Bunuel wrote:

Bunuel wrote:

What is the value of positive integer a?

(1) a! + a is a prime number

(2) \(\frac{a+2}{(a+2)!}\) is a terminating decimal

OFFICIAL SOLUTION:What is the value of 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, so it cannot be a prime. Thus 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 2's or/and 5's in its prime factorization, which is only possible if (a+1)! = 2! = 2 (all other factorials 3!, 4!, 5!, have primes other than 2 and 5 in them), making a = 1. Sufficient.

Answer: D.

Beautiful problem, beautiful solution.

Let me elaborate a bit about statement (2), so that the students will (I hope) understand better "what is going on" in this "terminating decimal business"...

1/(a+1)! is a terminating decimal if, and only if, there exists a positive integer N (sufficiently large) such that (10^N) * 1/(a+1)! is an integer.

(This is obvious if you think about the decimal expansion of any real number, and the fact that 10^N moves the decimal point, without changing any digits...)

We may assume (when it is the case) that we took

the minimum N in such conditions (*). Hence:

\(\left. {\operatorname{int} = \frac{{{2^N} \cdot {5^N}}}{{\left( {a + 1} \right)!}} = \left\{ \begin{gathered}

\frac{{{2^N} \cdot {5^N}}}{{2!}}\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,N = 1\,\,\,,\,\,\,{\text{when}}\,\,{\text{a}}\,{\text{ = }}\,{\text{1}}\,\, \hfill \\

\frac{{{2^N} \cdot {5^N}}}{{3 \cdot \operatorname{int} }}\,\,\,\, \Rightarrow \,\,\,{\text{impossible}}\,\,\,\,\,\,,\,\,\,{\text{when}}\,\,{\text{a}}\,\, \geqslant \,\,{\text{2}}\,\,\,\, \hfill \\

\end{gathered} \right.} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,a = 1\,\,\,\)

I hope you find all this interesting.

Regards,

fskilnik.

_________________

Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)

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