# 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. – greenturtle3141 May 25 at 19:58
• @greenturtle3141 No it isn't. It's perfectly allowed for someone to specialise in one particular tag or genre of puzzles. – Rand al'Thor Jun 4 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 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. – greenturtle3141 Jun 4 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. – greenturtle3141 Jun 4 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: $$\boxed{A,B,C,D,E,F,M,T = 2,0,9,8,7,3,4,1}$$