# College Math – Age Problem

by Irish Johnson

Word Problem - Age Problem

Last year Fred's age was a square.

Next year his age will be a cube.

How old is Fred?

HINT: he is older than 21, and there are 2 years between his age that is a square and his age is that is a cube.

### Comments for College Math – Age Problem

 Aug 28, 2012 Age Problem by: Staff Answer:A = Fred’s age in yearsWe assume Fred’s age must be a whole number of years, rather than a fraction.Last year’s age: (A – 1) = perfect square (or square number)       A perfect square is an integer which is the square of another integer. As an example, the number 4 is a perfect square because it is the square of the integer 2. 2² = 4, a perfect square.Next year’s age: (A + 1) = perfect cube (or cube number)       A perfect cube is an integer which is the cube of another integer. As an example, the number 8 is a perfect cube because it is the cube of the integer 2. 2³ = 8, a perfect cube.(A – 1) must be one of the numbers on a list of perfect squares.Perfect squares:1² = 12² = 43² = 94² = 165² = 256² = 367² = 498² = 649² = 8110² = 10011² = 12112² = 144Since A is greater than 21 (from the problems statement), (A - 1) can only be one of the following numbers:       25, 36, 49, 64, 81, 100 (A + 1) must be one of the numbers on a list of perfect cubes.Perfect Cubes:1³ = 12³ = 83³ = 274³ = 645³ = 1256³ = 2167³ = 3438³ = 5129³ = 72910³ = 1000Since A is greater than 21 (from the problems statement), (A + 1) can only be one of the following numbers:       27, 64, 125According the problem statement, the difference between Fred’s age next year (A + 1) and Fred’s age last year (A – 1) is 2 years.Fred’s age last year (A - 1) can only be: 25 (from the list of perfect squares)Fred’s age next year (A + 1) can only be: 27 (from the list of perfect cubes)Fred’s age next year - Fred’s age last year = 227 - 25 = 2 yearsSolve for A, using Fred’s age next year:(A + 1) = 27 A + 1 - 1 = 27 – 1A + 0 = 27 – 1A = 26final answer       Age = 26Thanks for writing. Staff www.solving-math-problems.com