2005 AMC 12B Problem 21

Attempt Problem 21 of the 2005 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 2005 AMC 12B solutions, or check the answer key.

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21.

A positive integer

n

has

60

divisors and

7n

has

80

divisors. What is the greatest integer

k

such that

7^k

divides

n?

0

1

2

3

4

Write

n = 7^k Q

with

Q

not divisible by

7

; let

d

be the number of divisors of

Q

Then

n

has

(k+1)d

divisors and

7n

has

(k+2)d

divisors

Write

n = 7^k Q

where

Q

is not divisible by

7,

and let

d

be the number of divisors of

Q.

Then

n

has

(k + 1)d = 60

divisors and

7n = 7^{k+1}Q

has

(k + 2)d = 80

divisors.

Dividing,

\dfrac{k + 2}{k + 1} = \dfrac{80}{60} = \dfrac43,

3(k + 2) = 4(k + 1),

giving

k = 2.

Thus, the correct answer is C.

Problem 21 in other years