The value of (sqrt((8!))+sqrt((9!)))^2 is an integer. What is the greatest integer n...

Question

The value of (sqrt((8!))+sqrt((9!)))^2 is an integer. What is the greatest integer n such that 2^n is a factor of (sqrt((8!))+sqrt((9!)))^2 ?

GMAT Panda · Accepted Answer

D We first expand the square of a sum using the formula (a+b)^2 = a^2 + 2ab + b^2. (sqrt(8!)+sqrt(9!))^2 = 8! + 9! + 2sqrt(8! 9!) We then simplify the factorial expressions and the square root term by recognizing that 9! = 9 8! and using factorial decomposition. 9! = 9 8! sqrt(8! 9!) = sqrt(9 (8!)^2) = 3 8! Next we combine like terms by factoring out 8! from each term. 8! + 9 8! + 2 3 8! = (1 + 9 + 6) 8! = 16 8! We now express 16 8! in prime-factor form to find the exponent of 2. We use prime factorization of 8!. 16 8! = 2^4 8! = 2^4 (2^7 odd) = 2^11 odd

Explanation	Calculations	Help
$We first expand the square of a sum using the formula (a + b)^{2} = a^{2} + 2 ab + b^{2} .$	$(8! + 9!)^2 = 8! + 9! + 2 8! \times 9!$
$We then simplify the factorial expressions and the square root term by recognizing that 9! = 9 \times 8! and using factorial decomposition.$	$9! = 9 \times 8! 8! \times 9! = 9 \times (8!)^{2} = 3 \times 8!$
$Next we combine like terms by factoring out 8! from each term.$	$8! + 9 \times 8! + 2 \times 3 \times 8! = (1 + 9 + 6) \times 8! = 16 \times 8!$
$We now express 16 \times 8! in prime-factor form to find the exponent of 2. We use prime factorization of 8! .$	$16 \times 8! = 2^{4} \times 8! = 2^{4} \times (2^{7} \times odd) = 2^{11} \times odd$

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