Let $N$ be a positive integer that ends in $9$, what are all the possibilities for the last two digits of $n^{10}$?

Question

The forms of the powers would be $(10x+9)^{10}$. This is congruent to $10×10x×9^9+9^{10}$, and this becomes $100×9^9+9^{10}$. We need the last two digits, so we only need to consider $9^{10}$, which ends in $01$. Am I correct?

Answer 1

You can do $(10x + 9)^9 \equiv 10*10x*9^9 + 9^{10}\mod 100$ to get the last two digits.

Or you can do $10x + 9 = 10y -1$ where $y = x+1$ and so

$(10y-1)^9 \equiv -10*10y + 1 \equiv -100y +1 \equiv 01 \mod 100$.

so last two digits are $01$.

==== Argh.... well if you ever want the last 3 digits ... read on ====

Hint: If $N$ ends with $9$ then $N = 10x -1$.

So $N^{10} = (10x)^{10} - ..... + \frac {10 2}(10x)^2 - 10(10x) + 1$

which modul $1000$ is

$N^{10} \equiv 4500x^2 - 100x + 1 \equiv 500*x^2 - 100x + 1 \mod 1000$.

If $x$ is odd then $500*x^2 \equiv 500 \mod 1000$ and if $x$ is even then $500x^2 \equiv 0 \mod 1000$.

And if $x \equiv k \mod 10$ then $-100k \equiv (10-k)*100\mod 1000$

If $k$ is even then last three digits are $(10-k)01$.

If $k$ is odd then the last three digits are $(5-k)01$.

So the last three digits are $m01$ for any even $m$.

In particular if the last two digits of $N$ are: then the last three of $N^3$ will be:

$99:: 001$

$09:: 401$

$19:: 801$

$29:: 201$

$39:: 601$

$49:: 001$

$59:: 401$

$69:: 801$

$79:: 201$

$89:: 601$.