Place Values of numbers

Question

I have distinct 3-digit numbers that, when multiplied by themselves, produced a number which ended in the same 3-digit number. What are the the distinct 3-digit numbers which have this property? I have tried using place value concept; but it's not working that way.

Answer 1

We have: $$ (100a+10b+c)^2\equiv 200ac+100b^2+20bc+c^2\pmod{1000} $$ So first of all we must have $c^2\equiv c\pmod{10}$: $$ \begin{align} \color{red}{0}^2&=\color{red}{0}\\ \color{red}{1}^2&=\color{red}{1}\\ 2^2&=4\\ 3^2&=9\\ 4^2&=16\\ \color{red}{5}^2&=2\color{red}{5}\\ \color{red}{6}^2&=3\color{red}{6}\\ 7^2&=49\\ 8^2&=64\\ 9^2&=81 \end{align} $$ So $c\in\{0,1,5,6\}$. Next we must divide into cases.

---

Suppose $c=0$: Then the $10$'s places becomes $$ [0]_{100}=[10b]_{100} $$ which is never the case.

---

Suppose $c=1$: Then the $10$'s places becomes $$ [20b]_{100}=[10b]_{100} $$ which is never the case.

---

Suppose $c=5$: Then the $10$'s places becomes $$ [100b+20]_{100}=[10b]_{100} $$ which implies $b=2$. Thus the $100$'s places must satisfy $$ [1000a+600]_{1000}=[100a]_{1000} $$ which implies $a=6$. Thus $625$ is a solution.

---

Suppose $c=6$: Then the $10$'s places becomes $$ [120b+30]_{100}=[10b]_{100}\\ \iff\\ [10b+30]_{100}=0 $$ since the $100$ of the $120$ vanishes modulo $100$ and $20b-10b=10b$. This implies $b=7$. Thus the $100$'s places must satisfy $$ [1200a+5700]_{1000}=[100a]_{1000}\\ \iff\\ [100a+700]_{1000}=0 $$ which implies $a=3$. Thus $376$ is a solution.

Answer 2

If we generalise this to finding $n$-digit numbers such that their square ends in the same $n$ digits as the original number, then we looking for solutions to the equation

$x^2 = x \mod 10^n \Rightarrow x(x-1) = 0 \mod 10^n$

If we exclude the trivial cases $x=0$ and $x=1$, then we have two possibilities:

(1) $x=0 \mod 5^n \text{ and } x=1 \mod 2^n$

(2) $x=0 \mod 2^n \text{ and } x=1 \mod 5^n$

The Chinese Remainder Theorem tells us that there is exactly one solution $\mod 10^n$ for each case, although it does not guarantee that both have $n$ digits with no leading zeros. For example, when $n=5$ we have

$90625 = 0 \mod 5^5 \text{ and } 90625 = 1 \mod 2^5$

$9376 = 0 \mod 2^5 \text{ and } 9376 = 1 \mod 5^5$

The symmetry of the remainders mod $10^n$ also explains why the sum of the two solutions is $10^n+1$.