2010 AMC 12B Problem 9

Question

Let

n

be the smallest positive integer such that

n

is divisible by

20,

n^2

is a perfect cube, and

n^3

is a perfect square. What is the number of digits of

n?

3

4

5

6

7

Write

n=2^a\cdot5^b;

any extra prime factor only makes

n

larger

n^2

a cube forces

a

and

b

to be multiples of

3;

n^3

a square forces

a

and

b

to be multiples of

2

To be smallest,

n

uses only the primes of

20,

so

n=2^a\cdot5^b

with

a\ge2

and

b\ge1.

Since

n^2=2^{2a}5^{2b}

is a perfect cube,

a

and

b

are multiples of

3.

Since

n^3=2^{3a}5^{3b}

is a perfect square,

a

and

b

are multiples of

2.

Hence

a

and

b

are multiples of

6.

The smallest choice is

a=b=6,

so

n=2^6\cdot5^6=10^6=1{,}000{,}000,

which has

7

digits.

Thus, the correct answer is E.

Problem 9 in other years