x!) / = 12,376 x is: a] 17 b] 34 c] 18 d] 13 e] 19 Is there

Question

(x!) /   = 12,376

x is:

a] 17
b] 34
c] 18
d] 13
e] 19

Is there a quick way to solve such problems?

Dellin · Answer

a=17
i went by trying the answer values, lucky that first one turned out right.
Even otherwise, we can get an idea about the value when you try two or more.

walker · Answer

A it is not so fast....

1. N=12,376=2^3*7*13*17 So, B,D,E are out. - B: N should include primes such as 31, 29 - D: N should not include primes such as 17 - E: N should include primes such as 19 2. X! for x=18 and x=17 contains two 7 (in 7 and 14). We subtract one 7 by 11! and (x-11)! should not include 7. at x=18: (x-11)!=7! - wrong; x=17: (x-11)!=6! - BINGO

walker · Answer

... or simpler way to distinguish between 17 and 18:

N(18)=18/7*N(17)=3^2*2/7*N(17). So, at x=18 N(18) should include prime 3 but 12,376 does not include 3. Therefore, x=17

sondenso · Answer

I can not get what are you reasoning! why should and why should not?

walker · Answer

For example, x=19
N=19!/(8!*11!)=  19  *18*  17  *16*15*14*  13  *12*11!/(8!*11!)

1. 19,17,13 are prime numbers. 
2. 8! and 11! don't include prime numbers19,17,13
3. 19,17,13 are prime factors of N.
4. N=12,376=2^3*7*13*17 does not include prime 19. therefore, x=19 is out.

Hope this help.

GMATBLACKBELT · Answer

Its actually not too bad.

starting with B. 34!/23!*11! u should be able to realize that this number is going to be really big. far to big for 12376

try D. 13!/2!*11! --> 13*12*11!/11!*2 --> to small

From here I noticed it was going to be either A or C. I tried A since it was smaller.

17!/6!*11! --> 17*16*15*14*13*12*11!/11! --> 17*16*15*14*13*12/6*5*4*3*2 cancel out the 12,15 and some of the 16. left with 17*4*13*14.

May look involved, but it took me less than 2minutes.

x!) / = 12,376 x is: a] 17 b] 34 c] 18 d] 13 e] 19 Is there

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