2018 AIME II Problem 3

Question

Find the sum of all positive integers

b \lt 1000

such that the base-

b

integer

36_{b}

is a perfect square and the base-

b

integer

27_{b}

is a perfect cube.

Answer

The conditions say

3b + 6

is a perfect square and

2b + 7

is a perfect cube

2b + 7

is odd and less than

2007,

so only a handful of odd cubes need to be checked

The conditions say

3b + 6

is a perfect square and

2b + 7

is a perfect cube. Since

2b + 7

is odd and

b \lt 1000

forces

2b + 7 \lt 2007,

the cube must be one of

1,

27,

125,

343,

729,

1331,

giving

b = -3,

10,

59,

168,

361,

662.

The corresponding values of

3b + 6

for the positive candidates are

36,

183,

510,

1089,

1992,

and only

36 = 6^2

(for

b = 10

) and

1089 = 33^2

(for

b = 361

) are perfect squares. The requested sum is

10 + 361 = 371.

Problem 3 in other years