GMAT Practice Question: 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} ?
- 3
- 6
- 8
- 11
- 14
Topics: factorials, divisibility, prime factorization, exponents, number theory
Solution
Step 1
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! + 2\sqrt{8!\times 9!}
Step 2
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! \sqrt{8!\times 9!} = \sqrt{9\times (8!)^2} = 3\times 8!
Step 3
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!
Step 4
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 \text{odd}) = 2^{11} \times \text{odd}
Answer
D
