DS question from princeton

[rakaisraka]

#m = ((m!)!). lf m is a positive integer, what is the
value of m ?
(1) 2! : (m - 1)!
(2 #m is six times the value of 5!


Can we cancel out the factorial from both sides?

[MartyMurray]

Did you write Statement 1 correctly?

Is Statement 1 what you are referring to with your question?

[rakaisraka]

Hi, Actually this is a DS question and this was the only info given. It is from princeton. I can get the answer but im not sure if we can cancel out factorials from both side. ?Thanks

[Brent@GMATPrepNow]

Hi, Actually this is a DS question and this was the only info given. It is from princeton. I can get the answer but im not sure if we can cancel out factorials from both side. ?Thanks

In MOST cases, you can cancel out the factorials.
That is, if k! = j!, then k = j
So, if 2! = (m - 1)!, then we can conclude that 2 = m-1, which means m = 3

HOWEVER, there is one proviso, which is based on the fact that 0! = 1! = 1
If either factorial is 0! or 1!, then the above rule does not apply.

For example, if we have (n - 2)! = 1!, then there are two possible solutions: n = 3 and n = 2

Cheers,
Brent

[Brent@GMATPrepNow]

I's assuming that statement 1 should read as follows:

#m = ((m!)!). lf m is a positive integer, what is the
value of m ?
(1) 2! = (m - 1)!
(2 #m is six times the value of 5!

Target question: What is the value of m?

Statement 1: 2! = (m - 1)!
This means that 2 = m-1, which means m = 3
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: #m is six times the value of 5!
So, (m!)! = (6)(5!)
(m!)! = (6)(5)(4)(3)(2)(1)
(m!)! =6!
So, m! = 6, which means m = 3
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent

[Max@Math Revolution]

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.


#m = ((m!)!). lf m is a positive integer, what is the
value of m ?
(1) 2! = (m - 1)!
(2) #m is six times the value of 5!

==> In the original condition we have 1 variable (m) and we need 1 equation to match the number of variables and equations. Since we have 1 each in 1) and 2), D is likely the answer.

In actual calculation,

In case of 1), 2=m-1, m=3 sufficient and in case of 2), (m!)!=6(5!)=6!, m!=6=3*2*1, m=3 therefore is sufficient. Thus the answer is D.

For 95% of the questions that have 1) =2), D is the answer.



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