Question

there are two positive integer x and y . when x is devided by 237 the remainder is 19 . when y is devided by 117 the quotient is the same but the remainder is 108 . find the remainder when the sum of x and y is devided by 118

Answer 1

Hello there!
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For this question, let's assume that the quotient we get when x is divided by 237 and when y is divided by 117 is "n"

Hence by the data given in the question we can say that:

$x = 237n + 19$
Also we can say that:

$y = 117n + 108$
Hence the sum of x and y is:

$x + y = 237n + 19 + 117n + 108 \\ x + y = 354n + 127$
We need to find the remainder when the sum of x and y is divided by 118.

Hence rearrange our equation:

$x + y = 118 \times 3 \times n + 118 \times 1 + 9$
Now from the right hand side we take 118 as the common factor in the following way:

$x + y = 118(3n + 1) + 9$
Let's assume that (3n+1) = k and write the equation again:

$x + y = 118k + 9$

In this form we can see that, when the sum of x and y is divided by 118, the quotient is k = (3n + 1) and the remainder is 9.

Hence the answer is 9.
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