# Purr-fection

What is the smallest perfect square that ends with the four digits
9009?

## Problem

What is the smallest perfect square that ends with the four digits 9009?

## Getting Started

Try to be systematic. How can you narrow down the cases to be tested?

If the last digit of a square number is 9 what can you say about the number?

Can you write down an expression for a number using its digits and powers of 10?

Possible support

Note: Other problems using modulus arithmetic: Dirisibly Yours , Obviously, Rational Round

An article that explains the basic ideas of modulus arithmetic: Modulus Arithmetic and Solution to Dirisibly Yours.

## Student Solutions

Peter of Madras College, St Andrew's employed an exhaustive search to find the smallest perfect square that ends in 9009 and came up with the answer 1503.

I first noted that the last digit has to be a 3 or a 7 for the square to end in 9. Noting that the last two digits of $x^2$ are only affected by the last two digits of $x$. I then systematically went through all the squares.

I kept a record of the numbers tried in two tree diagrams starting from the units digits 3 and 7. If any of these produced a number that ended in 09 then I marked that as the next branch point on the diagram.

I then went on to further generations looking for numbers ending in 009, and then finally the next generation looking for numbers ending in 9009. I found that there are no numbers with 3 digits or less whose squares end in 9009 and the four digit numbers are 1503, 6503, 2753, 7753, 2247, 7247, 3497 and 8497.

Alternatively suppose $x^2 = 100a + 10b + c$ where $a$, $b$ and $c$ are whole numbers, $a \geq 1$ and $b$ and $c$ are between 0 and 9 inclusive.

$$x^2 - 9 = (x - 3)(x + 3) = \star\star\star\star9000$$

As 10 divides the right hand side of this expression we know 10 divides $x - 3$ or $x + 3$. Thus $x$ ends in a 3 or a 7.

Case 1: c = 3

$(100a + 10b + 3)^2 = 10000a^2 + 200a(10b + 3) + 100b^2 + 60b + 9$ ends in 9009 Subtract 9, then take modulo 100.

$$\eqalign{ \Rightarrow 60b &\equiv 0 \qquad \mbox{(mod 100)}\\ \Rightarrow 3b &\equiv 0 \qquad \mbox{(mod 5)}\\ \Rightarrow b &= 0 \; or \; 5.}$$

If c = 3 and b = 0:

$(100a + 3)^2$ ends in 9009

$$\eqalign{ \Rightarrow 600a &\equiv 9000 \qquad \mbox{(mod 10000)} \\ \Rightarrow 6a &\equiv 90 \qquad \mbox{(mod 100)} \\ \Rightarrow 3a &\equiv 45 \qquad \mbox{(mod 50)} \\ \Rightarrow a &= 15 + 50k \\ \Rightarrow \mbox{smallest} \; a &= 15;\; x = 1503.}$$

If c = 3 and b = 5:

$(100a + 53)^2$ ends in 9009 $10000a^2 + 10600a + 2809$ ends in 9009 $100a^2 + 106a + 28$ ends in 90

$$\eqalign{ \Rightarrow 6a + 28 &\equiv 90 \qquad \mbox{(mod 100)} \\ \Rightarrow 3a &\equiv 31 \qquad \mbox{(mod 50)} \\ \Rightarrow 3a &= 31 + 50k \qquad \mbox{where k and a are non negative integers.} \\ \Rightarrow \mbox{smallest} \; a &= 27;\; x = 2753 \quad \mbox{which is not minimal.}}$$

Case 2: c = 7

$(100a + 10b + 7)^2 = 10000a^2 + 200a(10b + 7) + 100b^2 + 140b + 49$ ends in 9009.

$$\eqalign{ \Rightarrow 40b + 40 &\equiv 0 \qquad \mbox{(mod 100)} \\ \Rightarrow 2b + 2 &\equiv 0 \qquad \mbox{(mod 5)} \\ \Rightarrow b &= 4\; \mbox{or}\; 9}$$

In the cases $c$ = 7 and $b$ = 4 or 9 there are no solutions less than 1503.