[Fermat Math Contest] If 2^11 x 6^5 = 4^x × 3^y for some positive integers x and y, then the value of x + y is (A) 10 (B) 11 (C) 12 (D) 13 (E) 14

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Solution: 

$2^{11} . 6^{5} = 4^x . 3^y$

 $2^{11} . (2 +3)^{5} = 4^x . 3^y$                [$(a+b)^c = a^c + b^c$]

$2^{11} . 2^{5} . 3^{5} = 4^x . 3^y$

$2^{11+5} . 3^5 = 4^x . 3^y$

$2^{16} . 3^5 = 4^x . 3^y$

$(2^2)^8 . 3^5 = 4^x . 3^y$

$4^8 . 3^5 = 4^x . 3^y$

On comparing L.H.S. and R.H.S. 

x = 8

y = 5

x + y = 8 + 5 = 13

Option (D) = 13 , is correct.

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