Everything about Factorials on the GMAT

Consider 12!. Now, if we prime factorize 12! we'll get that the power of 2 in 12! will naturally be higher than the power of 5. To get one trailing zeros we need one 2 and one 5. Thus if we know what is the power of 5 in 12! we'll know how many trailing zeros 12! has.

Does this make sense?

If I go to excel and calculate factorial 50! and then divide by 900^6, i still have room for 900^7 which then still yields a positive result. Could you explain that further?

3.04141E+64 =50!
5.31441E+17 =900^6
5.722948210942210000000000E+46 =Quotient (50!/900^6)

Thank you very much

Use better calculator: https://www.wolframalpha.com/input/?i=50%21%2F900%5E6

Bunuel
could you please explain what this mean ? >> The formula actually counts the number of factors 5 in n!n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
Thank you!

We are getting 0 by multiplying 5 by 2. n! will have more 2's than 5's (in some cases equal number but you cannot have more 5's than 2's). So, if we count the number of 5's we;d be sure that there will be enough 2's for these 5's to make 0. So, the formula counts this number, the number of 5's in n!.

Hope it's clear.

This is a type of question you see on GMAT at times framed as follows:-
Q. What is the greatest integer k such that 2^k is a factor of 25!?

Bunuel chetan2u KarishmaB niks18 GMATPrepNow EMPOWERgmatRichC

What is the rationale behind choosing 5 and its powers in the denominator?
for e.g. in terminating decimals I need to know whether the denominator contains either of 2 or 5 or its powers, else the long division may go indefinitely.

If I do not recall this formula in exam under timed condition, is there any alternate approach to this?

I have explained why you need 2 and/or 5 in the denominator for a terminating decimal in this post:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/1 ... -the-gmat/

In number properties questions, if you are unable to recall the property, look for a pattern by using some simple numbers.
I am not sure to which question you are referring but if you point it out, I can tell you how to solve it without "knowing" the property.

Thank you KarishmaB for your two cents.

Below is a relevant quote from your blog:



Why do we not need to need to know whether b is greater than (or not) 1? I could not understand highlighed text in context of long division, Can you elaborate?

I would be grateful if you could explore an additional approach for factorial problem. TIA.

If b = 1, a/b is an integer and the question of terminating/non-terminating decimal doesn't arise. b cannot be 0 since then a/b is not defined for b = 0.
b would take values out of 2/3/4/5... etc

When you divide 1 by 3 by long division, how do you do it?

You will keep getting 0s after the decimal but it will never be fully divisible.
Only if you divide by 2 and/or 5 or some combination of them will it be fully divisible. 10/100/1000/10000 etc have only 2 and 5 as factors.

Hi, I have a question about the formula for trailing zeros. The example gives us something like 32, which cannot be wholly divided by 5. However, if our number is something like 350, where 5 is a factor of 350, can we still use 5 in the formula? In this case, wouldn't it be that the number of 2 is not equal to the number of 5 any more?

The factorial of a positive integer n, denoted by n!, is composed of the product of integers from 1 to n: n! = 1*2*3*4*5*... Since half of these integers are even, while only one-fifth of the integers are multiples of 5, the power of 2 in n! is always greater than the power of 5 (for n > 1). To calculate the number of trailing zeros in n!, we need to find the power of 5 in prime factorization of n!, since for each 5, there will always be a corresponding 2 in n!, to multiply with it and produce a trailing zero.

For example, the number of trailing zeros in 350! is 86, because the power of 5 in prime prime factorization of n! is 86.

350/5 + 350/25 + 350/125 = 70 + 14 + 2 = 86 (remember to consider only the quotient of the division).

However, the power of 2 in prime prime factorization of n! is 344:

350/2 + 350/4 + 350/8 + 350/16 + 350/32 + 350/64 + 350/128 + 350/256 = 175 + 87 + 43 + 21 + 10 + 5 + 2 + 1 = 344 (remember to consider only the quotient of the division).

Factoring 350! gives:

2^344*3^171*5^86*7^58*11^33*13^28*17^21*19^18*23^15*29^12*31^11*37^9*41^8*43^8*47^7*53^6*59^5*61^5*67^5*71^4*73^4*79^4*83^4*89^3*97^3*101^3*103^3*107^3*109^3*113^3*127^2*131^2*137^2*139^2*149^2*151^2*157^2*163^2*167^2*173^2*179*181*191*193*197*199*211*223*227*229*233*239*241*251*257*263*269*271*277*281*283*293*307*311*313*317*331*337*347*349

Notice that the power of 2 (344) is higher than the power of 5 (86). Additionally, note that the number of trailing zeros in 350! is equal to the power of 5 in its prime factorization.

Hope it helps.

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