Finding the number of powers of a number p in n!

Question

I have two questions on the topic, the  first  relates to finding the number of powers of a  prime  number p in n! and the  second  relates to finding the number of powers of a  non-prime  in n!.

1) According to the GMAT Club Math book, the formula is:

n/p + n/(p^2) + n/(p^3).. until (p^x)<n

However, this doesn't seem to work for the powers of 2 in 4! ?

We know that 4! = 4*3*2*1 = (2*2)*3*2*1 = (2^3)*3, so the answer we get via the above formula should be 3. However, if we go ahead and do this:

n/p = 4/2 = 2? If I take it one term further 4/(2^2) then we get 4/2 + 4/(2^2) = 2+1=3, which is indeed the right answer but this extra term violates the condition (p^x)<n ? The only thing I can think of is that p^x should be less than  or equal to  n and not just less than?

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2) The second question I have is regarding this (also from the GMAT Math Book)

https://i.ibb.co/p1FHKbM/zzzzzzzzzzzzzzz.png

I understand everything here apart from the last sentence. Can someone explain?

vijyband99 · Answer

It means that to find how many times can 900 divide into 50!..to find this we had figured out the prime factors of 900 those are 2,3and 5 in their square...so the last statement talks of 2,3&5 being represented twice (square power of these prime numbers). Minimum power out of 2,3&5 is for 5 which has 12th power in 50!. So power of 5 will limit the power of 900 that will be represented in 50! We are looking for square of 5 so 5 to the power 12 gives us 6power of 5 square. Hence our answer is 6.

Please give Kudos if you like the explanation.

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Please give Kudos if you like the explanation.

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jamalabdullah100 · Answer

I'm still confused. I don't follow the bold part if your explanation..

IanStewart · Answer

The method you're trying to learn is one I've never used on a real GMAT question, and I've seen several thousand official problems. In real GMAT questions, the numbers are always small enough that you don't need this particular method, which is only required when you have huge numbers. So I wouldn't worry about this at all - if you can answer a question like the OG question which asks something like "What is the largest integer k for which 3^k is a divisor of 30!", which you can do by hand (just by breaking down all the multiples of 3 up to 30), then you're prepared for any similar kind of question you might see on the real test.

jamalabdullah100 · Answer

Interesting. I've done all the OG questions and I did wonder why these sort of questions weren't asked in any of the three books. Not sure then why this is part of the GMAT CLUB Math book?

ccooley · Answer

You're correct, it should be less than or equal to. What you're doing in that formula, is counting the numbers in the factorial that have factors of p. If n itself includes one or more factors of p, it should be counted as well.

The prime factorization of 900 is 2*2*3*3*5*5. So, to divide a number by 900, you need to divide it by 2  twice , by 3  twice , AND by 5  twice . The explanation shows that you can divide 50! by 5 a total of 12 times. Therefore, you can divide 50! by 5*5 (what you need to do in order to divide by 900), a total of 6 times.

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Finding the number of powers of a number p in n!

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Finding the number of powers of a number p in n!

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