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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 $Additive$ $Relationships.$

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  • $\begingroup$ Why do you keep posting the exact same type of question? Some variety is in order. $\endgroup$
    – greenturtle3141
    Commented May 25, 2019 at 19:58
  • $\begingroup$ @greenturtle3141 No it isn't. It's perfectly allowed for someone to specialise in one particular tag or genre of puzzles. $\endgroup$
    – Rand al'Thor
    Commented Jun 4, 2019 at 9:41
  • $\begingroup$ @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 $\endgroup$
    – Uvc
    Commented Jun 4, 2019 at 9:56
  • $\begingroup$ @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. $\endgroup$
    – greenturtle3141
    Commented Jun 4, 2019 at 15:23
  • $\begingroup$ 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. $\endgroup$
    – greenturtle3141
    Commented Jun 4, 2019 at 15:36

1 Answer 1

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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: $\boxed{A,B,C,D,E,F,M,T = 2,0,9,8,7,3,4,1}$

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  • $\begingroup$ There are couple of other additive relations between the numbers and string parts of the square product $\endgroup$
    – Uvc
    Commented May 25, 2019 at 15:29

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