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P is the product of positive integers 1 through 30. How many consecutive 0’s are ther

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P is the product of positive integers 1 through 30. How many consecutive 0’s are ther

by Max@Math Revolution » Mon Jun 15, 2020 2:47 am

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[GMAT math practice question]

P is the product of positive integers 1 through 30. How many consecutive 0’s are there in the units digit of P?

A. 4
B. 5
C. 6
D. 7
E. 8
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Re: P is the product of positive integers 1 through 30. How many consecutive 0’s are ther

by Max@Math Revolution » Wed Jun 17, 2020 12:52 am
=>

Let P be the product of integers 1 through 30.
Then the prime factorization of P is P = 2^a3^b5^c7^d·…·29.
Define [x] as the greatest integer less than or equal to x.
Then the number of the prime factors of 2 is 26 = 15 + 7 + 3 + 1,
since [30/2] = 15, [30/2^2] = 7, [30/2^3] = 3 and [30/2^4] = 1.
The number of prime factors of 5 is 7 = 6 + 1,
since [30/5] = 6 and [30/5^2] = 1.

Then we have P = 2^26·3b·5^7·7d·…·29 = 2^{7+19}·3b·5^7·7d·…·29 =2^72^19·3b·5^7·7d·…·29 = 2^7·5^7·2^19·3b·7d·…·29 =10^7·2^19·3b·7d·…·29, and P has 7 consecutive zeros in the units digit.

Since we have 10 = 2*5 and more 2’s than 5’s, we can count the number of 5’s only.

Therefore, D is the answer.
Answer: D
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Re: P is the product of positive integers 1 through 30. How many consecutive 0’s are ther

by Scott@TargetTestPrep » Thu Jun 18, 2020 2:49 pm
Max@Math Revolution wrote: ↑
Mon Jun 15, 2020 2:47 am
 [GMAT math practice question]

P is the product of positive integers 1 through 30. How many consecutive 0’s are there in the units digit of P?

A. 4
B. 5
C. 6
D. 7
E. 8
Solution:

We are assuming the question intended to ask for the greatest number of consecutive 0’s at the end of P.

We see that P = 30! and we are looking for 2-and-5 pairs in 30!. Since there are fewer factors of 5 than 2, we need to determine the number of factors of 5 in 30! (since that is the number of consecutive 0’s at the end of the number P.)

We see that in 30!, the factors 5, 10, 15, 20, and 30 each contributes one factor of 5, whereas 25 contributes two factors of 5. Therefore, the total number of factors of 5 is 5 + 2 = 7. Since there are 7 factors of 5 (and also at least 7 factors of 2), there are 7 consecutive 0’s at the end of the number P.

Answer: D

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