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}

k \ln k = \ln(k^k),

e^p = \prod_{k=1}^{6} k^k

Only

2^2,

4^4,

and

6^6

contribute factors of

2

Since

k \ln k = \ln(k^k),

the sum gives

p = \ln\left(\prod_{k=1}^{6} k^k\right),

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

2 + 8 + 6 = 16.

Thus, the correct answer is C.

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