Riddler Classic for 8 June 2018

Spoiler for last week’s fivethirtyeight.com Riddler Classic:

From Ben Gundry via Eric Emmet, find and replace with a twist:

Riddler Nation has been enlisted by the Pentagon to perform crucial (and arithmetical) intelligence gathering. Our mission: decode two equations. In each of them, every different letter stands for a different digit. But there is a minor problem in both equations.

In the first equation, letters accidentally were smudged on their clandestine journey to a safe room within Riddler Headquarters and are now unreadable. (These are represented with dashes below.) But we know that all 10 digits, 0 through 9, appear in the equation.

What digits belong to what letters, and what are the dashes?

In the second equation, our mathematical spies have said that one of the letters in the equation is wrong. But they can’t remember which one. Which is it?

First puzzle:

The smudged letter business seems to be an obvious red herring; all 10 digits are used, but 6 letters appear and 4 dashes, so each dash is a different letter and might as well be replaced by four different letters, say: EXMREEK + EHKREKK = AKBHCXDE. Right away we know A = 1.

2E > 9 and 2K\mod 10 = E, so E = 6\textrm{ or } 8. But 2E\mod 10 is K = X-1 or X = K-1. So E = 6 with K = 3 and X = 2. D = 9.

Now 2+H\ge 9, so H = 7 or H = 8.

1+2R\mod 10 = C. The only possibility still available is R = 7 and C = 5 with H = 8. The two digits left are 0 and 4 and those are indeed what B and M work out to respectively. The sum is 6247663+6837633=13085296.

Second puzzle:

We’ll assume everything’s okay until we hit a contradiction. By exactly the same logic as before E = 1, Y= 6, D = 3. In column 7, D+Y=9 so T=9\textrm{ or }0. Since in column 5 2B\mod{10} cannot be 1, there must be a carry into that column, meaning T=9.

But in column 9, K+D\mod{10}=T so K=6 which is not available.

If one of the letters column 5 is wrong then there could be no carry into that column, in which case we could have T = 0, and then K=7. But from column 4 we’d still need B = 6\textrm{ or }7 neither of which is possible. So the problem lies elsewhere.

Maybe a letter in column 9 is wrong. Then R=0, and B=5. H=7. If there’s a carry into column 8 then M+7=P, which doesn’t work, so there’s no carry and M+6=P means M=2, P=8. We’re left with K=4 which gives the incorrect sum 695513243+673596633=1369109896; if the first two numbers are right the sum should be 1369109876. Or if the first number is right and the sum is right the second number should be 673596653. Or if the second number is right and the sum is right the first number should be 695513263. We can’t determine which letter is incorrect.

This doesn’t rule out other possibilities, such as that one of the letters in column 7 or column 10 is wrong.


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