If 60! is written out as an integer,

[katejohn7]

with how many consecutive 0's will that integer end?

I don't see this question posted up here so I am hoping someone can guide me through the explanation of the answer of 14. I read the explanation on the CAT and it is still not making sense to me.

I appreciate your help.

[kartikeya.payautomata]

hi there

in case of finding 10's present in any factorial we have to ind out how many 5's were present in the product of those integers present in the factorial
since 2 will be abundantly present in the product it is not worth finding it
now the question is why do we find 2's and 5's since only these two numbers produce a 10
so for finding number of zeros divide 60 by 5
=12
now 25 will also be present so 60/25(only interested in quotient)=2

now 125 could not be present since quotient will be 0 and remainder is 60

so simply add the quotients=12+2=14

[gnetiq]

with how many consecutive 0's will that integer end?

Hi, as kartikeya mentioned, we will need to find out how many multiples of 10 (60, 58*55, 50, etc.) are present in 60!. Furthermore, 10 has two prime factors, 2 and 5. We are essentially looking for the number of multiples of the bigger prime number i.e. 5 in 60!

The number of multiples of p in the numbers 1 to n is given by

n/(p)^1 + n/(p)^2 + n/(p)^3 + ........

(for each term above, consider the greatest integer value equal to or less than the term)

a) 60 / (5)^1 = 60/5 => greatest integer value is 12, therefore 12 multiples of 5 in 60!
b) 60/(5)^2 = 60/25 => greatest integer value (quotient) is 2, therefore 2 multiples of 25 in 60!
We cannot go any further because any higher power of 5 will not give an integer value (quotient).

Adding the above results, we get 12 + 2 = 14 possible multiples of 5 in 60! The same stands true for 10. If 10 has 14 multiples in the factorization of 60!, that implies:

60! = some integer x (10)^14

therefore, 60! will have 14 consecutive zeros in the end.

[kean.allison]

Can someone explain why we need to count the 5's in 25 and 50 too? I don't understand that part of the explanation - it seems like those multiples of 5 would already have been included in the multiples of 5 that make up 60.

[gnetiq]

Can someone explain why we need to count the 5's in 25 and 50 too? I don't understand that part of the explanation - it seems like those multiples of 5 would already have been included in the multiples of 5 that make up 60.

Hi, 60/5 gives us 12 multiples of 5 in 60!, but that does not mean that 5 has total 12 multiples only in factorization of 60!

For example, 2 is divisible by 2. So is 4. But the difference with 4 is that when we divide it with 2, it gives us a 2 which is again divisible by 2. Therefore, if I want to find the highest power of prime number 2 that divides (4x3x2), I will need to account for the fact that 4 has two factors of 2.

Similarly, when trying to find the highest power of 5 that divides 60!, we need to count all those numbers too which have two (or higher) factors of 5.

Between 1 and 60, there are 12 numbers divisible by 5 (these are 60, 55, 50, 45, 40, 35, 30, 25, 20, 15, 10, and 5).

But when we divide 50 and 25 by 5, they leave 10 and 5 respectively which are also divisible by 5. We need to account for these two as well. Hence, the total number of multiples becomes 12 + 2 = 14.

~~~~
There's actually another formula too to determine the exact power of a given prime p which divides n! (but you will need to know or learn how to convert a decimal to a different base -- it's pretty simple though!)

[n - (digit sum of n in base p)] / (p - 1)

In this case it becomes => (60 - digit sum of 60 in base 5) / (5-1)

60 converted to base 5 is 220. The digit sum, hence, is 2+2+0 = 4.

Therefore, (60 - 4)/(5-1) gives us 56/4 = 14
~~~~

[RonPurewal]

post33056.html#p33056

[JaveriaK678]

Why do we consider 5's and not 2's?

I understand 5's gives 14 zeros

And 2's give 56 zeros.

But why do we only consider 5?

[RonPurewal]

each zero comes from a 5 AND a 2. so, whichever one runs out first (= whichever one you have fewer of) is the only one that matters.

if this doesn't make sense, just try multiplying together different numbers of 5's and 2's, and pay attention to the behavior of the results.