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

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

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26 Mar 2017, 00:06
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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03 Sep 2017, 02:47
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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Re: How many trailing Zeroes does 49! + 50! have? [#permalink]

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03 Sep 2017, 03:38
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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16 Sep 2017, 09:21
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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16 Sep 2017, 12:41
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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How many trailing Zeroes does 49! + 50! have?   [#permalink] 16 Sep 2017, 12:41

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