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16 Sep 2014, 00:44
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What is the largest integer \(n\) such that \(3^n\) is a factor of 21! ?
A. 7
B. 9
C. 10
D. 12
E. 15
A. 7
B. 9
C. 10
D. 12
E. 15
16 Sep 2014, 00:45
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Official Solution:
What is the largest integer \(n\) such that \(3^n\) is a factor of 21! ?
A. 7
B. 9
C. 10
D. 12
E. 15
The largest integer \(n\) for which \(3^n\) divides 21! can be determined by counting the number of times the prime divisor 3 appears in the prime factorization of \(21! = 1*2*3*...*21\). The numbers 3, 6, 12, 15, and 21 each contribute one factor of 3. Meanwhile, \(9 = 3^2\) and \(18 = 2*3^2\) each contribute two factors of 3. In total, there are \(1 + 1 + 1 + 1 + 1 + 2 + 2 = 9\) factors of 3. Therefore, 21! is divisible by \(3^9\) but not by \(3^{10}\). Thus, \(n\) equals 9.
Alternative Explanation
To find the highest integer power of a prime number \(p\) in \(n!\), we use the formula:
\(\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}\), where \(k\) must be chosen such that \(p^k \leq n\)
For example, find the power of 2 in 25!:
\(\frac{25}{2}+\frac{25}{2^2}+\frac{25}{2^3}+\frac{25}{2^4}=12+6+3+1=22\). Note that the last denominator (\(2^4\)) must be less than 25. Also, note that we only consider the quotient of the division, e.g., \(\frac{25}{2^4}=1\). Therefore, the highest integer power of 2 in 25! is 22, which means that 25! is divisible by \(2^{22}\) but not by \(2^{23}\).
Therefore, the highest power of 3 in 21! is \(\frac{21}{3}+\frac{21}{3^2}=7+2=9\) (again, keep in mind that we only consider the quotient of each term, for instance, 21/9 equals 2).
Answer: B
What is the largest integer \(n\) such that \(3^n\) is a factor of 21! ?
A. 7
B. 9
C. 10
D. 12
E. 15
The largest integer \(n\) for which \(3^n\) divides 21! can be determined by counting the number of times the prime divisor 3 appears in the prime factorization of \(21! = 1*2*3*...*21\). The numbers 3, 6, 12, 15, and 21 each contribute one factor of 3. Meanwhile, \(9 = 3^2\) and \(18 = 2*3^2\) each contribute two factors of 3. In total, there are \(1 + 1 + 1 + 1 + 1 + 2 + 2 = 9\) factors of 3. Therefore, 21! is divisible by \(3^9\) but not by \(3^{10}\). Thus, \(n\) equals 9.
Alternative Explanation
To find the highest integer power of a prime number \(p\) in \(n!\), we use the formula:
\(\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}\), where \(k\) must be chosen such that \(p^k \leq n\)
For example, find the power of 2 in 25!:
\(\frac{25}{2}+\frac{25}{2^2}+\frac{25}{2^3}+\frac{25}{2^4}=12+6+3+1=22\). Note that the last denominator (\(2^4\)) must be less than 25. Also, note that we only consider the quotient of the division, e.g., \(\frac{25}{2^4}=1\). Therefore, the highest integer power of 2 in 25! is 22, which means that 25! is divisible by \(2^{22}\) but not by \(2^{23}\).
Therefore, the highest power of 3 in 21! is \(\frac{21}{3}+\frac{21}{3^2}=7+2=9\) (again, keep in mind that we only consider the quotient of each term, for instance, 21/9 equals 2).
Answer: B
27 Apr 2017, 23:40
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high quality question..
