All your base are belong to us (cryptarithm extension)

If you don't know Futility Closet, I suggest you take a look. It is a fun and quirky combination of math puzzles, chess puzzles, and historical anecdotes

This recent post had a nice puzzle, Hidden sum, that led to a fun conversation with J2 and J1.

This was a fun puzzle on its own that I knew would appeal to J2, since one of his familiar number friends, 111, is lurking in the solution.

The base

Before I got a chance to discuss with J2, however, I spent some time considering a small clause in the question: "in base 10." Strangely, if this clause hadn't been included, I probably would never have thought to investigate in other bases. This restriction, though, seemed like an invitation to go exploring in other bases. Since the older J's had recently done some work in non-decimal bases, I thought they would enjoy this extra exploration.

I told J2 this puzzle. First, he worked through the base 10 version, including seeing an old friend (and familiar factorization) along the way, I asked what he thought about doing it in other bases. He was interested, so we started with binary. Luckily, his first idea was to consider possibilities for TTT. In binary, the only three digit TTT is 111, aka 7 in decimal. He saw that was prime, so couldn't be factored into two 2-digit factors. That proved to be the first key insight of the exploration.

We moved on to base 3, 4, 5, 6, 7, 8, 9, 10, and 11. At some point, J1 joined the game. Along the way, they made the following observations and conjectures:

1. if 111 is prime, there is no solution. This is because TTT will have to have a 3 digit factor.
2. If 111 is not prime, it will have one 1-digit and one 2-digit factor (why?)
3. If 111 is not prime, neither factor will end with a 0 in the ones place (why?)
4. Given a 2-digit number (ME) with a non-zero ones digit in the ones place (E not 0), and a (non-zero) one digit number X, there is a single digit value T such that multiple of T x X is of the form YE (a 2-digit number sharing the earlier value in the ones place). 

For the first three conjectures, "(why?" means that most of you should be able to prove these. For the fourth, this conjecture isn't true! However, there is something extra that happens in the scenario for the puzzle that gives extra information and makes it true when ME x X = 111

The sum

One other point is lingering for me: why does the original puzzle ask for the sum of E, M, T, and Y? Sometimes, this form of question is a clue that there is some interesting relationship that allows us to calculate the answer without finding values for all the variables. Though it is common, I still get a kick out of this, probably because there is such a strong instinct to solve for all the variables.

In this case, I really don't see a way to get the sum directly, without finding values for E, M, T, and Y. Any ideas?

If there isn't a direct path, why did they phrase the question this way? Without seeing the exam, my guess is that this is an information reduction operation that allows this to be a multiple choice question.