Everything about Factorials on the GMAT

Question

FACTORIALS This post is a part of GMAT MATH BOOK ] created by: Bunuel edited by: bb , Bunuel -------------------------------------------------------- Definition The factorial of a non-negative integer n , denoted by n! , is the product of all positive integers less than or equal to n . For example: 4!=1*2*3*4=24 . Properties Factorial of a negative number is undefined. 0!=1 , zero factorial is defined to equal 1. n!=(n-1)!*n , valid for n\geq{1} . Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow. For example: 125,000 has 3 trailing zeros; The number of trailing zeros in the decimal representation of n! , the factorial of a non-negative integer n , can be determined with this formula: \frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k} , where k must be chosen such that 5^k\leq{n} Example: How many zeros are in the end (after which no other digits follow) of 32!? \frac{32}{5}+\frac{32}{5^2}=6+1=7 . Notice that the denominators must be less than or equal to 32 also notice that we take into account only the quotient of division (that is \frac{32}{5}=6 not 6.4). Therefore, 32! has 7 trailing zeros. The formula actually counts the number of factors 5 in 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. Finding the powers of a prime number p , in the n! The formula is: \frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k} , where k must be chosen such that p^k\leq{n} Example: What is the power of 2 in 25!? \frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22 . _________________________________________________________________________________________________ Questions to practice: https://gmatclub.com/forum/if-60-is-wri ... 01752.html https://gmatclub.com/forum/how-many-zer ... 00599.html https://gmatclub.com/forum/find-the-num ... 08249.html https://gmatclub.com/forum/find-the-num ... 08248.html https://gmatclub.com/forum/if-n-is-the- ... 01187.html https://gmatclub.com/forum/if-m-is-the- ... 08971.html https://gmatclub.com/forum/if-p-is-a-na ... 08251.html https://gmatclub.com/forum/if-10-2-5-2- ... 06060.html https://gmatclub.com/forum/p-and-q-are- ... 09038.html https://gmatclub.com/forum/question-abo ... 08086.html https://gmatclub.com/forum/if-n-is-the- ... 06289.html https://gmatclub.com/forum/what-is-the- ... 05746.html https://gmatclub.com/forum/if-d-is-a-po ... 26692.html https://gmatclub.com/forum/if-10-2-5-2- ... 06060.html https://gmatclub.com/forum/how-many-zer ... 42479.html

Bunuel · Accepted Answer

Yes, it is. I've separated them as for GMAT generally we need only trailing zeros and almost never other prime's power.

It's better to illustrate it on the example:
How many powers of 900 are in 50!
 900=2^2*3^2*5^2

Find the power of 2: 
 \frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47

=  2^{47}

Find the power of 3: 
 \frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22

= 3^{22}

Find the power of 5: 
 \frac{50}{5}+\frac{50}{25}=10+2=12

= 5^{12}

We need all of them (2,3,5) to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!
900^6

noboru · Answer

Point 1 is just point 2 for k=5, isnt it?

bb · Answer

This post has been split off the  original discussion  and cleaned up for reference.

shalva · Answer

Bunuel

Valuable post! +1

Can you please post this one too? It's still interesting, though may not be usable for GMAT

juukkk · Answer

If you actually go and check 32! in Excel the result will be 263130836933694000000000000000000000

So more like 21 zeros... I really hope Excel is making a mistake because of how neat is your formula but someone please explain!?

Bunuel · Answer

32! = 26313083693369353016721801216 0000000  This is what 32!  really  equals to.
32!= 263130836933694 000000000000000000000  Accoroding to Excell. Don't worry it's just rounded, so formula is correct.

juukkk · Answer

Kudos given

and formula memorized already. From where did you got the what "32! really equals to"?

bb · Answer

Wow. I want to know how you calculated it too

Bunuel · Answer

Wiki has the the answer to 32! on trailing zeros article.

aim-high · Answer

Hi,

I just went through this thread. I understand point# 1 about trailing zeroes. However, for the life of me, I cant understand Point#2. 
Can anyone please explain me what are we really trying to solve in "2. Finding the number of powers of a prime number k, in the n!."

For the question "What is the power of 2 in 25!", the answer is given as 22. What does it mean ?

Thanks

Bunuel · Answer

25! is some number, let's say x. Power of 2 (highest power, 2 will have) in 25!, means the power of 2 in prime factorization of x.

