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Reading from the right, how many consecutive zeros, starting

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by Max@Math Revolution » Thu Mar 08, 2018 11:13 pm
[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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by mbawisdom » Fri Mar 09, 2018 8:03 am
Max@Math Revolution 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 will be determined by the number of 10's we can make from the factors. The number of 10's will be determined by how man 2's and 5's we have as factors (10 = 2*5). Since there will be many more 2's than 5's in 29! we only need to count the number of 5's we have as factors.

5, 10, 15 and 20 give us 1 factor of 5 each

5*1 = 5
5*2 = 10
5*3 = 15
5*4 = 20

25 gives us 2 factors of 5
5*5 = 25

6 factors of 5's means we have six factors of 10 and therefore 6 zeros at the end of 29!

Answer is B. 6
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by Max@Math Revolution » Sun Mar 11, 2018 5:25 pm
=>
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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