Problem 1. Four friend families each have two children, and all the children were born after 1989. The birth year of the middle children is the same, with the sum of the digits equal to the product of the non-zero digits. The age difference between the siblings in the same family is a natural number of years expressed by a non-zero perfect square. Find the birth year of the older siblings from each family, knowing that they cannot have the same age.
Solution. We show that the birth year of the twins cannot be of the form $\overline{199 a}$. Assuming that the birth year has the form $\overline{199 a}$, it must be that $1+9+9+a=1 \cdot 9 \cdot 9 \cdot a$, i.e., $19+a=81 \cdot a$; this relation is impossible under the condition that $a$ is a digit. Therefore, the birth year of the twins has one of the forms $\overline{200 a}$ or $\overline{201 a}$.

If it has the form $\overline{200 a}$, from the condition $2+a=2 \cdot a$ we deduce $a=2$, and if it has the form $\overline{201 a}$, from

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The cases 1990, 2000, and 2010 do not work. ................................................................ 1p

The age difference between the siblings is the same as the difference between their birth years. For 2002, these differences are: $2002-x, 2002-y, 2002-z, 2002-t$, where $x, y, z, t$ are the birth years of the older siblings.

The differences must be equal to 1, 4, 9, and 16. Assuming $x \geq 16$ we get $2002-x \leq$ 1986. But all the children were born after 1989, so this case is not possible.......... 2p

For 2013, these differences are: $2013-x, 2013-y, 2013-z, 2013-t$, where $x, y, z, t$ are the birth years of the older siblings.

The differences must be equal to 1, 4, 9, and 16. Assuming $x \geq 25$ we get $2013-x \leq 1988$, but all the children were born after 1989.

Therefore, we can have $2013-x=1, 2013-y=4, 2013-z=9, 2013-t=16$, from which $x=2012$, $y=2009, z=2004, t=1997$