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)

\(6 + 3 + 1 = 10\)

ANSWER IS "A"

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 \((2*2*3 = 12)\)

\(13..\). noneThe 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.