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

Answer: 371

Concepts:number base perfect square perfect power bounding to limit cases

Difficulty rating: 2170

Solution:

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 $10 + 361 = 371.$

2018 AIME II Problem 3

Answer: 371

Difficulty rating: 2170

Solution:

Problem 3 in Other Years