[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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Reading from the right, how many consecutive zeros, starting
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Max@Math Revolution
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mbawisdom
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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.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
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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Max@Math Revolution
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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.
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.
Math Revolution
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Score an excellent Q49-51 just like 70% of our students.
[Free] Full on-demand course (7 days) - 100 hours of video lessons, 490 lesson topics, and 2,000 questions.
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