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What is the value of positive integer a?

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

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New post 28 May 2015, 06:02
1
13
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

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

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

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New post 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.

Answer: D.
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What is the value of positive integer a?  [#permalink]

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New post 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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J

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

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New post 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.

so answer is D.
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Re: What is the value of positive integer a?  [#permalink]

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New post 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.

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