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# M32-01

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Math Expert
Joined: 02 Sep 2009
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21 May 2016, 03:21
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How many trailing zeroes does $$49! + 50!$$ have?

A. 9
B. 10
C. 11
D. 12
E. 22
[Reveal] Spoiler: OA

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21 May 2016, 03:21
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Official Solution:

How many trailing zeroes does $$49! + 50!$$ have?

A. 9
B. 10
C. 11
D. 12
E. 22

Factor out 49! from the expression: $$49! + 50!=49!(1+50)=49!*51$$.

51 won't contribute to the number of zeros at the end of the number, therefore all zeros will come from 49!.

Trailing zeros in 49!: $$\frac{49}{5}+\frac{49}{5^2}=9+1=10$$

THEORY:

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 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 $$5^{(k+1)} \gt n$$

It's easier if we consider an example:

How many zeros are in the end (after which no other digits follow) of 32!?

$$\frac{32}{5}+\frac{32}{5^2}=6+1=7$$. Notice that the last denominator ($$5^2$$) must be less than 32. Also notice that we take into account only the quotient of the division, that is $$\frac{32}{5}=6$$.

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

Another example, how many trailing zeros does 125! have?

$$\frac{125}{5}+\frac{125}{5^2}+\frac{125}{5^3}=25+5+1=31$$,

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.

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28 May 2016, 09:10
Hello,
49! has 10 trailing zeros OK.
But 49!(10*5+1)? doesn't the 10 in the expression add an extra zero to the trailing zeros of 49! ?

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28 May 2016, 09:19
1
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Avigano wrote:
Hello,
49! has 10 trailing zeros OK.
But 49!(10*5+1)? doesn't the 10 in the expression add an extra zero to the trailing zeros of 49! ?

Hi,
no, 49!(10*5+1) = 49!(50+1) = 49!*51...
Now 51 does not contain any 5 in it, so the term depends on ONLY 49! and as you have said 49! has 10 5s
so ans is 10
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Manager
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28 May 2016, 09:38
chetan2u wrote:
Avigano wrote:
Hello,
49! has 10 trailing zeros OK.
But 49!(10*5+1)? doesn't the 10 in the expression add an extra zero to the trailing zeros of 49! ?

Hi,
no, 49!(10*5+1) = 49!(50+1) = 49!*51...
Now 51 does not contain any 5 in it, so the term depends on ONLY 49! and as you have said 49! has 10 5s
so ans is 10

Ok! got it! thanks

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29 Dec 2016, 13:13
chetan2u wrote:
Avigano wrote:
Hello,
49! has 10 trailing zeros OK.
But 49!(10*5+1)? doesn't the 10 in the expression add an extra zero to the trailing zeros of 49! ?

Hi,
no, 49!(10*5+1) = 49!(50+1) = 49!*51...
Now 51 does not contain any 5 in it, so the term depends on ONLY 49! and as you have said 49! has 10 5s
so ans is 10

Hi All - I don't 100% understand the rationale behind 51 not contributing to the number of trailing zeros. Does 51 not contain 10^1? Thanks.

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29 Dec 2016, 21:28
Dear Bunuel,

I know the concept and deduced the final step 49! * 51. But 49/5 = 9.8 ~ 10 and 49/25 =1.96 ~ 2 .So I marked the answer as 12 instead of 10.Please clarify,if we need to select the lowest rounded integer or do we need to consider the nearest rounded integer to answer these kind of questions.

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29 Dec 2016, 23:37
GouthamNandu wrote:
Dear Bunuel,

I know the concept and deduced the final step 49! * 51. But 49/5 = 9.8 ~ 10 and 49/25 =1.96 ~ 2 .So I marked the answer as 12 instead of 10.Please clarify,if we need to select the lowest rounded integer or do we need to consider the nearest rounded integer to answer these kind of questions.

Yes, you should take only the quotient into the account, that is 49/5 = 9 and 49/25 = 1.

Check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.

Hope it helps.
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04 Jan 2017, 13:55
chetan2u wrote:
Avigano wrote:
Hello,
49! has 10 trailing zeros OK.
But 49!(10*5+1)? doesn't the 10 in the expression add an extra zero to the trailing zeros of 49! ?

Hi,
no, 49!(10*5+1) = 49!(50+1) = 49!*51...
Now 51 does not contain any 5 in it, so the term depends on ONLY 49! and as you have said 49! has 10 5s
so ans is 10

Hi. How does 51 not contain any 5s?

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05 Jan 2017, 02:23
Cez005 wrote:
chetan2u wrote:
Avigano wrote:
Hello,
49! has 10 trailing zeros OK.
But 49!(10*5+1)? doesn't the 10 in the expression add an extra zero to the trailing zeros of 49! ?

Hi,
no, 49!(10*5+1) = 49!(50+1) = 49!*51...
Now 51 does not contain any 5 in it, so the term depends on ONLY 49! and as you have said 49! has 10 5s
so ans is 10

Hi. How does 51 not contain any 5s?

51 = 3*17, so he means that 51 does not have 5 as a factor.
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20 Jan 2017, 09:16
I think this is a high-quality question.

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17 Mar 2017, 21:05
Sorry i have a question. what is wrong if you factor 49 and 50 factorial and add??
So you will get 22

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18 Mar 2017, 01:09
muthappashivani wrote:
Sorry i have a question. what is wrong if you factor 49 and 50 factorial and add??
So you will get 22

Please elaborate what you mean. Write a solution. In it's current form it's not clear at all what you mean.
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15 Jul 2017, 12:08
Hi,

I solved with the trailing zero method only but I calculated for 49! and 50! separately and I got 22 as answer. I want to know how that is a wrong method of solving. Thanks!
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16 Jul 2017, 03:14
ashikaverma13 wrote:
Hi,

I solved with the trailing zero method only but I calculated for 49! and 50! separately and I got 22 as answer. I want to know how that is a wrong method of solving. Thanks!

With that logic the number of trailing zeros of 10 + 100 would be 1 + 2 = 3 but in fact 10 + 100 = 110 and 110 has 1 trailing zero.

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30 Sep 2017, 09:04
This was confusing for me too, but the logic lies in the fact that 49! is a factorial. 1x2x3x4....40x41x...49. vs. 51 which is just 17x3. There are no 5s in 51. The formula allows you to count how many 5s are in factorials. If it was 51! (factorial) then yes it would have 12 fives, but the answer is 10 because 51 is NOT a factorial in this problem (after you simplify it.)

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Re: M32-01   [#permalink] 30 Sep 2017, 09:04
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# M32-01

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