## Integer Q

Find the greatest integer $q$ such that $475 and the last three digits of $q^3$ (ones, tens, and hundreds) are the same as integer $q$ itself.
Source: mathcontest.olemiss.edu 4/29/2013

SOLUTION
Let $abc$ be the three digits of $q$ where $a$ is the hundreds digit, $b$ the tens digit, and $c$ the ones digit.
We want $c^2$ to end in 1 so that $c^2\times c=c$.
$1^2=1$
$9^2=81$
The ones digit $c$ is either $1$ or $9$.
$(abc)^2$ ending in $001$ is a necessary condition. For example,
$249^2=62001$
Multiply $249^2\times 249$
$\quad\;62001$
$\times\;249$
$\text{-------------}$
$00558009$
$0248004$
$124002$
$\text{-------------}$
$15438249$

$249^2=62001\quad\;\;249^3=15438249$
$251^2=63001\quad\;\;251^3=15813251$
$499^2=249001\quad 499^3=124251499$
$501^2=251001\quad 501^3=125751501$
$749^2=561001\quad 749^3=420189749$
$751^2=564001\quad 751^3=423564751$
$999^2=998001\quad 999^3=997002999$

$751$ is the largest integer such that $475<751<875$ and the last three digits of $751^3$ match the digits of $751$

Answer: $751$