was silly question. Got it, thanks:-)

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