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Reading from the right, how many consecutive zeros, starting with the

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Math Revolution GMAT Instructor
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Reading from the right, how many consecutive zeros, starting with the [#permalink]

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New post 09 Mar 2018, 00:23
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[GMAT math practice question]

Reading from the right, how many consecutive zeros, starting with the units digit, appear in 29!?

A. 5
B. 6
C. 7
D. 8
E. 9

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Re: Reading from the right, how many consecutive zeros, starting with the [#permalink]

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New post 09 Mar 2018, 02:16
MathRevolution wrote:
[GMAT math practice question]

Reading from the right, how many consecutive zeros, starting with the units digit, appear in 29!?

A. 5
B. 6
C. 7
D. 8
E. 9


The number of zeros at the right end of the number is determined by the number of 10s in its factorization.
That is, we can use basic divisibility properties to answer this question, a Logical approach.

We'll count the number of 5s in the prime factorization of 29! :
5 - 1
10 - 1
15 - 1
20 - 1
25 - 2
Total of 6 5s and certainly more than 6 2s meaning that we have 6 pairs of 5*2 = 10
Then there are 6 zeros at the right end of 29!
(B) is our answer.
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Re: Reading from the right, how many consecutive zeros, starting with the [#permalink]

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New post 11 Mar 2018, 03:01
MathRevolution wrote:
[GMAT math practice question]

Reading from the right, how many consecutive zeros, starting with the units digit, appear in 29!?

A. 5
B. 6
C. 7
D. 8
E. 9


A zero will be formed by \(2*5\). So we need to find out number of 5s in \(29!\). this can be calculated as

\(\frac{29}{5}+\frac{29}{5^2}=5+1=6\)

Option B
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Re: Reading from the right, how many consecutive zeros, starting with the [#permalink]

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New post 11 Mar 2018, 18:34
=>

To find the number of \(0’s\) ending \(29!\), we need to count the numbers of \(2’s\) and \(5’s\) in the prime factorization of \(29!\).
Since the number \(2’s\) is greater than the number of \(5’s\), we only need to count the number of \(5’s\) in the prime factorization of \(29!\).
The factors of \(5\) are contributed by \(5, 10, 15, 20\) and \(25\). Each of \(5, 10, 15\) and \(20\) contributes one \(5\), while \(25\) contributes two \(5s\) to the prime factorization.
Thus, there are \(6\) copies of \(5\) in the prime factorization of \(29!\), giving rise to \(6\) consecutive \(0’s\) at the end of \(29!\).

Therefore, the answer is B.
Answer : B

Note: The actual value of \(29!\) is \(8841761993739701954543616000000.\)
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Re: Reading from the right, how many consecutive zeros, starting with the   [#permalink] 11 Mar 2018, 18:34
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