Question

please help me ! A number being successively divided by 3, 5 and 8 leaves remainders 1, 4 and 7 respectively. Find the remainders if the order of divisors be reversed explain it with correct procedure.

Answer 1

say the starting number is \(x\)

Answer 2

then we have :

\(
x = 3i + 1 \\
i = 5j + 4 \\
j = 8k + 7

\)

Answer 3

eliminating \(j\) from first two equations gives :
\(x = 3(5j+4) + 1 = 15j + 13\)

Answer 4

eliminating \(j\) gives :
\(x = 15(8k+7) + 13 = 120k + 118\)

Answer 5

\(x = 120k + 118\)
divide this in reverse order and see what u get