3
$\begingroup$

I came across a question in an exam today (don't worry, I have already completed the test), and was wondering whether there was a better solution to mine.

The question, as I remember, is as follows:

The teacher gives 12 students squares of side lengths 1, 2, 3, 4 ... 11, 12. All students receive a distinct square. Then, he asks the students to cut up the squares into unit squares (of side length 1). He challenges the students to arrange their squares adjacently to create a larger square, with no gaps. Of course, they find that it is impossible.

Alice has a square of side length $a$. She exclaims that if she doesn't use any of her unit squares, then the class is able to create a square.

Similarly, Bill has a square of side length $b$, and says the same thing.

Given neither Alice nor Bill is lying, what is the value of $ab$?

That was a mouthful of a question, and I have attempted it, with very messy results. This is how I started:

\begin{align} 1^2 + 2^2 + 3^2 ... 11^2 + 12^2 -a^2 & = x^2 \\ 1^2 + 2^2 + 3^2 ... 11^2 + 12^2 -b^2 & = y^2 \\ \end{align} Then, by subtracting the latter equation from the first,

\begin{align} b^2 - a^2 & = x^2 - y^2 \\ (b+a)(b-a) & = (x+y)(x-y) \\ \end{align}

From there, my brain switched off and I just turned to trial & error (which is one of my favourite problem solving techniques).

We know: $1^2 + 2^2 + 3^2 ... 11^2 + 12^2 = 650$, therefore the square created by the class must be $650 - 12^2 < x^2 < 650$.

I got the answers 5 and 11 which multiply for 55, which I believe is correct, but can someone please show me a 'proper' way of completing this question.

$\endgroup$
4
$\begingroup$

Hint. Note that $650=2\cdot 5^2\cdot 13$ (it is not a perfect square) can be written as a sum of $2$ squares in three ways: $$650=5^2+25^2=11^2+23^2=17^2+19^2.$$ (actually, in your case, it suffices to check for representations $x^2+y^2$ such that $1\leq x\leq 12\leq y$). Since $5$ and $11$ are $\leq 12$, it follows that $a\cdot b=5\cdot 11=55$.

$\endgroup$
  • $\begingroup$ How did the prime factorization help you to split it as sums of squares? Or you are using the factorization just to say 650 is not prime. Is there a systematic way to get those splits, like the OP is asking? $\endgroup$ – Mathemagical Sep 12 '17 at 9:52
  • 1
    $\begingroup$ I wrote the factorization just to say that $650$ is not a perfect square. Anyway note that $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$. $\endgroup$ – Robert Z Sep 12 '17 at 9:56
  • $\begingroup$ Thanks. That's neat. Btw, I meant "650 is not square" (not "prime") in my original comment. $\endgroup$ – Mathemagical Sep 12 '17 at 12:58
1
$\begingroup$

For this particular problem, I think the easiest thing is to note that there are $650$ unit squares, which is not a square number. Then subtract $1^2$, $2^2$, etc, from $650$ until the results are square numbers. For $650-5^2$ you get $625$, a square number $25^2$, and for $650-11^2$ you get $529$, a square number $23^2$, so one has $5$ and one has $11$, a product of $55$.

$\endgroup$
  • $\begingroup$ Please read Mathjax, so other users can properly understand your answer. $\endgroup$ – Mr Pie Sep 15 '17 at 0:41
  • $\begingroup$ Agreed. Sorry, I don't know Mathjax (never heard of it before) - I'll look for a tutorial. $\endgroup$ – Jud McCranie Sep 15 '17 at 3:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.