Question

There are two positive integers X and Y. When X is divided by 237, the remainder is 192. When Y is divided by 117, the quotient is same but the remainder is 108. The remainder when the sum of X and Y is divided by 118 is

Answer 1

Answer:

There are two positive integers X & Y. When X is divided by 237, the remainder is 192, and when Y is divided by 117 the quotient is the same, but the remainder is 108. How do I find the remainder when the sum of X & Y is divided by 118?

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Let the quotient obtained in both the cases be n.

X = 237 n + 192

Y = 117 n + 108

Adding both the equations,

X + Y = 354 n + 300

= 118 ( 3 n ) + 118 ( 2 ) + 64

=118 ( 3 n + 2 ) + 64.

Therefore, the remainder obtained when sum of X & Y is divided by 118 is 64.

As per the given question,

X = 237k+192 (k is 0,1,2,3...)

Y = 117k+108 (the quotient is same)

Adding both, X+Y = (273+117)k+192+108

X+Y = 354k + 300

= (118*3)k + 118*2+64

= 118(3k+2)+ 64

When this number X+Y is divided by 118, the quotient is 3k+2 and the reminder is 64.

Thanks for A2A.