\section*{Problem 14}

a) Find the smallest square with last digit not 0 which becomes another square (not zero) by the deletion of its last two digits. \#(b) Find all squares, not containing the digits 0 or 5, such that if the second digit is deleted the resulting number divides the original one.


\section*{Solution}

(a) This one must have slipped through: 121 !

(b) Answer: \(16,36,121,484\). Suppose the number has more than 2 digits. Write it as \((10 \mathrm{~m}+\) n) \(10^{\mathrm{r}}+\mathrm{s}\), where \(11\).

s does not contain the digits 0 or 5 , so 5 does not divide s. Hence 5 divides k-1, and so \(\mathrm{k}\) must be 6,11 , or 16 (if \(\mathrm{k}\) was 21 or more, then the rhs would be negative). Since 25 does not divide \(\mathrm{k}-1\), we must have \(\mathrm{r}=1\) and \(\mathrm{s}\) is a single digit.

We look at each possibility for \(\mathrm{k}\) in turn. \(\mathrm{k}=6\) gives no solutions. \(\mathrm{k}=11\) gives about two dozen multiples of 11 from 121 to 891 . By inspection the only squares are 121 and \(484 . \mathrm{k}=\) 16 gives 192 , which is not a square.

In addition, there is the possibility of 2 digit solutions, which I had overlooked. It is easiest to check each of the 2 digit squares, thus finding the additional solutions 16,36 .

