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# M31-43

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Math Expert
Joined: 02 Sep 2009
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20 Jun 2015, 11:28
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75% (hard)

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58% (02:14) correct 42% (02:15) wrong based on 31 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
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20 Jun 2015, 11:28
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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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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.
Math Expert
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21 Jun 2015, 10:59
vik09 wrote:
Hello,
Can you please explain the first statement when a>1. I din't get how you derived the equation from a!+a.

$$a! = (a-1)!*a$$, so you can factor $$a$$ out of $$a!+a$$ to get $$a((a−1)!+1)$$.
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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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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 ! ! !

Hope it helps.
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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?

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

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