2014 AMC 12B Problem 15

Attempt Problem 15 of the 2014 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 12B solutions, or check the answer key.

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15.

When $p = \sum_{k=1}^{6} k \ln k,$ the number $e^p$ is an integer. What is the largest power of $2$ that is a factor of $e^p?$

$2^{12}$

$2^{14}$

$2^{16}$

$2^{18}$

$2^{20}$

Answer: C

Concepts:logarithm prime factorization

Difficulty rating: 1950

Solution:

Since $k \ln k = \ln(k^k),$ the sum gives $p = \ln\left(\prod_{k=1}^{6} k^k\right),$ so $e^p = 1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4 \cdot 5^5 \cdot 6^6.$

The factors of $2$ come from $2^2$ (giving $2$ ), $4^4 = 2^8$ (giving $8$ ), and $6^6 = 2^6 \cdot 3^6$ (giving $6$ ). In total the exponent of $2$ is $2 + 8 + 6 = 16.$

Thus, the correct answer is C.

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