# Lessons from teaching gifted elementary school students (Part 1)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received:

When playing with my calculator, I noticed the following pattern:

$256 \times 256 = 65,5\underline{36}$

$257 \times 257 = 66,0\underline{49}$

$258 \times 258 = 66,5\underline{64}$

Is there a reason why the last two digits are perfect squares? I know it usually doesn’t work out this way.

I leave a thought bubble in case you’d like to think this. One way of answering this question appears after the bubble.

The answer is: This always happens as long as the tens digits is either 0 or 5.

To see why, let’s expand $(50n + k)^2$, where $n$ and $k$ are nonnegative integers and $0 \le k \le 9$. If $n$ is odd, then the tens digit of $50n+k$ will be a 5. But if $n$ is even, then the tens digit of $50n+k$ will be 0.

Whether $n$ is even or odd, we get

$(50n+k)^2 = 2500n^2 + 100nk + k^2 = 100(25n^2 + nk) + k^2$

The expression inside the parentheses is not important; what is important is that $100(25n^2 + nk)$ is a multiple of 100. Therefore, the contribution of this term to the last two digits of $(50n+k)^2$ is zero. We conclude that the last two digits of $(50n+k)^2$ is just $k^2$.

Naturally, elementary-school students are typically not ready for this level of abstraction. That’s what I love about this question: this is a completely natural question for a curious grade-school child to ask, but the teacher has to have a significantly deeper understanding of mathematics to understand the answer.

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