Square value from a rectangle

Question

This question seems like it should be simple to solve but I'm struggling to find an answer.

If I have a rectangle with dimensions 68.8 x 103.5, how can I find the largest common square value? Finding this value means I can completely fill the rectangle with squares and leave nothing spare. The number of rows and columns is irrelevant as is the total number of squares. Also the value isn't likely to be an integer.

Thank you

Anthony

Answer 1

To make the shapes squares, the side length must be a divisor both of $688$ and $1035$. The greatest common divisor of these two numbers is $1$, so the largest possible value is $0.1$ units.

Replying to the comment, suppose there was a greater, non-integer common divisor. For example, $1.2$ is a common divisor of $6$ and $7.2$. But when you multiply by $10$, $12$ becomes a common divisor of $60$ and $72$.

Here's the problem: no matter how many times you multiply by $10$, a larger GCD than the original will not magically appear. $ \gcd(10a,10b) = 10 \gcd(a, b)$, and $ \gcd(100a, 100b) = 100 \gcd(a, b)$. This is because if you multiply both sides by $10$, both sides will have a factor of $10$.

But after removing the factor of $10$ from both sides, you are left with the same numbers, and hence the same GCD. This argument works for any power of $10$ you choose.

Answer 2

You can apply Euclid's Algorithm without converting your numbers to integers.

\begin{align} 103.5&=1\times68.8+34.7\\ 68.8&= 1\times 34.7+34.1\\ 34.7&= 1\times 34.1+0.6\\ 34.1&=56\times0.6+0.5\\ 0.6&= 1\times 0.5+0.1\\ 0.5&=5\times0.1 \end{align}

So the answer is $0.1.$

As already discussed elsewhere, it's convenient to multiply by ten so you are dealing with integers; in fact, the first step in grade-school long division, which you can use to find the integer quotient at each step, is to multiply by ten to get integers. Multiplying the two dimensions by an integer neither causes solutions to magically appear nor disappear, any more than you gain or lose solutions to a geometric problem by measuring the figure in millimeters rather than centimeters.