Henry Ernest Dudeney/Puzzles and Curious Problems/167 - The Cancelled Cheque/Solution

by : $167$

 * The Cancelled Cheque

Solution

 * $81 \, 64 \, 25 = 113 \times 85^2$

Proof
Let $n$ be the $6$-digit number on the cheque.

We have that $n = 113 m$ where $m$ is a square number.

Because $100 \, 000 \le n \le 999 \, 999$ we have that:
 * $885 \le m \le 8849$.

From Square Modulo 5: Corollary, no square number ends with $2$, $3$, $7$ or $8$.

So $m$, being square, ends in $0$, $1$, $4$, $5$, $6$ or $9$.

If $m$ ends in $1$, then $n$ ends in $3$, so is not a square.

If $m$ ends in $4$, then $n$ ends in $2$, so is not a square.

If $m$ ends in $6$, then $n$ ends in $8$, so is not a square.

If $m$ ends in $9$, then $n$ ends in $7$, so is not a square.

Suppose $m$ ends in $0$.

Then $n$ also ends in $0$.

But then the last two digits form a square ending in $0$.

Hence the last two digits of $n$ would have to be $00$.

In order for $n$ to end in $00$, we find that $m$ must also end in $00$.

Suppose $m$ ends in $5$.

Then $n$ also ends in $5$, and then in fact must end in $25$.

In order for $n$ to end in $25$, we find that $m$ must also end in $25$.

Thus $m$ is a square number which lies between $900$ and $8825$ inclusive, and ends in either $00$ or $25$.

It remains to investigate all square numbers of this form.

Thus we need to investigate the squares of all numbers ending in $0$ and $5$ between $30$ and $90$, as $25^2 = 526$ and $95^2 = 9025$.

This is not arduous, so we proceed:

Note that 's analysis dispenses of square numbers ending in $00$ before he even starts, suggesting that he may not be comfortable with defining $00$ itself as a square.

He also jumps straight from the fact that $m$ lying between $885$ and $8845$ eliminates all possibilities except $1225$ and $7225$ without considering the other numbers, which suggests he has further techniques to eliminate them.