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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## When m and n are positive integers, is m!*n! an integer squa tagged by: Max@Math Revolution ##### This topic has 2 expert replies and 0 member replies ### GMAT/MBA Expert ## When m and n are positive integers, is m!*n! an integer squa ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult [Math Revolution GMAT math practice question] When m and n are positive integers, is m!*n! an integer squared? 1) m = n + 1 2) m is an integer squared _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only Worldâ€™s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$149 for 3 month Online Course
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### GMAT/MBA Expert

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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

When m and n are positive integers, is m!*n! an integer squared?

1) m = n + 1
2) m is an integer squared
$m,n\,\, \geqslant \,\,1\,\,\,{\text{ints}}\,\,\,\left( * \right)$
$m!\,\,n!\,\,\,\mathop = \limits^? \,\,{K^{\,2}}\,\,\,,\,\,\,K\mathop \geqslant \limits^{\left( * \right)} 1\,\,\,\operatorname{int}$

$\left( 1 \right)\,\,\,m = n + 1\,\,\,\,\,\, \Rightarrow \,\,\,\,m! = \left( {n + 1} \right)n!\,\,\,\,\,\, \Rightarrow \,\,\,\,m! \,n! = \left( {n + 1} \right){\left( {n!} \right)^2}$
$\left\{ \begin{gathered} \,{\text{Take}}\,\,n = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\left[ {\,2 \cdot 1\,\,{\text{is}}\,\,{\text{not}}\,\,{\text{a}}\,\,{\text{perfect}}\,\,{\text{square}}\,} \right]\,\, \hfill \\ \,{\text{Take}}\,\,n = 3\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\left[ {\,4 \cdot {6^2}\,\, = \,\,{{12}^2}\,\,{\text{is}}\,\,{\text{a}}\,\,{\text{perfect}}\,\,{\text{square}}\,} \right]\,\, \hfill \\ \end{gathered} \right.$

$\left( 2 \right)\,\,m = {J^2}\,\,\,,\,\,\,J\mathop \geqslant \limits^{\left( * \right)} 1\,\,\,\operatorname{int}$
$\left\{ \begin{gathered} \,{\text{Take}}\,\,\left( {m,n} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\ \,{\text{Take}}\,\,\left( {m,n} \right) = \left( {1,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\ \end{gathered} \right.$

$\left( {1 + 2} \right)\,\,\,n + 1 = {J^2}\,\,\, \Rightarrow \,\,\,\,m!\,n! = {J^2} \cdot {\left( {n!} \right)^2}\,\, = {\left( {J \cdot n!} \right)^2}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\left[ {\,K = \,J \cdot n!\,} \right]\,\,\,\,$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (m and n) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
By condition 1),
m!*n! = (n+1)!*n! = (n+1)(n!)(n!) = (n+1)(n!)^2 = m(n!)^2.
Since (n!)^2 is an integer squared, and m is an integer squared by condition 2), m!*n! is an integer squared.
Thus, both conditions together are sufficient.

Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
If m = 4 and n = 3, then 4!*3! = 4*3!*3! = (2(3!))^2 and the answer is â€˜yesâ€™.
If m = 3 and n = 2, then 3!*2! = 6*2 = 12, and the answer is â€˜noâ€™.
Since it does not give a unique answer, condition 1) is not sufficient on its own.

Condition 2)
If m = 4 and n = 3, then 4!*3! = 4*3!*3! = (2(3!))^2 and the answer is â€˜yesâ€™.
If m = 4 and n = 2, then 4!*2! = 24*2 = 48 and the answer is â€˜noâ€™.
Since it does not give a unique answer, condition 2) is not sufficient on its own.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

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