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

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Joined: 14 Oct 2015
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If 13!/2^x is an integer, which of the following represents all possib  [#permalink]

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Updated on: 19 Jul 2017, 08:28
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72% (01:30) correct 28% (01:31) wrong based on 28 sessions

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If $$\frac{13!}{2^x}$$ is an integer, which of the following represents all possible values of $$x$$?

A - $$0\leq{x} \leq{10}$$
B - $$0 < x < 9$$
C - $$0\leq{x} < 10$$
D - $$1\leq{x} \leq{10}$$
E - $$1 < x < 10$$

Original question didn't have it but I think $$x$$ has to be a non-negative integer for this to be a viable question.

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Originally posted by jedit on 19 Jul 2017, 08:03.
Last edited by Bunuel on 19 Jul 2017, 08:28, edited 1 time in total.
Renamed the topic.
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Re: If 13!/2^x is an integer, which of the following represents all possib  [#permalink]

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19 Jul 2017, 08:19
jedit wrote:
If $$\frac{13!}{2^x}$$ is an integer, which of the following represents all possible values of $$x$$?

A - $$0\leq{x} \leq{10}$$
B - $$0 < x < 9$$
C - $$0\leq{x} < 10$$
D - $$1\leq{x} \leq{10}$$
E - $$1 < x < 10$$

Original question didn't have it but I think $$x$$ has to be a non-negative integer for this to be a viable question.

The value on the lower end should be zero, because $$2^0$$ = 1, and $$13!/1$$ is an integer.

To find the value on the higher end, I like to use Euclid.

$$13/2 =$$6 (discard the remainder)
$$6/2 =$$3
$$3/2 =$$1 (discard the remainder)

$$6 + 3 + 1 = 10$$

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Re: If 13!/2^x is an integer, which of the following represents all possib  [#permalink]

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19 Jul 2017, 09:29
1
jedit wrote:
Original question didn't have it but I think $$x$$ has to be a non-negative integer for this to be a viable question.

You're absolutely right - not only is the expression an integer for any negative integer value of x, it's also an integer for several values of x up to the base 2 logarithm of 13!, so the question is wrong on both the lower and upper bounds on x, as the question is written.
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Re: If 13!/2^x is an integer, which of the following represents all possib  [#permalink]

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19 Jul 2017, 12:18
None of the answer choices are negative, so obviously the solution set is only the whole numbers.
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Re: If 13!/2^x is an integer, which of the following represents all possib  [#permalink]

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19 Jul 2017, 12:39
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.
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Re: If 13!/2^x is an integer, which of the following represents all possib  [#permalink]

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19 Jul 2017, 13:34
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.
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Re: If 13!/2^x is an integer, which of the following represents all possib  [#permalink]

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19 Jul 2017, 19:09
jedit wrote:
If $$\frac{13!}{2^x}$$ is an integer, which of the following represents all possible values of $$x$$?

A - $$0\leq{x} \leq{10}$$
B - $$0 < x < 9$$
C - $$0\leq{x} < 10$$
D - $$1\leq{x} \leq{10}$$
E - $$1 < x < 10$$

Original question didn't have it but I think $$x$$ has to be a non-negative integer for this to be a viable question.

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

Sent from my LG-LS980 using GMAT Club Forum mobile app
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Re: If 13!/2^x is an integer, which of the following represents all possib  [#permalink]

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19 Jul 2017, 19:34
grimm1111 wrote:
jedit wrote:
If $$\frac{13!}{2^x}$$ is an integer, which of the following represents all possible values of $$x$$?

A - $$0\leq{x} \leq{10}$$
B - $$0 < x < 9$$
C - $$0\leq{x} < 10$$
D - $$1\leq{x} \leq{10}$$
E - $$1 < x < 10$$

Original question didn't have it but I think $$x$$ has to be a non-negative integer for this to be a viable question.

The value on the lower end should be zero, because $$2^0$$ = 1, and $$13!/1$$ is an integer.

To find the value on the higher end, I like to use Euclid.

$$13/2 =$$6 (discard the remainder)
$$6/2 =$$3
$$3/2 =$$1 (discard the remainder)

$$6 + 3 + 1 = 10$$

Woah, can you please explain in a bit more detail how you did this? Thanks in advance!
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Re: If 13!/2^x is an integer, which of the following represents all possib  [#permalink]

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19 Jul 2017, 19:38
yiyanf wrote:
grimm1111 wrote:
jedit wrote:
If $$\frac{13!}{2^x}$$ is an integer, which of the following represents all possible values of $$x$$?

A - $$0\leq{x} \leq{10}$$
B - $$0 < x < 9$$
C - $$0\leq{x} < 10$$
D - $$1\leq{x} \leq{10}$$
E - $$1 < x < 10$$

Original question didn't have it but I think $$x$$ has to be a non-negative integer for this to be a viable question.

The value on the lower end should be zero, because $$2^0$$ = 1, and $$13!/1$$ is an integer.

To find the value on the higher end, I like to use Euclid.

$$13/2 =$$6 (discard the remainder)
$$6/2 =$$3
$$3/2 =$$1 (discard the remainder)

$$6 + 3 + 1 = 10$$

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.
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Re: If 13!/2^x is an integer, which of the following represents all possib  [#permalink]

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24 Jul 2017, 17:52
1
yiyanf wrote:
grimm1111 wrote:
jedit wrote:
If $$\frac{13!}{2^x}$$ is an integer, which of the following represents all possible values of $$x$$?

A - $$0\leq{x} \leq{10}$$
B - $$0 < x < 9$$
C - $$0\leq{x} < 10$$
D - $$1\leq{x} \leq{10}$$
E - $$1 < x < 10$$

Original question didn't have it but I think $$x$$ has to be a non-negative integer for this to be a viable question.

The value on the lower end should be zero, because $$2^0$$ = 1, and $$13!/1$$ is an integer.

To find the value on the higher end, I like to use Euclid.

$$13/2 =$$6 (discard the remainder)
$$6/2 =$$3
$$3/2 =$$1 (discard the remainder)

$$6 + 3 + 1 = 10$$

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

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.
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Re: If 13!/2^x is an integer, which of the following represents all possib  [#permalink]

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25 Jul 2017, 10:51
jedit wrote:
If $$\frac{13!}{2^x}$$ is an integer, which of the following represents all possible values of $$x$$?

A - $$0\leq{x} \leq{10}$$
B - $$0 < x < 9$$
C - $$0\leq{x} < 10$$
D - $$1\leq{x} \leq{10}$$
E - $$1 < x < 10$$

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.

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Re: If 13!/2^x is an integer, which of the following represents all possib   [#permalink] 25 Jul 2017, 10:51
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