For a positive integer n, suppose n! is written in the form

Question

For a positive integer  k , suppose  n!  is written in the form  n! = 2^k · m  ,where  m  is an odd integer. Which of the following could be the value of  k ?

I. 4
II. 5
III. 6

A. I only
B. II only
C. III only
D. I and II only
E. I and III only

suscipitlabore · Answer

Is this question missing some information?  kevincan

kevincan · Answer

Remember that m is an odd integer

Edskore · Answer

Good question — this is testing Counting Prime Factors in Factorials, also known as Legendre's Formula, and it's a concept that comes up on harder GMAT Focus Number Properties problems.

The key insight: k equals the total power of 2 in n!, which is found by repeatedly dividing n by 2 and summing the floor values.

Formula: k = floor(n/2) + floor(n/4) + floor(n/8) + ... (add terms until they reach 0)

The big trap: students assume every integer value of k is achievable. It isn't. When n crosses a power of 2, k jumps by more than 1, so some values are skipped entirely.

Step 1 — Check Statement I: k = 4.
Try n = 6: floor(6/2) + floor(6/4) = 3 + 1 = 4. This works.
So k = 4 IS achievable (with n = 6). Statement I is possible. CHECK.

Step 2 — Check Statement II: k = 5.
From n = 6: k = 4 (computed above).
From n = 7: floor(7/2) + floor(7/4) = 3 + 1 = 4 as well.
From n = 8: floor(8/2) + floor(8/4) + floor(8/8) = 4 + 2 + 1 = 7.
k jumps from 4 (at n = 7) directly to 7 (at n = 8), skipping 5 and 6.
So k = 5 is NOT achievable. Statement II is not possible. X

Step 3 — Check Statement III: k = 6.
Same jump: at n = 7 we get k = 4, and at n = 8 we get k = 7. k = 6 is skipped.
So k = 6 is NOT achievable. Statement III is not possible. X

Answer: A (I only)

Why does this happen? Adding n = 8 contributes three new factors of 2 at once (because 8 = 2^3 contributes to the floor(n/2), floor(n/4), and floor(n/8) terms simultaneously). So k jumps by 3 in a single step. This is the same reason why k values like 2, 5, 6 cannot be produced by any factorial.

Takeaway: on "which of the following COULD be k" questions for factorials, build a small table of k values for consecutive n. The first time you see a value or skip, you'll know immediately which statements work.

Dereno · Answer

n! = 2^k · m, where m is an odd integer.

if n=3, then the number of 2s in 3! = 2^1

3! = (2^1)*3

so, k can be 1.

if n =5, then 5! = 5*4*3*2*1

Since, between 3 and 5, we have a 4 = 2^2 is added to the value.

then, 5! = (2^3) * (3*5). Thus, k =3.

Then n =6, adds an extra 2. So the power of 2 is 4. (2^4).

when n = 7, there is no 2, the power of 2 is (2^4).

if n=8, then we add 2^3 into the system

n! = 8! = 2*7 * (multiple of odd number).

So,  only 4 qualifies.

Option A

vasu1104 · Answer

if n=4
then 4! = 4*3*2*1 = 2^3 * 3
here k= 3 and m=3 (possible)
but since 3 is not in choice we move forward.

5!= 2^3 * 5*3
again k=n so not possible

6!= 2^4* 15
k=4 so this can be value of k.

from now onwards power of 2 will be always greater than 7.
so only option A is correct

For a positive integer n, suppose n! is written in the form

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GMAT Problem Solving (PS) Questions

For a positive integer n, suppose n! is written in the form

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