Last two digits of $x^y$, when the units digit of $x$ is $1$

Question

I want to know why the statement below is true.

Let $x$ and $y$ be two positive integers. And consider that: the units digit of $x$ is $1$; the ten's place digit of $x$ is $t$; and the units digit of $y$ is $u$. Then, the last two digits of $z=x^y$ can be obtained as follows:

• The unit's place digit of $z$ is $1$
• The ten's place digit of $z$ is the unit's place digit of $t\times u$

Answer 1

Let $A$ be the number given by the digits: $a_n,...,a_2,a_1,a_0$. That is, $A = a_0 + 10\,a_1 + 100\,a_2 + ... + 10^n\,a_n$.

Let $B$ be the number given by the digits: $b_m,...,b_2,b_1,b_0$. That is, $B = b_0 + 10\,b_1 + 100\,b_2 + ... + 10^m\,b_m$.

If $\boxed{a_0 = b_0 = \mathbf{1}}$ (the unit digit of both numbers is $1$), the product $A \times B$ will be given by:

$C = A\times B = (\mathbf{1} + 10\,a_1 + 100\,a_2 + ... + 10^n\,a_n)\times B$

$C = [B] + [10\,(a_1+10\,a_2+...+10^{n-1}\,a_n)\times B]$

$C = [\mathbf{1} + 10\,(b_1 + 10\,b_2 + ... + 10^{m-1}\,b_m)] + [10\,(a_1+10\,a_2+...+10^{n-1}\,a_n)\times (\mathbf{1} + 10\,b_1 + 100\,b_2 + ... + 10^m\,b_m)]$

$\boxed{C = [1] + 10\,[a_1+b_1] \\+100\,[a_2 + ... + 10^{n-2}\,a_n + b_2+...+10^{m-2}\,b_m + (a_1+...+10^{n-1}\,a_n)(b_1+...+10^{m-1}\,b_m)]}$

This means that the units digit of $C$ is $1$ and the ten's place digit of $C$ is the units digit of $a_1+b_1$.

• If $y=1$ the statement is trivial;
• If $y=2$, we have $A=B=x$, so $a_1+b_1 \text{ mod }10 \equiv 2\,a_1 \text{ mod } 10$;
• If $y=3$, we have $A=x$ and $B=x^2$, which results in $a_1+b_1 \text{ mod }10 \equiv 3\,a_1 \text{ mod } 10$

For the general case, when the units digit of $x$ is $1$, the ten's place digit of $x^y$ will be the units digit of the product between $y$ and the ten's place digit of $x$. Since only the units digit of the product is relevant, we can operate only with the units digit of $y$.