Congruence Problem 2

We can rewrite the expression as $x!+y!+x!y!=(x!+1)(y!+1)-1$

So $(x!+1)(y!+1) \equiv 4 \bmod 10$

Hence either $x!$ or $y!$ must be odd. The only odd factorials are $0!=1!=1$ . So (at least) one of the numbers is $0$ or $1$ . To make it easier to count solutions, let's say this number is $x$ , so $x!=1$ .

Substituting back in, $2(y!+1) \equiv 4 \bmod 10$ . This leads to two possible congruences; either $y!\equiv 1$ or $y!\equiv 6$ . Both of these have solutions; they are $y=0,1$ , or $y=3$ . (There are very few possibilities here - note that $y!\equiv0\bmod 10$ for $y\ge5$ )

Combining these (and remembering to count ordered pairs), the full set of solutions is $\{(0,0),(0,1),(0,3),(1,0),(1,1),(1,3),(3,0),(3,1)\}$ ; there are $\boxed8$ of these.

Since $x! \bmod 10 = 0$ for all $x \ge 5$ , we need only to consider $x, y \in [0, 5]$ . We have:

$\begin{cases} 0! = 1 & \implies 0! \bmod 10 = 1 \\ 1! = 1 & \implies 1! \bmod 10 = 1 \\ 2! = 2 & \implies 2! \bmod 10 = 2 \\ 3! = 6 & \implies 3! \bmod 10 = 6 \\ 4! = 24 & \implies 4! \bmod 10 = 4 \\ 5! = 120 & \implies 5! \bmod 10 = 0 \end{cases}$

For $x = 5$ , then $x! \bmod 10 = 0$ and

$\begin{aligned} x! + y! + x!y! & \equiv 3 \text{ (mod 10)} \\ 0 + y! + 0 & \equiv 3 \text{ (mod 10)} \\ y! & \equiv 3 \text{ (mod 10)} \end{aligned}$

Since $y! \bmod 10 \ne 3$ for all $y$ , there is no solution. Of course, if there is then there will be infinitely many solutions to this problem.

For $x = 0, 1$ , then $x! \bmod 10 = 1$ and

$\begin{aligned} x! + y! + x!y! & \equiv 3 \text{ (mod 10)} \\ 1 + y! + y! & \equiv 3 \text{ (mod 10)} \\ 2y! & \equiv 2 \text{ (mod 10)} \\ y! & \equiv 1 \text{ (mod 10)} \\ \implies y & = 0, 1 \end{aligned}$

There are $\color{#D61F06}4$ solutions: $(x,y) = (0,0), (0,1), (1,0), (1,1)$ .

For $x = 2$ , then $x! \bmod 10 = 2$ and

$\begin{aligned} x! + y! + x!y! & \equiv 3 \text{ (mod 10)} \\ 2 + y! + 2y! & \equiv 3 \text{ (mod 10)} \\ 3y! & \equiv 1 \text{ (mod 10)} \end{aligned}$

Since $3y! \bmod 10 \ne 1$ for all $y$ , there is no solution.

For $x=3$ , then $x! \bmod 10 = 6$ and

$\begin{aligned} x! + y! + x!y! & \equiv 3 \text{ (mod 10)} \\ 6 + y! + 6y! & \equiv 3 \text{ (mod 10)} \\ 7y! & \equiv -3 \equiv 7 \text{ (mod 10)} \\ y! & \equiv 1 \text{ (mod 10)} \\ \implies y & = 0, 1 \end{aligned}$

There are $\color{#D61F06}4$ solutions: $(x,y) = (0,3), (1,3), (3,0), (3,1)$ .

For $x = 4$ , then $x! \bmod = 4$ and

$\begin{aligned} x! + y! + x!y! & \equiv 3 \text{ (mod 10)} \\ 4 + y! + 4y! & \equiv 3 \text{ (mod 10)} \\ 5y! & \equiv -1 \equiv 9 \text{ (mod 10)} \end{aligned}$

Since $5y! \bmod 10 \ne 9$ for all $y$ , there is no solution.

Therefore, there are $\color{#D61F06} \boxed 8$ solutions.

At first, I missed the fact 0 for x or y was legitimate. I realized that the maximum value for x or y was 4 as anything higher would result in a 0 value from the modulus function. This left 25 cases. First I checked the x=y cases (4 cases) and found two cases where the modulus was equal to 3. Then, I examined the cases with y>x, of which there are 10, and found 3. Those cases need to be doubled by switching x and y. This gives a grand total of 8 passing cases.