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What is the value of the positive integer \(a\)?
(1) \(a! + a\) is a prime number.
(2) \(\frac{a+2}{(a+2)!}\) is a terminating decimal.
(1) \(a! + a\) is a prime number.
(2) \(\frac{a+2}{(a+2)!}\) is a terminating decimal.
20 Jun 2015, 11:28
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Official Solution:
What is the value of the positive integer \(a\)?
(1) \(a! + a\) is a prime number.
If \(a = 1\), then \(a! + a = 2 = prime\). If \(a > 1\), then \(a! + a = a((a - 1)! + 1)\) is a product of two integers, each of which is greater than 1; thus, it cannot be a prime number. Therefore, \(a = 1\). Sufficient.
(2) \(\frac{a+2}{(a+2)!}\) is a terminating decimal.
\(\frac{a+2}{(a+2)!} = \frac{a+2}{(a+1)!*(a+2)} = \frac{1}{(a+1)!}\). For \(\frac{1}{(a+1)!}\) to be a terminating decimal, the denominator, \((a+1)!\), must have only 2s and/or 5s in its prime factorization. This is only possible if \((a+1)! = 2! = 2\) (all other factorials 3!, 4!, 5!, etc., have primes other than 2 and 5 in them), making \(a = 1\). Sufficient.
Answer: D
What is the value of the positive integer \(a\)?
(1) \(a! + a\) is a prime number.
If \(a = 1\), then \(a! + a = 2 = prime\). If \(a > 1\), then \(a! + a = a((a - 1)! + 1)\) is a product of two integers, each of which is greater than 1; thus, it cannot be a prime number. Therefore, \(a = 1\). Sufficient.
(2) \(\frac{a+2}{(a+2)!}\) is a terminating decimal.
\(\frac{a+2}{(a+2)!} = \frac{a+2}{(a+1)!*(a+2)} = \frac{1}{(a+1)!}\). For \(\frac{1}{(a+1)!}\) to be a terminating decimal, the denominator, \((a+1)!\), must have only 2s and/or 5s in its prime factorization. This is only possible if \((a+1)! = 2! = 2\) (all other factorials 3!, 4!, 5!, etc., have primes other than 2 and 5 in them), making \(a = 1\). Sufficient.
Answer: D
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