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16 Sep 2014, 00:40
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If \(n\) is a positive integer, what is the smallest value of \(n\) for which \(n!\) is divisible by 1,000?
A. 8
B. 10
C. 15
D. 20
E. 25
A. 8
B. 10
C. 15
D. 20
E. 25
16 Sep 2014, 00:40
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Official Solution:
If \(n\) is a positive integer, what is the smallest value of \(n\) for which \(n!\) is divisible by 1,000?
A. 8
B. 10
C. 15
D. 20
E. 25
Trailing zeros:
Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow. For example, 125,000 has 3 trailing zeros.
The number of trailing zeros in \(n!\), the factorial of a non-negative integer \(n\), can be determined with this formula:
\(\frac{n}{5} + \frac{n}{5^2} + \frac{n}{5^3} + ... + \frac{n}{5^k}\), where \(k\) must be chosen such that \(5^{(k+1)} > n\).
It's easier if we consider an example:
How many zeros are at the end (after which no other digits follow) of \(32!\)?
\(\frac{32}{5} + \frac{32}{5^2} = 6 + 1 = 7\). Notice that the last denominator (\(5^2\)) must be less than 32. Also note that we take into account only the quotient of the division, that is \(\frac{32}{5} = 6\). So there are 7 zeros at the end of \(32!\).
Another example: how many trailing zeros does \(125!\) have?
\(\frac{125}{5} + \frac{125}{5^2} + \frac{125}{5^3} = 25 + 5 + 1 = 31\). So there are 31 zeros at the end of \(125!\).
The formula actually counts the number of factors of 5 in \(n!\), but since there are at least as many factors of 2, this is equivalent to the number of factors of 10, each of which contributes one more trailing zero.
Back to the original question:
According to the above, for \(n!\) to be divisible by 1,000, it must have at least 3 trailing zeros. To achieve this, \(n\) must have at least 3 factors of 5. Therefore, the answer is 15, since \(\frac{15}{5} = 3\).
Answer: C
If \(n\) is a positive integer, what is the smallest value of \(n\) for which \(n!\) is divisible by 1,000?
A. 8
B. 10
C. 15
D. 20
E. 25
Trailing zeros:
Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow. For example, 125,000 has 3 trailing zeros.
The number of trailing zeros in \(n!\), the factorial of a non-negative integer \(n\), can be determined with this formula:
\(\frac{n}{5} + \frac{n}{5^2} + \frac{n}{5^3} + ... + \frac{n}{5^k}\), where \(k\) must be chosen such that \(5^{(k+1)} > n\).
It's easier if we consider an example:
How many zeros are at the end (after which no other digits follow) of \(32!\)?
\(\frac{32}{5} + \frac{32}{5^2} = 6 + 1 = 7\). Notice that the last denominator (\(5^2\)) must be less than 32. Also note that we take into account only the quotient of the division, that is \(\frac{32}{5} = 6\). So there are 7 zeros at the end of \(32!\).
Another example: how many trailing zeros does \(125!\) have?
\(\frac{125}{5} + \frac{125}{5^2} + \frac{125}{5^3} = 25 + 5 + 1 = 31\). So there are 31 zeros at the end of \(125!\).
The formula actually counts the number of factors of 5 in \(n!\), but since there are at least as many factors of 2, this is equivalent to the number of factors of 10, each of which contributes one more trailing zero.
Back to the original question:
According to the above, for \(n!\) to be divisible by 1,000, it must have at least 3 trailing zeros. To achieve this, \(n\) must have at least 3 factors of 5. Therefore, the answer is 15, since \(\frac{15}{5} = 3\).
Answer: C
earnit
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21 Oct 2014, 05:16
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Bunuel
Could we find more questions on the above types? Trailing zeroes/ no. of series in 72! etc.
