# A Gardner-esque puzzle

One of my favourite sources of puzzles at the moment is @WWMGT - What Would Martin Gardner Tweet? (Martin Gardner, in case you’re not up on the greats of popular maths writing, was one of the greats of popular maths writing - and is indirectly responsible for Big MathsJam.)

Recently, it was decreed that Martin Gardner would have tweeted:

Show that no square of two or more digits can have only odd digits.

Must be easy, I thought. Let’s do it by contradiction, and try to find a square number - $k^2$ with only odd digits.

$k$ has to be odd, because otherwise it would end in an even digit - so it must end in 1, 3, 5, 7 or 9.

If $k$ ends in 1, it can be written as $k = 10n+1$, so $k^2 = 100n^2 + 20n + 1$; its last digit is 1, but the one before it is even ($\frac{k^2 - 1}{10} = 10n^2 + 2n$).

A similar argument accounts for $k$ ending in 3: if $k = 10n + 3$, $k^2 = 100n^2 + 60n + 9$, which again has an even penultimate digit.

It was at exactly this point that inspiration struck.

You see, I’d been worried about how I was going to deal with awkward numbers like somethingty-seven and somethingty-nine - but then it struck me: I can write those as $10n - 3$ and $10n - 1$ and apply the same reasoning as before! That just leaves me with somethingty-five.

If $k = 10n + 5$, then $k^2 = 100n^2 + 100n + 25$, in which case the penultimate digit has to be 2.

And we’re done! None of the possibilities hold up, so our assumption that there was such a number must have been mistaken.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

### 5 comments on “A Gardner-esque puzzle”

• ##### Andrew Old

Is it actually the case that every square number of two or more digits has at least one even number in its last two digits? If so, it strikes me that this might be easier to prove and imply the result.

• ##### Colin

I think that’s what I ended up proving, which is a stronger statement than the puzzle set.

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