M31-43

Question

What is the value of the positive integer  a ?

(1)  a! + a  is a prime number.
 
(2)  \frac{a+2}{(a+2)!}  is a terminating decimal.

Bunuel · Accepted Answer

Official Solution:

What is the value of the 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; thus, it cannot be a prime number. Therefore,  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 2s and/or 5s in its prime factorization. This is only possible if  (a+1)! = 2! = 2  (all other factorials 3!, 4!, 5!, etc., have primes other than 2 and 5 in them), making  a = 1 . Sufficient.

Answer: D

vik09 · Answer

Hello,
Can you please explain the first statement when a>1. I din't get how you derived the equation from a!+a.

Bunuel · Answer

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

xspongebobx · Answer

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.

Thanks.

Bunuel · Answer

THEORY: 
 
A reduced fraction  \frac{a}{b}  (meaning that the fraction is already in its simplest form, so reduced to its lowest term) can be expressed as a terminating decimal if and only if the denominator  b  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 the denominator  250  equals  2*5^3 . The fraction  \frac{3}{30}  is also a terminating decimal, as  \frac{3}{30}=\frac{1}{10}  and the denominator  10=2*5 .
 
 Note that if the denominator already consists of only 2s and/or 5s, 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 a terminating decimal.
 
(We need to reduce the fraction in case the denominator has a prime other than 2 or 5, to see whether it can be reduced. For example, the fraction  \frac{6}{15}  has 3 as a prime in the denominator, and we need to know if it can be reduced.)
 
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.

Kartalyst · Answer

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,

ImranBinkhaled · Answer

5! = 5*4*3*2*1.

In this case, since the denominator would have factors other than 2 and 5, the fraction would never result in a terminating decimal. The thread contains the relevant theory explained by  Bunuel . Hope it clears the confusion.

jayditya · Answer

what if a=0 then ?
0 is also a integer

Bunuel · Answer

a  cannot be 0 because we are told that  a  is a positive integer ( What is the value of  positive  integer a? ), while 0 is neither positive nor negative.

ZERO:

1. Zero is an  INTEGER .

2. Zero is an  EVEN  integer.

3. Zero is  neither positive nor negative  (the only one of this kind)

4. Zero is  divisible by EVERY integer except 0 itself  ( \frac{x}{0} = 0 , so 0 is a divisible by every number, x).

5. Zero is a  multiple of EVERY integer  ( x*0 = 0 , so 0 is a multiple of any number, x)

6. Zero is  NOT a prime number  (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed:  anything/0 is undefined .

8. Any non-zero number  to the power of 0 equals 1  ( x^0 = 1 )

9.  0^0  case is  NOT tested on the GMAT .

10. If the exponent n is positive (n > 0),  0^n = 0 .

11. If the exponent n is negative (n < 0),  0^n  is undefined, because  0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0} , which is undefined.  You CANNOT take 0 to the negative power .

12.  0! = 1! = 1 .

2. Properties of Integers  
     Theory   
 Math: Number Theory 
 Number Properties: Tips and hints 
 Advanced Number Properties on GMAT (All 5 parts) 
 Do you make these 3 mistakes in GMAT Even-Odd Questions?  
 Everything about Factorials on the GMAT 
 Solving Complex Problems Using Number Line 
     Questions   
 Tough New Set: Number Properties Set 
 Power of a Number in a Factorial Problems 
 Number Properties Ability Quiz 
 Special Numbers and Sequences Problems 
 The E-GMAT Number Properties Knockout: 10 Great 700+ Ques for you! 
 Trailing Zeros Problems 
 DS Questions 
 PS Questions

For more check  Ultimate GMAT Quantitative Megathread

Hope it helps.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

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