National Mathematics Olympiad Contest Question 2000(by DAMT)

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If x and y are integers such that
(x+2y)² + (x+4y) = 710
The value of x is
(A) 13
(B) 15
(C) 18
(D) 26

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5 Answers

1

mentor_Adil Hayat ·2009-12-08 21:47:37

take (x+2y)=a
then the given equation reduces to

a² + a + 2y=710.

clearly x+2y is an even number which implies x is even.

now we if we have to make sum equal to 710.
taking a=26 we get 2y=8.
now (x+2y)=26 â†’ x=18.
Hence the answer is (c)

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1

b_k_dubey ·2009-12-08 23:56:27

x² + 4xy + 4y² + x + 4y = 710

x² + (2y)² + 1² + 2(x)(2y) + 2(2y)(1) + 2(1)(x) = 711 + x

(x+2y+1)² = 711 + x

so, 711+x must be a perfect square for which x=18 from given options

(x,y) = (18,4)

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1

Fermat ·2009-12-09 08:10:24

How did you take a=26 to get 2y=8?

Is it possible to solve without using the options?

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62

Lokesh Verma ·2009-12-09 09:10:39

@adil.. how did you conclude that x+2y is an even number?!

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Avinava Datta ·2009-12-09 10:35:01

if x+2y is not even then what would hav been the problem in @dil sirs solution .

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Your Answer

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