The Real 10 Years Challenge!

Let $P=M^2I$ . Then we have $PX+10=PX^2\implies P+\frac{10}{X}=PX$ . Since $PX$ and $P$ are both integers, $\frac{10}{X}$ must also be an integer and $X\in\{1,2,5,10\}$ .

• If $X=1$ then $P+10=P$ , a contradiction.
• If $X=5$ then $5P+10=25P\implies P=\frac{1}{2}$ , a contradiction.
• If $X=10$ then $10P+10 = 100P\implies P=\frac{1}{9}$ , a contradiction.

Thus $X=2$ and $P = 5$ . Since $P=M^2I$ and the only perfect square that divides $5$ is $1$ , it must be true that $M^2=1\implies M=1$ and $I=5$ .

So $M+I+X = 1+5+2=\boxed{8}$ .

First, let's focus on X, so treat the "Roman numbers" as:

$(some number) * X + 10 = (some number) * X^2$

In other words, introducing factor X once more would increase the product by 10. This is only feasible if:

• $X$ is 10, or an integer divisor of 10. (So: 1, 2, 5, 10).

• $X$ is not 1 (as inserting a factor of 1 wouldn't change the product) (Remained: 2, 5, 10)

• $X^2-X$ is 10 at most (assuming the extreme case where all other factors are 1) (Remained: 2)

So, X=2

Now, for M and I:

$(MMI) * 2 + 10 = (MMI) * 4$

$10 = (MMI) * 2$

$(MMI)=5$

5, being a prime number, can only be composed of factors 1 and 5. Therefore M should be 1 and X should be 5.

Conclusion:

X=2

M=1

I=5

Solution: $5+2+1=8$