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

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

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New post 26 Mar 2017, 01:06
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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New post 03 Sep 2017, 03:47
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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New post 03 Sep 2017, 04:38

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

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New post 16 Sep 2017, 10:21
chetan2u wrote:
How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22


Self made



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 ...?

thanks in advance ..

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

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New post 16 Sep 2017, 13:41
chetan2u wrote:
How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22

Self made

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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Kudos [?]: 593 [0], given: 16

How many trailing Zeroes does 49! + 50! have?   [#permalink] 16 Sep 2017, 13:41

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