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M31-43

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M31-43  [#permalink]

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New post 20 Jun 2015, 11:28
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

58% (02:14) correct 42% (02:15) wrong based on 31 sessions

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Re M31-43  [#permalink]

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New post 20 Jun 2015, 11:28
1
4
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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Re: M31-43  [#permalink]

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New post 21 Jun 2015, 10:33
Hello,
Can you please explain the first statement when a>1. I din't get how you derived the equation from a!+a.
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Re: M31-43  [#permalink]

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New post 21 Jun 2015, 10:59
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Re: M31-43  [#permalink]

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New post 14 Jan 2019, 17:10
Would someone be able to explain statement 2? I am unsure how " (a+1)!(a+1)!, must have only 2's or/and 5's in its prime factorization, which is only possible if (a+1)!=2!=2(a+1)!=2!=2 (all other factorials 3!, 4!, 5!, have primes other than 2 and 5 in them)" was derived

Thanks.
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Re: M31-43  [#permalink]

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New post 14 Jan 2019, 21:55
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xspongebobx wrote:
Would someone be able to explain statement 2? I am unsure how " (a+1)!(a+1)!, must have only 2's or/and 5's in its prime factorization, which is only possible if (a+1)!=2!=2(a+1)!=2!=2 (all other factorials 3!, 4!, 5!, have primes other than 2 and 5 in them)" was derived

Thanks.


Theory:
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\). Other examples of terminating decimals: fractions \(\frac{1}{8}\), \(\frac{1}{25}\), ...

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal.

Check below for more:
Math: Number Theory
How to Identify Terminating Decimals on the GMAT
Terminating Decimals in Data Sufficiency on the GMAT
Terminating and Recurring Decimals


Even more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread


Hope it helps.
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Re: M31-43  [#permalink]

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New post 29 Aug 2019, 05:21
Thanks for the explanation.

I only got lost at (a+1)! = 2!. What happens if (a+1)! = 5!? Since the prime factorization of the denominator must have 2 or/and 5?

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Re: M31-43   [#permalink] 29 Aug 2019, 05:21
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