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Orange08
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Whenever you want to find the number of zeros in a N! then do the following :

devide N by 5 such that
(N/5) + (N/5^2) + (N/5^3) + ..... unless 5^p where p= 1, 2,3 ... is more than N ..

Eg : let's say you want to find the number of Zero's in 125! so
divide 125/5 = 25 then
divide 125/5^2 =125/25= 5 then
divide 125/5^3= 125/125=1 ,
so a total of 25 +5+1 trailing zeros will be present. Always consider the rounded figures .In the original example :
60/5 = 12
60/5^2 = 60/25=2.4 , however you are not concerned with the decimal values here, so take this as 2
next would be 60/5^3 = 60/125 , so this would be (.some number) so stop your division here.
Whenever the denominator exceeds numerator , stop the process. Add the values to get the answer.

what would you do if the question asks to find the maximum power of 3 in 50! ?
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Bunuel
Orange08
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
6
12
14
42
56

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 .....
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Bunuel
Orange08
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
6
12
14
42
56

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.

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
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Bunuel
Orange08
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
6
12
14
42
56

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 .....

We take into account only the quotient of the division, that is 32/5=6.

keiraria
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

The last denominator (5^2) must be less than numerator (60).

Hope it's clear.
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Bunuel
Orange08
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
6
12
14
42
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,

This method makes complete sense but question for you regarding the accounting for "2" part.

You say that "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."

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!
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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!

The point is that you need both a 2 and a 5 to make a 10. If I have 100 2s but only 3 5s, I can make only 3 10s. No number of 2s alone can make a 10. So even if there are many more 2s, they are useless to us because we have limited number of 5s.
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Orange08
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

60! = (1)(2)(3)(4)(5)(6) . . . (57)(58)(59)(60)

KEY CONCEPT: For every pair of one 2 and one 5, we get a product of 10, which accounts for one zero at the end of the integer.
So, the question is "How many pairs of one 2 and one 5 are "hiding" in the product?"

Well, there is no shortage of 2's hiding in the product. In fact, there are FAR MORE 2's than 5's. So, all we need to do is determine how many 5's are hiding in the product.
60! = (1)(2)(3)(4)(5)(6) . . . (57)(58)(59)(60)
= (1)(2)(3)(4)(5)(6)(7)(8)(9)(2)(5)(11)...(3)(5)...(4)(5)...(5)(5)...(6)(5)...(7)(5)...(8)(5)...(9)(5)...(2)(5)(5)...(11)(5)...(56)(57)(58)(59)(12)(5)
In total, there are 14 5's hiding in the product.
And there are MORE THAN 14 2's hiding in the product.

So, there are 14 pairs of 2's and 5'2, which means the integer ends with 14 zeros

Answer: C

Cheers,
Brent
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Orange08
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

To determine the number of trailing zeros in a number, we need to determine the number of 5-and-2 pairs within the prime factorization of that number.

Since we know there are fewer 5s in 60! than 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.

To determine the number of 5s within 60!, we can use the following shortcut in which we divide 60 by 5, then divide the quotient of 60/5 by 5 and continue this process until we no longer get a nonzero quotient.

60/5 = 12

12/5 = 2 (we can ignore the remainder)

Since 2/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 60!.

Thus, there are 12 + 2 = 14 factors of 5 within 60!

Since there are 14 factors of 5 within 60!, there are 14 5-and-2 pairs and thus 14 trailing zeros.

Answer: C
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Orange08
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

Let's first make a few observations:

1300 = (2)(2)(5)(5)(13)
8000 = (2)(2)(2)(2)(2)(2)(5)(5)(5)
140,000 = (2)(2)(2)(2)(2)(5)(5)(5)(5)(7)

Notice that for each PAIR of one 2 and one 5, we get a zero at the end of the number.

So, our question really becomes "How many PAIRS of one 2 and one 5 are "hiding" in the prime factorization of 60!"

ASIDE: There are wayyyyyy more 2's hiding in the prime factorization of 60! than there are 5's hiding in the prime factorization.
So, if we know the number of 5's hiding in the prime factorization, then we can rest assured knowing that there will be enough 2's to pair up with each 5.

So, let's determine how many 5's are "hiding" in the prime factorization of 60!

60! = (60)(59)(58)......(3)(2)(1)

60 = (2)(2)(3)(5) [one 5]
55 = (5)(11) [one 5]
50 = (2)(5)(5) [two 5's]
45 [one 5]
40 [one 5]
35 [one 5]
30 [one 5]
25 [two 5's]
20[one 5]
15 [one 5]
10 [one 5]
5 [one 5]

Add them all to get: 14 fives

Answer: C

Cheers,
Brent
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Orange08
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

We need to find trailing zeros in 60!

Every consecutive zero = Multiple of 10

i.e. Total Number of zeros in 60! = Total power of 10 in 60!

But every 10 = 2*5
i.e. Total power of 10 in 60! = Total Pairs of 2 and 5 in 60!

Power of 2 in 60! > power of 5 in 60! bigger the prime number i.e. lower it's frequency and thereby lower power in an factorial

i.e. Power of 5 will be detrimental of power of 10 in 60 !

Power of 5 in 60! = [60/5] + [60/5^2] + [60/5^3] = 12+2+0 = 14
Power of 2 in 60! = [60/2] + [60/2^2] + [60/2^3] + [60/2^4] + [60/2^5]+ [60/2^6] = 30+15+7+3+1= 56

i.e. \(60! = 2^56*5^14*\)....

i.e. \(60! = 10^14*\)...

i.e. 14 consecutive Zeros

Answer: Option C
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Solution



To find
    • Number of zeroes at the end of 60!.

Approach and Working out
    • Number of zeroes = [60/5]+[60/25]
    • =12 +2 = 14
Hence, option c is the correct answer.

Correct Answer: Option C
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Bunuel
Orange08
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
6
12
14
42
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
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Bunuel
Orange08
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
6
12
14
42
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

https://gmatclub.com/forum/if-60-is-wri ... l#p2446781

Hi Farina

The method is applicable only to find out the exponent of Prime Number so you can't use any non-prime number as divisor to find it's power in any factorial

Since 10 = 2*5 hence it's not a Prime number therefore we don't divide factorial by 10 to find the power of 10 in any factorial.
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"So there are 7 zeros in the end of 32!"

32! = 263,130,836,933,694,000,000,000,000,000,000,000
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"So there are 7 zeros in the end of 32!"

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

Use better calculator:

32! = 263,130,836,933,693,530,167,218,012,160,000,000
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Bunuel
Orange08
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
6
12
14
42
56

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 .....



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

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Alexandrea176456
No, please just to clarify must I use 5. What’s the maths behind using 5

Please review the discussion above carefully, and follow the links to theory and similar questions from this post

Hope it helps.
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