What is the value of the positive integer a?

Ok,

Statement 1 says a! + a is prime number.

take a=2, this will make 2! + 2 = 4, not a prime
take a=3 this will make 3!+3 = 9, not a prime.

take any higher number and it will not be a prime number.

now take a=1, this will make 1! + 1 = 2, prime number. so a=1 is our solution.


Statment 2 =>

a+2 / (a+2)!, is a terminatin decimal, this means (a+2)! is either 2 or 5 or both , take same number a=1,2,3 and only a=1 will satisfy the equation.

ex a= 2, this will make 4/4! => 1/3! => 1/6, not a terminating decimal as factors of 6 are 2 and 3.

a= 1 => 3/3! => 1/2! , terminating decimal.

so answer is 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.

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