# We are Unique Friends. Our Special Squares Share a Common Bond...Figure us out

Use logical deduction only and document your steps in detail.

$$Given$$:

A, B, C, D, E, F, M , T are distinct digits(0 to 9).

ABC, EBF, DDABC, MCMABC, TBBB are all concatenated Numbers.

$$Relationships$$:

$$ACE$$ X $$ACE$$ = $$DDABC$$

$$EBF$$ X $$EBF$$ = $$MCMABC$$

$$ACE$$ + $$EBF$$ = $$TBBB$$

$$Find$$all the numbers and $$figure out$$ the hidden $$Additive$$ $$Relationships.$$

• Why do you keep posting the exact same type of question? Some variety is in order. May 25, 2019 at 19:58
• @greenturtle3141 No it isn't. It's perfectly allowed for someone to specialise in one particular tag or genre of puzzles. Jun 4, 2019 at 9:41
• @Randal'Thor..thx..I strongly believe one should play to their strengths and this community has such a diversity that they will attract likeminded people to the puzzle
– Uvc
Jun 4, 2019 at 9:56
• @Randal'Thor I respectfully disagree, but only in this specific instance. My main concern is what appears to be a value of quantity over quality. This is evidenced by the extreme frequency in the posting of these puzzles, which suggests to me that not a ton of creative thought is being put into each one. For most of these "puzzles", nothing remarkable seems to happen, and IMO they're easy to make. A few of these puzzles actually ended up being good in that they were aesthetically pleasing, e.g. the satan one, but this and the most recent one are not. Jun 4, 2019 at 15:23
• To clarify, yes this is allowed, but I do not think this should be encouraged. Suppose, for instance, that I started posting three What is a Word? style puzzles a day, and each puzzle was either awfully obscure or dully unremarkable. Jun 4, 2019 at 15:36

Let $$ACE = x$$ and $$EBF = y$$.
So, $$y^2-x^2 = (x+y)(y-x) = TBBB(y-x) = 1000(MCM-DD)$$.
It's very tempting to let $$B = 0$$ already, but let's see why the other numbers don't work. It's clear that if $$B = 1,3,7,9$$ then $$1000$$ won't share any factors with $$TBBB$$, so $$TBBB$$ would have to divide $$MCM-DD$$ but the latter is a smaller positive number. If $$B = 5$$, then $$y-x$$ is even whereas $$TBBB$$ is odd. Yet $$y-x$$ and $$x+y$$ must share the same parity. If $$B = 2,4,6,8$$, we know that $$y-x$$ contains all of the factors of 5, as well as share the same parity as $$x+y$$ (that is, even). So $$y-x$$ is divisible by 10. However, this means that $$ACE \equiv EBF \bmod 10$$, that is, $$E \equiv F \bmod 10$$, contradicting our distinct digits condition earlier.
Great, so $$B = 0$$. Therefore, $$C = 9$$ as $$ACE+E0F = T000$$. Oh right, $$ACE+EBF \leq 1998$$, so $$T = 1$$. We have $$E0F^2 = MCMA09$$. This means that $$F = 3$$, and therefore $$E = 7$$ and therefore $$A = 2$$. So $$ACE = 297$$ and $$EBF = 703$$. We can now square these numbers to get the other digits: $$297^2 = (300-3)^2 = 90000-1800+9 = 88209$$, and $$703^2 = (700+3)^2 = 490000+4200+9 = 494209$$, so $$M = 4$$ and $$D = 8$$.
Therefore, the digits we want are: $$\boxed{A,B,C,D,E,F,M,T = 2,0,9,8,7,3,4,1}$$