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The ages of the brothers

The ages of the brothers

Two brothers write their ages, one after the other and get a four-digit number that is exactly the square of their father's age. Nine years later they rewrite their ages, in the same way, again obtaining a four-digit number that is the square of their father's age.

What is the age difference between the two brothers?

Solution

Sean ab Y CD the ages of the children and n the age of the father

According to the statement, the following two equations are fulfilled:
abcd = n²
abcd + 909 = (n + 9) ²

We solve the system as follows:
n² + 909 = n² + 9² + 2 · 9n
n² + 909 = n² + 81 + 18n
18n = 909 - 81
n = 828/18 = 46 years
abcd = 46² = 2116

The ages of the children are initially 21 and 16 and the difference is 21-16 = 5 years. Nine years later, the ages will be 30 and 25 that form the number 3025 which is the square of the father's age, 55 years.