Find the largest 4 digit positive integer n such that 10 divides $n^{19}+99^n$

Question

I've solved this problem and got the answer of $9991$. I manually proved how some digits could or couldn't be the ones digit of $n$, but I feel there is a faster way

Answer 1

$n^{19}+99^n=n^{19}+(100-1)^n\equiv n^{19}+(-1)^n\bmod 10$ by Newton's formula.

We therefore need $n^{19}\equiv -1^n \bmod 10$.

Clearly both expressions cycle $\bmod 10$. So you need only solve $\bmod 10$, and the only solution is $1\bmod 10$ by inspection.

Answer 2

We first note that $\gcd(n,10)=1$ (clearly). In particular, $n$ is odd. Thus $99^n\equiv (-1)^n\equiv -1\pmod {10}$.

Now, $\varphi(10)=4$ so $n^{16}\equiv 1 \pmod {10}$ Thus your expression is $$n^3-1\pmod {10}$$ It is easy to see that the only cube root of $1 \pmod {10}$ is $1$ so we just want the largest $4$ digit number congruent to $1\pmod {10}$, and that's $9991$.

Answer 3

Consider this number $n$ as $XXXY$. Divisibility on $10$ of $n^{19}+99^n$ requires that last digit of it is $0$. So, what really matters here is last digits of $n^{19}$ and $99^n$.

Last digit of $99^n$ has periodic structure: $\{9,1,9,1,9,1,\dots\}\to9$ for odd $n$, $1$ for even $n$.
Last digit of $n^{19}\to XXXY^{19}\to Y^{19}$.

Together it lead to constraints: $$ \begin{cases} Y^{19}+9\equiv 0\mod{10},\quad \text{ if Y odd}\\ Y^{19}+1\equiv 0\mod{10},\quad \text{ if Y even} \end{cases} $$ where $0\le Y\le 9$.

Notice that $19$th power of even number is still even number, even $+$ odd is odd, hence second congruence has no solutions.

Then only thing left is to examine cases $Y=\{1,3,5,7,9\}$ for $Y^{19}+9\equiv 0\mod{10}$, from where we conclude that $1$ is only suitable choice and as far as $X$ are arbitrary, we can conclude that final result should be $9991$.