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If the letters T*(RBJBR)=VPLNT each represented a unique digit, and "RBJBR" was a five digit number, what are possible values for the letters? (Or ONE possible value.)

Can we do this in a way that can be done on paper without "guessing and plugging numbers" (or computers, for that matter)?

Got this question as a math challenge from a club, not sure how do go about it.

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This is not so much "cryptography" as it is a "cryptarithm". It's a puzzle. – Arturo Magidin Apr 19 '12 at 4:41
@BrettFrankel I have no idea what that means. New here. :D Apologies about the tag misplacement, I'm not sure where I should put it. – badreferences Apr 19 '12 at 4:46
If you ask a question and somebody solves it to your satisfaction, you can click the check mark underneath the arrows. The answerer will be awarded with a few reputation points, and other users will see that you are in need of a solution. – Brett Frankel Apr 19 '12 at 4:51
Whoops, now I should go back through my questions and click those marks. – badreferences Apr 19 '12 at 5:33

Here is a start...

If you check the last digit, you'll see that $T*R$ ends in $T$. Also, by looking at the first digit, $T*R$ is at most $9$, otherwise the right side has $6$ digits.

Then $T*R=T$ which means that either $T=0$ (not possible) or $R=1$. Thus $R=1$.


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which reduces the problem to $T\cdot(1BJB)=VPLN$. – alex.jordan Apr 19 '12 at 18:25

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