2014 AMC 12B Problem 15

Below is the professionally curated solution for Problem 15 of the 2014 AMC 12B, 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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Concepts:logarithmprime factorization

Difficulty rating: 1950

15.

When p=k=16klnk,p = \sum_{k=1}^{6} k \ln k, the number epe^p is an integer. What is the largest power of 22 that is a factor of ep?e^p?

2122^{12}

2142^{14}

2162^{16}

2182^{18}

2202^{20}

Solution:

Since klnk=ln(kk),k \ln k = \ln(k^k), the sum gives p=ln(k=16kk),p = \ln\left(\prod_{k=1}^{6} k^k\right), so ep=112233445566. e^p = 1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4 \cdot 5^5 \cdot 6^6.

The factors of 22 come from 222^2 (giving 22), 44=284^4 = 2^8 (giving 88), and 66=26366^6 = 2^6 \cdot 3^6 (giving 66). In total the exponent of 22 is 2+8+6=16.2 + 8 + 6 = 16.

Thus, the correct answer is C.

Problem 15 in Other Years