For example:  5!=120=2^3*3*5 , so the power of 2 in 5! is 3.

fatihaysu · Answer

We need all of them (2,3,5) to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!
900^6

bruel i just dont understand that part. CAn you explain it plz

Bunuel · Answer

50!=900^xa=(2^2*3^2*5^2)^x*a , where  x  is the highest possible value of 900 and  a  is the product of other multiples of  50! .

50!=2^{47}*3^{22}*5^{12}*b=(2^2*3^2*5^2)^6*(2^{35}*3^{10})*b=900^{6}*(2^{35}*3^{10})*b , where  b  is the product of other multiples of  50! . So  x=6 .

Below is another example:

Suppose we have the number  18!  and we are asked to to determine the power of  12  in this number. Which means to determine the highest value of  x  in  18!=12^x*a , where  a  is the product of other multiples of  18! .

12=2^2*3 , so we should calculate how many 2-s and 3-s are in  18! .

Calculating 2-s:  \frac{18}{2}+\frac{18}{2^2}+\frac{18}{2^3}+\frac{18}{2^4}=9+4+2+1=16 . So the power of  2  (the highest power) in prime factorization of  18!  is  16 .

Calculating 3-s:  \frac{18}{3}+\frac{18}{3^2}=6+2=8 . So the power of  3  (the highest power) in prime factorization of  18!  is  8 .

Now as  12=2^2*3  we need twice as many 2-s as 3-s.  18!=2^{16}*3^8*a=(2^2)^8*3^8*a=(2^2*3)^8*a=12^8*a . So  18!=12^8*a  -->  x=8 .

Hope it's clear.

yasar434 · Answer

can anyone tell me how to calculate the no.of factors for 20!. well this was a small number if i would have to answer a bigger number like no.of factors for 720! what will be the solution.

Bunuel · Answer

You won't need this for GMAT but still let's see if we can do it:

First we should make prime factorization of 20!. 20! will have all primes from 0 to 20, so we should find the powers of these primes in 20!.

Powers of 2 -->  \frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18 ;
Powers of 3 -->  \frac{20}{3}+\frac{20}{9}=6+2=8 ;
Powers of 5 -->  \frac{20}{5}=4 ;
Powers of 7 -->  \frac{20}{7}=2 ;
Powers of 11 -->  \frac{20}{11}=1 ;
Powers of 13 -->  \frac{20}{13}=1 ;
Powers of 17 -->  \frac{20}{17}=1 ;
Powers of 19 -->  \frac{20}{19}=1 .

So  20!=2^{18}*3^8*5^4*7^2*11^1*13^1*17^1*19^1 .

Next: How to Find the Number of Factors of an Integer.

First make prime factorization of an integer  n=a^p*b^q*c^r , where  a ,  b , and  c  are prime factors of  n  and  p ,  q , and  r  are their powers.
 
The number of factors of  n  will be expressed by the formula  (p+1)(q+1)(r+1) .  NOTE:  this will include 1 and n itself.

Example:  Finding the number of all factors of 450:  450=2^1*3^2*5^2

Total number of factors of 450 including 1 and 450 itself is  (1+1)*(2+1)*(2+1)=2*3*3=18  factors.

So, the # of positive factors of  20!=2^{18}*3^8*5^4*7^2*11^1*13^1*17^1*19^1  will be  (18+1)(8+1)(4+1)(2+1)(1+1)(1+1)(1+1)(1+1)=19*9*5*3*2*2*2*2=41040 .

The same way we can find for 720!, but we'll need much more time.

Hope it helps.

theptrk · Answer

Im getting a problem trying to use the formula. 
When looking for power of 3 in 35! i do
35/3 + 35/9 + 35/27 = 11+3+1 = 15

But I've multiplied out the factorial 32! and get 18 3's including the square for 9 and the cubed for 27. Am i doing something wrong?

Bunuel · Answer

Yes, as formula is correct then it must be that you have miscalculated.

By the way here is complete factorization of 35!:  35!=2^{32}*3^{15}*5^8*7^5*11^3*13^2*17^2*19*23*29*31 , so you can see that the power of 3 is indeed 15.

Impenetrable · Answer

How come I get completely different results when I use my calculator to get 25! and 2^22?

Did I understand it wrong?

Bunuel · Answer

25! is a huge number, not many calculators can handle it. Check whether it gives you the following result: 25!=15,511,210,043,330,985,984,000,000.

Everything about Factorials on the GMAT

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