If 13!/2^x is an integer, which of the following represents all possib

None of the answer choices are negative, so obviously the solution set is only the whole numbers.

You're absolutely right about the intentions of the question writer - he or she meant to restrict the solutions to non-negative integers - and your solution is a perfect one to that question. But the answer to a math question doesn't change just because of the answer choices some prep company author has decided upon. Real GMAT questions never require you to divine the intentions of the question designer; their wording is always perfectly precise.

Totally agree. And, that was a good catch by the OP.

By the way, there's a world-famous mathematician and author named Ian Stewart. I like his books. So...great name. Cheers.

Good job. Aspiring to be at a level to identify mistakes in questions.

Sent from my LG-LS980 using GMAT Club Forum mobile app

Woah, can you please explain in a bit more detail how you did this? Thanks in advance!

Here is Bunuel 's post on number theory.

https://gmatclub.com/forum/math-number- ... ml#p666609

Search for the string "Finding the number of powers of a prime number p". You will see the method detailed there.

I'm happy to go in to more detail.

The question asks what is the range of possible values for “\(x\)” when \(\frac{13!}{2^x}\) is an integer.


In other words, you’re looking to find the range of values of "x" that make \(2^x\) divide \(13!\) evenly.

This is possible only when the number of factors of \(2\) in the numerator is greater than or equal to the factors of \(2\) in the denominator.


Finding the lower end of the range is easy. We simply observe that if "x" is zero, then we have \(2^0 = 1\), and any integer divided by 1 is an integer. \(13!/1\) is an integer. Therefore, zero is the lower end of the range.

Now we have to find the upper value in the range.



Given what we've said above (that the number of factors of \(2\) in the numerator must be greater than or equal to the factors of \(2\) in the denominator), we see that what we’re really trying to find is how many factors of “\(2\)” does \(13!\) have?


To begin, remember that \(13!\) is just the sequence: \((13*12*11*10* … *4*3*2*1.)\)



So first, how many multiples of “\(2\)” are on that list? Well, if we take \(13\) divided by \(2\), we get the count of numbers that are a multiple of \(2\).
\(13/2 = 6\). So there are six multiples of \(2\) in our sequence above. The numbers are: \((2,4,6,8,10,12)\).


But we’re not done yet. Notice that some of the numbers on the list we just created have MORE than one multiple of two in them (for example, \(4 = 2*2,\) and \(8 = 2*2*2\)). We need to count those extra two’s as well, because we're trying to count ALL of the factors of 2 in 13!.



So we next look for all the multiples of \(2^2\), or in other words, we look for all the multiples of \(4\).

\(13/4 = 3\). So we’ve counted three more 2’s. (\(4, 8, 12\)) We've counted one two in each of these numbers already, and now we are counting another one.


And finally, we look for all the multiples of \(2^3\), or in other words, all the multiples of \(8\).

\(13/8 = 1\) (\(8\) is the only number that works.) We've counted two of the multiples of two in 8, and now we need to count the last one.


And we’re almost done. Add up the number of \(2\)’s obtained in the three steps above (\(6 + 3 + 1\)) and you get the number of factors of “2” in the factorial \(13!\). (the answer is 10)




Here is a table of what you've just counted, which when you look at it, might clear up that we're just counting factors of 2.


\(1...\) none
\(2...\) = ONE FACTOR OF TWO \((2*1 = 2)\)
\(3...\) none
\(4...\) = TWO FACTORS OF TWO \((2*2 = 4)\)
\(5...\) none
\(6...\) = ONE FACTOR OF TWO \((3*2 = 6)\)
\(7...\) none
\(8...\) = THREE FACTORS OF TWO \((2*2*2 = 8)\)
\(9...\) none
\(10...\) = ONE FACTOR OF TWO \((5*2 = 10)\)
\(11...\) none
\(12...\) = TWO FACTORS OF TWO \((2*2*3 = 12)\)
\(13..\). none


The key concepts being tested in this question are: 1.) That a denominator will divide a numerator evenly (and result in an integer) if the numerator has at least ALL of the prime factors of the denominator. 2.) Method of finding the number of factors of (any number) in a factorial.

Let’s determine the maximum number of factors of 2 within 13!. It would be very time consuming to list out each multiple of 2 in 13!. Instead, we can use the following shortcut in which we divide 13 by 2, and then divide the quotient of 13/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient.

13/2 = 6 (we can ignore the remainder)

6/2 = 3

3/2 = 1 (we can ignore the remainder)

Since 1/2 does not produce a nonzero quotient, we can stop.

The next step is to add our quotients; that sum represents the number of factors of 2 within 13!.

Thus, there are 6 + 3 + 1 = 10 factors of 2 within 13!.

So, x can be between zero and 10 inclusive.

Answer: A