National Mathematics Olympiad Contest Question 2000(by DAMT)

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

5 Answers

1
mentor_Adil Hayat ·

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 → x=18.
Hence the answer is (c)

1
b_k_dubey ·

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)

1
Fermat ·

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

Is it possible to solve without using the options?

62
Lokesh Verma ·

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

1
Avinava Datta ·

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

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