A number theory problem

Number Theory Level 3

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 the same but the remainder is 108.

Find the remainder when the sum of X and Y is divided by 118.

58 cannot say 70 64

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Himanshi Sethi by himanshi sethi , 18, India

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2 solutions

Viki Zeta
Oct 17, 2016

x = 237 a + 192 y = 117 a + 108 x + y = 354 a + 300 = 118 × 3 a + 118 × 2 + 64 = 118 ( 3 a + 2 ) + 64 = 118 b + 64 ⟹ r = 64 x = 237a + 192 \\ y = 117a + 108 \\ x + y = 354a + 300 \\ = 118 \times 3a + 118 \times 2 + 64 \\ = 118 (3a + 2) + 64 \\ = 118b + 64 \\ \implies r = 64

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Himanshi Sethi
Oct 17, 2016

Given that

X=192 mod 237

⟹X=237k+ 192

Y=108 mod 117Y ⟹Y=117k+ 108

Adding the numbers we have

X+Y=(237+117)k+(192+108)

⟹X+Y=354k+300

⟹X+Y=118(3k)+118×2+64

⟹X+Y=118(3k+2)+64

⟹X+Y≡64 mod 118

Remainder is 64

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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