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# What is the value of positive integer a?

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Math Expert
Joined: 02 Sep 2009
Posts: 50729
What is the value of positive integer a?  [#permalink]

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28 May 2015, 06:02
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Difficulty:

75% (hard)

Question Stats:

51% (01:53) correct 49% (02:07) wrong based on 172 sessions

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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

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Joined: 02 Sep 2009
Posts: 50729
Re: What is the value of positive integer a?  [#permalink]

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01 Jun 2015, 02:40
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.

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What is the value of positive integer a?  [#permalink]

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Updated on: 28 May 2015, 08:14
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Hello all

My attempt:

Looks like a tricky question

$$a$$ is a positive integer which means $$a={1,2,3,4.....}$$

Statement 1:

$$a! + a$$ is $$prime$$
$$a*[ (a-1)! + 1 ]$$
for $$a = 1$$
$$1 * [ (1-1)! + 1 ] = 2$$ which is $$prime$$
for all other values of $$a$$ the expression reduces to a product of two positive integers and hence cannot be prime. For instance $$a=3$$
$$3 * [ (3-1)! + 1 ] = 3*3$$ thus not a $$prime$$
so statement has only one solution which is $$a=1$$. Hence solvable from this.

Statement 2:

$$\frac{(a+2)}{(a+2)!}$$ is a terminating decimal
Expression can be reduced to $$\frac{1}{(a+1)!}$$
For a fraction to be a terminating decimal it must be in the form of $$\frac{1}{(2^x * 5^y)}$$
There is only one value of $$a$$ at which it is in the terminating decimal form that is at $$a=1$$ at others such as $$a=2,3,4..$$ it is not in terminating decimal form.
Thus we can find out the value of $$a$$ from this statement also.

Hence answer is $$D$$

p.s. corrected the earlier solution which was incorrect due to a silly mistake
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Regards
J

Do consider a Kudos if you find the post useful

Originally posted by Jackal on 28 May 2015, 07:04.
Last edited by Jackal on 28 May 2015, 08:14, edited 1 time in total.
##### General Discussion
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Joined: 17 Oct 2013
Posts: 54
Re: What is the value of positive integer a?  [#permalink]

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28 May 2015, 07:45
1
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

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.

GMATH Teacher
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Re: What is the value of positive integer a?  [#permalink]

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03 Sep 2018, 13:07
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.

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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Re: What is the value of positive integer a? &nbs [#permalink] 03 Sep 2018, 13:07
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