Here's a $6$th-grade exam problem:
$$OUI \times OUI = OOUUI + OOUUI$$
$O, U$ and $I$ are digits. e.g $365 \times 365 = 33665 + 33665$ (not the case)
Thus, because the digits are into account I believe there's no algebraic equation. Also, $6$th graders don't even know algebra anyway.
With some extremely clever insight, someone could end up with the conclusion that $I=2$, pretty fast, and that it has to be $212 > OUI > 102$, giving only $10$ possibilities to test out.
There's a catch though; They don't have calculators. Can you tell me what kind of thinking or if there's an analytical solution?
Someone who has done puzzles like this will get $I=2$ quickly. We need $I^2$ to end with the same digit as $2I$. Next would be to see $O=1$. If $O=2$ it conflicts with $I$ and if $O$ is greater $O^2 \gt 2\cdot O$. We are now down to eight possibilities for $U$. Multiplying by hand to check if the tens digit of $UI^2$ is the same as the tens digit of $2\cdot UI$ is not too much work. It is only a two digit multiply and you don't have to do the hundreds column.