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Intern  B
Joined: 04 Jan 2017
Posts: 7
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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Hi bunuel,
What if the question was 48!+49!. Please correct me if i'm wrong with my calculation;
Factorization leads to 48!(1+49) => 48!*50 =>No of trailing zeroes in 48! is 9
Since 50=2*2*5; one more trailing zero will be added since we have either a 2 or a 5
Math Expert V
Joined: 02 Sep 2009
Posts: 59095
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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wwambugu14 wrote:
Hi bunuel,
What if the question was 48!+49!. Please correct me if i'm wrong with my calculation;
Factorization leads to 48!(1+49) => 48!*50 =>No of trailing zeroes in 48! is 9
Since 50=2*2*5; one more trailing zero will be added since we have either a 2 or a 5

48!*50

48! will have 48/5 + 48/25 = 9 + 1 = 10 zeros.

50 = 10*5. 10 there will add 1 more zero. So, will 5 because 48! definitely has an extra 2 for that 5 to make one more trailing zero. Therefore, 48!*50 will have 10 + 1 + 1 = 12 trailing zeros.
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22

49!+50!=49!(1+50)==49!(51)

now for zero to be the end there has to be a 5 therefore 49/5+9/5== 9+1=10 B
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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Bunuel

We calculate trailing zeroes using 5 and not 2. In this, do we assume that the number of two's will be at least equal to the 5's?
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Math Expert V
Joined: 02 Sep 2009
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Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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Shiv2016 wrote:
Bunuel

We calculate trailing zeroes using 5 and not 2. In this, do we assume that the number of two's will be at least equal to the 5's?

Yes, since 2 < 5, then in n! there will be at leas as many 2's as 5's.
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How many trailing Zeroes does 49! + 50! have?  [#permalink]

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chetan2u wrote:
How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22

hi

great ....

during multiplication, zeros are added ...
during addition, the fewer zeros of the two or (more??) are counted....
is there any rule, however, for subtraction ...?

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

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chetan2u wrote:
How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22

Rule : n! is divided by prime number m to leave trailing zeroes.
The number of trailing zeroes is the sum of $$\frac{n!}{(m)}, \frac{n!}{(m^2)}, till \frac{n!}{(m^x)}$$ such that $$m^x < n$$

$$49! + 50! = 49! + 50*49! = 49!(1+50) = 49!*51$$

Therefore, 49! has $$(\frac{49}{5})9+(\frac{49}{5^2})1 = 10$$ zeroes (Option B)

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Joined: 09 Mar 2016
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How many trailing Zeroes does 49! + 50! have?  [#permalink]

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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

$$9+2 =11$$ VP  D
Joined: 09 Mar 2016
Posts: 1229
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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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 Director  G
Joined: 20 Feb 2015
Posts: 737
Concentration: Strategy, General Management
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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Is there a glitch ?
How come my post went to page 1 ?
VP  D
Joined: 09 Mar 2016
Posts: 1229
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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CounterSniper wrote:
Is there a glitch ?
How come my post went to page 1 ? i marked it as the best community reply  Non-Human User Joined: 09 Sep 2013
Posts: 13602
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

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_________________ Re: How many trailing Zeroes does 49! + 50! have?   [#permalink] 24 Oct 2019, 03:31

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