If 60! is written out as an integer, with how many

Trailing zeros:
Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 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+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

So there are 7 zeros in the end of 32!

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.

BACK TO THE ORIGINAL QUESTION:

According to above 60! has \(\frac{60}{5}+\frac{60}{25}=12+2=14\) trailing zeros.

Answer: C.

For more on this issues check Factorials and Number Theory links in my signature.

Hope it helps.

Trailing zeros:
Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 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+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

So there are 7 zeros in the end of 32!

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.

BACK TO THE ORIGINAL QUESTION:

According to above 60! has \(\frac{60}{5}+\frac{60}{25}=12+2=14\) trailing zeros.

Answer: C.

For more on this issues check Factorials and Number Theory links in my signature.

Hope it helps.

a question Bunuel (just for sure): 32/25 is 1.28 ....BUT if the result was for instance a number / another number = 1,764 we round it to 2 (the next integer) ??

I hope to be clear with my question .....

Hello Brunuel
Thanks for this great answer
but I am not familiar at all with trailing zeros
How did you determined the limit to raise the power
up to K ? how did you get the K

BEST regards

keiraria

Trailing zeros:
Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow.

125000 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+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

So there are 7 zeros in the end of 32!

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.

BACK TO THE ORIGINAL QUESTION:

According to above 60! has \(\frac{60}{5}+\frac{60}{25}=12+2=14\) trailing zeros.

Answer: C.

For more on this issues check Factorials and Number Theory links in my signature.

Hope it helps.

How do we know that there aren't more factors of 2 vs. 5? If there were more factors of 2, would we modify the equation to account for powers of 2 in the denominator?

Thanks!

If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56

If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56

If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56

If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56

Trailing zeros:
Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow.

125000 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+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

So there are 7 zeros in the end of 32!

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.

BACK TO THE ORIGINAL QUESTION:

According to above 60! has \(\frac{60}{5}+\frac{60}{25}=12+2=14\) trailing zeros.

Answer: C.

For more on this issues check Factorials and Number Theory links in my signature.

Hope it helps.

Hi Bunuel,

I know the method of trailing zeros. I have just one confusion about the divisor.
if 2^k is a factor of 10! we divide by 2. Agreed
if 6^k is a factor of 40! we divide by 3 not 2. Again agreed
In this question it is asking about zeros so why 5 and not 10 as divisor?
The divisor is the confusing part

"So there are 7 zeros in the end of 32!"

32! = 263,130,836,933,694,000,000,000,000,000,000,000

a question Bunuel (just for sure): 32/25 is 1.28 ....BUT if the result was for instance a number / another number = 1,764 we round it to 2 (the next integer) ??

I hope to be clear with my question .....

No, please just to clarify must I use 5. What’s the maths behind using 5