Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.0 AIME, harder Prove it Romania

Four families have two children each and all eight children are born after the year 1989. All four youngest siblings are born in the same year, and the sum of the digits of the year is equal to the product of the non zero digits. The differences of the ages between the siblings of each family is a perfect square. Find the birth years of the four eldest siblings in each family, given that their ages are all different.

Solution

There exists no digit $a$ for which $1 + 9 + 9 + a = 1 \cdot 9 \cdot 9 \cdot a$ , hence the youngest are born after $2000$ , say in $200a$ or $201a$ . In the first case, the condition $2 + a = 2a$ gives $a = 2$ . The second case leads to $a = 3$ . The years are $2002$ or $2013$ .

The age differences between the siblings of each family are at least $1$ , $4$ , $9$ and $16$ . Since $2002 - 1989 = 13 < 16$ , we conclude that the youngest are all born in $2013$ , whence the eldest are born in $2012$ , $2009$ , $2004$ and $1997$ , respectively.

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