Question about p prime in to n factorial!

Question

So I was using the flash cards that Dreamy made for Quant, and had a question on the topic

Can you see if I am doing it right?

I made this up, it is not a real problem....

What is the highest power of 6 in 291! ?

by using quotient method this is how i worked:

291/6 = 48
48/6 = 8
8/6 = 1
2/6 =/ anything

48+8+1 = 57

Therefore, there are 57 powers of 6 in 291!?

Did I do this correctly?

Bunuel · Accepted Answer

6 is not a prime number. When determining the highest power of non-prime in n! the approach should be slightly different. Check this: https://gmatclub.com/forum/everything-ab ... 85592.html (main thread about this issue with examples of determining the highest power of primes as well as non-primes in n!); Other examples: https://gmatclub.com/forum/gmat-club-m12-100599.html https://gmatclub.com/forum/if-n-is-the-p ... 06289.html https://gmatclub.com/forum/facorial-ps-105746.html As for your question: what is the highest power of 6 in 291!? 6=2*3 so you'll need as many 3-s as 2-s, so the power of 3 in 291! will determine the highest power of 6 in 291! (as highest power of 3 will be obviously less than the highest power of 2. Or in other words there will be less 3-s than 2-s in 291!, so the power of 3 will be limiting factor): 291/3+291/9+291/27+291/81+291/243=97+32+10+3+1=143 (the formula actually counts the number of factors 3 in 291!, but since there are at least as many factors 2, this is equivalent to the number of factors 6). So the highest power of 6 in 291! is 143. P.S. I doubt that you'll be asked to determine the highest power of non-prime (except the power of 10) in n! on the GMAT.

Wishbone · Answer

as soon as i posted this i knew what i did wrong, that's why the example was power of 2 in 25!. must be prime. thanks for clearing that up

pgmatk · Answer

Hi,

So in the below, I don't follow why "3" will be the limiting factor? Also, if highest power of 3 will be less that that of 2 how does that mean more 3's than 2's in 291!? Maybe it is obvious but I am not able to figure this out.

"...so the power of 3 in 291! will determine the highest power of 6 in 291! (as highest power of 3 will be obviously less than the highest power of 2. Or in other words there will be more 3-s than 2-s in 291!..."

Thanks!

pgmatk · Answer

Ok never mind, it is because 3 is a larger number , it needs fewer powers to express a number!

testcracker · Answer

if there are more 3s, then how cannot be the power highest ....?

ccooley · Answer

There are actually fewer 3s than 2s. (I think Bunuel may have made a typo in the very old post above).

To see why, try a smaller example (291! is way too big to write out all of the factors.) Let's look at 10!.

10! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10

The numbers in that product that contain a 2 are 2, 4, 6, 8, and 10.

2, 6, and 10 each contribute one 2. 
4 contributes two 2s, since it's actually 2*2.
8 contributes three 2s, since it's actually 2*2*2.

In total, there are eight 2s.

The numbers that contain a 3 are 3, 6, and 9. Notice that this is every third number, rather than every other number. That's why there are fewer 3s than there are 2s.

3 and 6 each contribute one 3. 
9 contributes two 3s.

In total, there are four 3s.

So, the 3s are the limiting factor. You need one 2 and one 3 to make a 6. How many 6s can you make? Well, you're limited to no more than four, because after that point, you run out of 3s. So, you can divide 6 out of 10! four times.

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Question about p prime in to n factorial!

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