If 15!/m^3 is a multiple of 5, which of the following cannot be the

Question

If 15!/m^3 is a multiple of 5, which of the following cannot be the value of m?

A. 6
B. 8
C. 9
D. 10
E. 12

vitaliyGMAT · Answer

15! = 2^{11}*3^6*5^3*7^2*11*13

10^3 = 2^3*5^3

If we divide our  15!  by  10^3  all factors of  5  will be cancelled out and our fraction  \frac{15!}{m^3}  won't be a multiple of 5.

All other options satisfy our requirement.

Answer D

Felipe28g · Answer

Is there a short cut to doing this? Or am I expected to breakdown 15! by all of its multiples during the exam and then crossing out every answer until one fits?

ScottTargetTestPrep · Answer

Solution:
 
Since 15! = 2 x 3 x 2^2 x 5 x 2 x 3 x 7 x 2^3 x 3^2 x 2 x 5 x 11 x 2^2 x 3 x 13 x 2 x 7 x 3 x 5 = 2^11 x 3^6 x 5^3 x 7^2 x 11 x 13 = (2^3 x 3^2)^3 x 5^3 x 2^2 x 7^2 x 11 x 13, therefore, as long as m is a factor of 2^3 x 3^2, then 15!/m^3 will be a multiple of 5.

Of the answer choices given, since 10 is not a factor of 2^3 x 3^2, 10 can’t be a value of m.

Answer: D

zindahi · Answer

15!/m^3 = 5 * Some Integer

15! has 5,10, and 15, which are multiples of 5. We need to ensure that, dividing by the numbers given in the option, not all these multiples vanish.

We see that Option D i.e 10 when performed m^3 which is 10^3 will cancel out the entire multiples of 5 present in 15!, which then will not fetch us any multiple of 5 via the Question Stem equation 15!/m^3.

Answer: D

paudomingoo · Answer

A number is a  multiple of 5  if it has  at least one factor of 5 .

So we need: 
 - 15!/m^3 has at least one 5 
 --> That means m^3 must not “use up” all the 5s in 15! 
 
Step 1: Count how many 5s are in 15!

Use:   v5(15!)=  +  --> v5(15!)=3+0=3

So 15! contains exactly three 5s  .

Step 2: What must be true about m^3?

As 15! has exactly three 5s, m^3 should contain fewer than three 5s for it to return a multiple of 5 after the division

Let v5(m) be the number of 5s in m.

Then m^3 contains 3*v5(m)

So  m cannot contain any factor of 5 . 
 
That is the whole shortcut. 
 
Step 3: Check answer choices for a factor of 5 
 
  6 = 2⋅3 → no 5 ✓ can work  
  8 = 2^3 → no 5 ✓ can work  
  9 = 3^2 → no 5 ✓ can work  
  10 = 2⋅5 → contains a 5 ✗ cannot work  
  12 = 2^2⋅3 → no 5 ✓ can work
   
 So the value of mm that  cannot  be is:   10 (because it contains a 5, 10^3 contains three 5s)

Correct answer: D.

If 15!/m^3 is a multiple of 5, which of the following cannot be the

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

If 15!/m^3 is a multiple of 5, which of the following cannot be the

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards