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# How many trailing Zeroes does 49! + 50! have?

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How many trailing Zeroes does 49! + 50! have?  [#permalink]

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27 Apr 2016, 06:58
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How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22

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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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27 Apr 2016, 17:14
2
10
debbiem wrote:
i think the answer is 9,
50!+49!= 49!(1+50)
51*49!,
49 factorial has 9 trailing zeroes, sor 51*49! also should have 9 trailing zeroes.

(A).

Please let me know if my technique is wrong.

49! has 10 and not 9 trailing zeroes.

Number of trailing zeroes for n! -->INTEGER sum of n/5 + n/(5^2) + ....+ n/(5^k), where k is a positive integer such that n<5^k.

Thus for 49! ---> 49/5 + 49/(5^2) [just take the integer values here] ---> 9+1 = 10 trailing zeroes.

We do this as any zero will be made by a combination of a 2 and a 5. In bigger number factorials, there will be more 2s than 5s. Thus 5s will become the critical factor.

Hope this helps.
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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27 Apr 2016, 08:22
3
2
chetan2u wrote:
How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22

OA in 1 day

49! + 50!

= (49*48*47........ ) + (50*49*48*47.....)

= 49! ( 1 + 50 )_____________ ( Take 49*48*47........common )

= 49! * 51

No of zeroes in 49! is

49/5 = 9
9/5 = 1

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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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28 Apr 2016, 08:43
3
Engr2012 wrote:
Abhishek009 wrote:
Health Discussion going on around here Would like to share something interesting !!!

I hope all of you agree till this part - 51*49! ( maipenrai it won't be 49! + 50!)

Now the game begins

As others have correctly said we need to find the no of 5's in 49! ( Which will be 10)

Here is why we don't need to find the number of 5's in 51 -

49! * 51 => ( 49 * 48 * 47......13.....7 ....3 * 2 * 1) ( 13 * 7 )

Here is where maipenrai went wrong

49! Factorial ( the red part ) has 10 zeroes , but the blue part doesn't have a 5 neither a 2...

Had the number been 50 instead , we would have one additional zero ( 10+ 1) ; but thats not the issue here

The most straightforward way is to directly see how many trailing zeroes 49! and 50! will have and then pick the lower number. You dont have to then worry about number of 5s in 51 or any other number this way. But yes, if you do go the way you are mentioning, then yes, you need to take care of the additional zeroes that might be added.

Hi Engr2012 and Abhishek009,

here this may be true and most of the time will be TRUE but in some cases it may NOT..
I'll just explain with an example..
53!+54! ..
53! will have 53/5 +53/25 = 10+2 = 12..
54! will also have 54/5 +54/25 = 12..
so our answer should be the LOWER or 12 as both are the SAME..
But is it so -- NO..
53!+54! = 53!( 1+54) = 53!*55..
so here we have another 5 in 55 ..
our answer will be 12+1 = 13..

we can tru it wityh smaller number
3!+4!.. should be NONE..
3!=6 and 4! =24 so 3!+4! = 6+24 =30.. We have ONE trailing ZERO..

we should be careful with numbers..
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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21 May 2016, 03:27
2
2
chetan2u wrote:
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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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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27 Apr 2016, 17:25
1
So in order to find a number of zeros in X! we need to divide the number by $$5^n$$ as long as $$5^n<=X$$

So for 49!+50!:

$$49/5+49/25=9+1=10$$
$$50/5+50/25=10+2=12$$

10+12=22

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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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27 Apr 2016, 17:32
1
1
maipenrai wrote:
So in order to find a number of zeros in X! we need to divide the number by $$5^n$$ as long as $$5^n<=X$$

So for 49!+50!:

$$49/5+49/25=9+1=10 50/5+50/25=10+2=12$$

10+12=22

You are missing an important point here. If the question would have been 49!*50! , then yes your approach would have been correct. But since it is addition, you can not simply add the number of trailing zeroes.

Example,

Say, A = 100, B = 1000, A+B =1100 (2 trailing zeroes and not 5 trailing zeroes).

A=10000, B = 100000 , A+B = 110000 (4 trailing zeroes and not 9 trailing zeroes) etc. You see that the number of trailing zeroes = lower number of trailing zeroes out of the 2.

Hope this helps.
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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28 Apr 2016, 08:03
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2
Health Discussion going on around here Would like to share something interesting !!!

I hope all of you agree till this part - 51*49! ( maipenrai it won't be 49! + 50!)

Now the game begins

As others have correctly said we need to find the no of 5's in 49! ( Which will be 10)

Here is why we don't need to find the number of 5's in 51 -

49! * 51 => ( 49 * 48 * 47......13.....7 ....3 * 2 * 1) ( 13 * 7 )

Here is where maipenrai went wrong

49! Factorial ( the red part ) has 10 zeroes , but the blue part doesn't have a 5 neither a 2...

Had the number been 50 instead , we would have one additional zero ( 10+ 1) ; but thats not the issue here
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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28 Apr 2016, 08:09
1
Abhishek009 wrote:
Health Discussion going on around here Would like to share something interesting !!!

I hope all of you agree till this part - 51*49! ( maipenrai it won't be 49! + 50!)

Now the game begins

As others have correctly said we need to find the no of 5's in 49! ( Which will be 10)

Here is why we don't need to find the number of 5's in 51 -

49! * 51 => ( 49 * 48 * 47......13.....7 ....3 * 2 * 1) ( 13 * 7 )

Here is where maipenrai went wrong

49! Factorial ( the red part ) has 10 zeroes , but the blue part doesn't have a 5 neither a 2...

Had the number been 50 instead , we would have one additional zero ( 10+ 1) ; but thats not the issue here

The most straightforward way is to directly see how many trailing zeroes 49! and 50! will have and then pick the lower number. You dont have to then worry about number of 5s in 51 or any other number this way. But yes, if you do go the way you are mentioning, then yes, you need to take care of the additional zeroes that might be added.
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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07 Aug 2018, 23:51
1
dave13 wrote:
Bunuel wrote:
chetan2u wrote:
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.

Bunuel why 51 won't contribute to the number of zeros at the end of the number ? i dont get

$$\frac{49}{5} = 9$$

$$\frac{51}{5^2}$$ = 2

$$9+2 =11$$

The formula works only when we have a factorial !
when we have an integer , it has to be factorised.

consider 6!
6!= 6*5*4*3*2*1
no of zeroes = no of 5*2 pairs
we have only one in this case , so no of trailing zeroes = 1
using the formula
6/5=1

now consider 6
6=2*3
doesn't have a five in it , and therefore doesn't have a trailing zero

similarly 51 is an integer
factors of 51 are 1,3,17,51
no 5's or 2's , so no trailing zeroes
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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08 Aug 2018, 00:24
1
dave13 wrote:
CounterSniper many thanks for great explanation i have one question if after factoring out we have 49!(1+50) = 49! *51

why Bunuel uses 49! here twice ---> Trailing zeros in 49!: $$\frac{49}{5}+\frac{49}{5^2}=9+1=10$$ we have only one 49! left after factoring out

have a good day

Thats how the formula works .

you keep on dividing the numerator with increasing powers 5 .

you might find this useful !!

https://gmatclub.com/forum/gmat-math-book-87417.html
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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27 Apr 2016, 17:03
i think the answer is 9,
50!+49!= 49!(1+50)
51*49!,
49 factorial has 9 trailing zeroes, sor 51*49! also should have 9 trailing zeroes.

(A).

Please let me know if my technique is wrong.
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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27 Apr 2016, 18:29
maipenrai wrote:
So in order to find a number of zeros in X! we need to divide the number by $$5^n$$ as long as $$5^n<=X$$

So for 49!+50!:

$$49/5+49/25=9+1=10$$
$$50/5+50/25=10+2=12$$

10+12=22

Hi,
Since it is sum of two factorials, each of these will have some trailing zeroes. But their SUM will have the lower of the two..
Say you have 23400 and 560000..
Here when you add two, the zeroes will be the lower of the two, which is 2..
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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27 Apr 2016, 18:33
Engr2012 wrote:
maipenrai wrote:
So in order to find a number of zeros in X! we need to divide the number by $$5^n$$ as long as $$5^n<=X$$

So for 49!+50!:

$$49/5+49/25=9+1=10 50/5+50/25=10+2=12$$

10+12=22

You are missing an important point here. If the question would have been 49!*50! , then yes your approach would have been correct. But since it is addition, you can not simply add the number of trailing zeroes.

Example,

Say, A = 100, B = 1000, A+B =1100 (2 trailing zeroes and not 5 trailing zeroes).

A=10000, B = 100000 , A+B = 110000 (4 trailing zeroes and not 9 trailing zeroes) etc. You see that the number of trailing zeroes = lower number of trailing zeroes out of the 2.

Hope this helps.

Yes, thanks a lot, finally got this.
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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28 Apr 2016, 08:20
Engr2012 wrote:

The most straightforward way is to directly see how many trailing zeroes 49! and 50! will have and then pick the lower number.You dont have to then worry about number of 5s in 51 or any other number this way. But yes, if you do go the way you are mentioning, then yes, you need to take care of the additional zeroes that might be added.

Yes !! Very true , I would definitely adopt the said approach, it looks much better...

Expanding the discussion further for others as well ( Hope you don't object to it - And I am sure you won't )

Engr2012's point is very useful for such problems..

The highlighted part states , to take the number of zero's of the lower number , here is why -

100 has 2 trailing zero's & 10 has 2 trailing zeros...

If we want to add those numbers to find the trailing zero's of the result , it will be = Only 1 Trailing Zero { Since the lower number has only one trailing zero }

The reason is , if you add 100 & 10 , the result will be 110 ( Having only one trailing Zero )

However this rule of trailing zero's is applicable only for addition ONLY

If You Multiple Or Divide then it won't Work, here is why -

100 * 10 = 1000
1000/10 = 100

Engr2012 hope I am correct, please add if I am missing something !!

PS :I think subtraction will also work , 100 - 10 = 90
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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28 Apr 2016, 09:11
chetan2u

Very true Chetan , I was trying to form a pattern...

After your post I am convinced that it's better go by the fundamentals ( Concepts ) for these problems
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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28 Apr 2016, 21:48
Thanks for the explanations. Fully understand it now.

Lousy mistake would have cost me a point on the real exam!
The good news is that if I ever get the trailing zeros question I will nail it

And I will leave my earlier post so the others can learn from my mistake.
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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02 May 2016, 16:53
Abhishek009 wrote:
Health Discussion going on around here Would like to share something interesting !!!

I hope all of you agree till this part - 51*49! ( maipenrai it won't be 49! + 50!)

Now the game begins

As others have correctly said we need to find the no of 5's in 49! ( Which will be 10)

Here is why we don't need to find the number of 5's in 51 -

49! * 51 => ( 49 * 48 * 47......13.....7 ....3 * 2 * 1) ( 13 * 7 )

Here is where maipenrai went wrong

49! Factorial ( the red part ) has 10 zeroes , but the blue part doesn't have a 5 neither a 2...

Had the number been 50 instead , we would have one additional zero ( 10+ 1) ; but thats not the issue here

Easiest Approach.>!
I used the same..!!
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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17 Jul 2016, 10:50
1
Bunuel wrote:
chetan2u wrote:
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.

Hi Bunuel,

Could you please let me know if this approach is correct?

I calculated the trailing zeroes for both numbers, we get 10 for 49! and 12 for 50!
So if we sum both numbers we would get a number with only 10 zeroes because those 2 extra zeroes from 50! would become numbers when added 49!.
Therefore, we should take the smallest zeroes in both numbers.

Is this correct? Can this be applied for example to 49! + 50! + 30! Getting 7 trailing zeroes?

Thanks!
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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21 Jul 2016, 10:00
49!+50!= 51*49!

No of trailing 0's = no of 5*2

no of 2's = 49/2+49/4+49/8+49/16+49/32 = D

no of 5's = 49/5+49/25=10=P

now since

P < D

no of trailing 0's = 10
Re: How many trailing Zeroes does 49! + 50! have?   [#permalink] 21 Jul 2016, 10:00

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