duahsolo wrote:
If x = a!, which of the following values of a is the least possible value for which the last 8 digits of the integer x will all be zero?
A) 21
B) 29
C) 38
D) 40
E) 100
We are given that x = a! and need to determine the least value for a that will create 8 trailing zeros in integer x. We must remember that a trailing zero is created with each 5-and-2 pair within the prime factorization of a!. Furthermore, since we know there are fewer 5s in a! (or any factorial) than 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.
Let’s test answer choice A:
To determine the number of 5s within 21!, we can use the following shortcut in which we divide 21 by 5, and then divide the quotient of 21/5 by 5 and continue this process until we no longer get a nonzero quotient:
21/5 = 4 (we can ignore the remainder)
Since 4/5 does not produce a nonzero quotient, we can stop.
21! contains four 5-and-2 pairs, and thus 21! does not contain 8 trailing zeros.
Next we can test answer choice B:
29/5 = 5 (we can ignore the remainder)
5/5 =1 (we can ignore the remainder)
Since 1/5 does not produce a nonzero quotient, we can stop.
The final step is to add up our quotients; that sum represents the number of factors of 5 within 29!. Thus, there are 5 + 1 = 6 factors of 5 within 29!
29! contains six 5-and-2 pairs, and thus 29! does not contain 8 trailing zeros.
Next we can test answer choice C:
38/5 = 7 (we can ignore the remainder)
7/5 =1 (we can ignore the remainder)
Since 1/5 does not produce a nonzero quotient, we can stop.
The final step is to add up our quotients; that sum represents the number of factors of 5 within 38!. Thus, there are 7 + 1 = 8 factors of 5 within 38!
38! contains eight 5-and-2 pairs, and thus 38! does contain 8 trailing zeros.
Answer: C
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