What Do Digits Mean, Anyway?

Puzzle

I found this puzzle in the G+ Mathematics community, courtesy of Paul Cooper. Solve the final addition:

  1. 50 + 60 + 90 = 380
  2. 30 + 40 + 60 = 330
  3. 90 + 60 + 70 = 350
  4. 50 + 90 + 30 = 10
  5. 70 + 30 + 20 = 370
  6. 40 + 50 + 90 = 300
  7. 60 + 70 + 30 = 380
  8. 20 + 40 + 10 = 600
  9. 60 + 45 + 15 = ?

My first reaction was to dismiss this problem as another case of the equality sign being misused. However, Mr. Cooper insisted that the addition and equality signs were behaving themselves as expected. He also said that the numbers on each side of the equality sign were behaving in the same way.

I considered a few approaches, such as unusual bases and number as variables, but couldn’t quite crack the nut. It turns out that the solution is much simpler than that, and serves as an excellent display of why I don’t like those “88 = 4; 91 = 1” problems. “88 = 4” clearly indicates that we’ve thrown out some convention or other, but what convention?

If you haven’t yet tried your hand at the puzzle and would like to (spoilers…), take a bit of time before I work it out below the header line.

Solution

The addition sign and equality sign behave as expected, and the numbers are in base ten. The bits that aren’t behaving as expected are the numerals themselves.

The assignment of values to characters is completely arbitrary. In this case, as with a cryptogram, the number symbols have their numeric values mixed up. I’m used to solving Word Math problems, where letters stand for numbers, so once I knew what was going on, the problem itself was pretty easy.

Let’s take a look at it again, replacing the digits 0..9 with the letters A..J, respectively:

  1. FA + GA + JA = DIA
  2. DA + EA + GA = DDA
  3. JA + GA + HA = DFA
  4. FA + JA + DA = BA
  5. HA + DA + CA = DHA
  6. EA + FA + JA = DAA
  7. GA + HA + DA = DIA
  8. CA + EA + BA = GAA

This is now a matter of finding values for each of the letters so that each letter has a different single-digit numerical value and each addition is valid. It’s possible that there are solutions where different letters can stand for the same digit; I didn’t look for such solutions.

First, notice that 3A = xA. The only values for which this is true are 0 and 5, so A = 0 or A = 5. I’ll assume that each digit has been mapped differently than it ought to be, and since A started out as 0, I’ll try A = 5. It’s possible there are solutions where some, but not all, of the digits are mapped correctly, but I didn’t look for such solutions there, either. This gives us:

  1. F5 + G5 + J5 = DI5
  2. D5 + E5 + G5 = DD5
  3. J5 + G5 + H5 = DF5
  4. F5 + J5 + D5 = B5
  5. H5 + D5 + C5 = DH5
  6. E5 + F5 + J5 = D55
  7. G5 + H5 + D5 = DI5
  8. C5 + E5 + B5 = G55

Line 4 tells us that F + J + D + 1 = B. Three two-digit numbers added up cannot be greater than 99 * 3 = 297, so D and G are 1 and 2, in some order. Lines 1 and 4 tell us that D5 < G5 (since the other two addends are the same), so D = 1 and G = 2.

Hence, F + J + 1 + 1 = B. F and J both have to be at least 3; since 3 + 4 + 2 = 9, F and J are 3 and 4 in some order, and B is 9. Let’s update what we have:

  1. F5 + 25 + J5 = 1I5
  2. 15 + E5 + 25 = 115
  3. J5 + 25 + H5 = 1F5
  4. F5 + J5 + 15 = 95
  5. H5 + 15 + C5 = 1H5
  6. E5 + F5 + J5 = 155
  7. 25 + H5 + 15 = 1I5
  8. C5 + E5 + 95 = 255

Line 2 tells us that 115 – 40 = 75, so E = 7. We know that either F5 = 35 and J5 = 45, or vice versa, so line 1 tells us that 35 + 25 = 45 = 105, so I = 0. Line 5, meanwhile, tells us that C + H + 1 + 1 = 1H, so C + 2 = 10; hence, C = 8. Updating again:

  1. F5 + 25 + J5 = 105
  2. 15 + 75 + 25 = 115
  3. J5 + 25 + H5 = 1F5
  4. F5 + J5 + 15 = 95
  5. H5 + 15 + 85 = 1H5
  6. 75 + F5 + J5 = 155
  7. 25 + H5 + 15 = 105
  8. 85 + 75 + 95 = 255

This leaves F, J, and H. Since we have possible values for F and J, H must equal 6. From line 3, we get J5 + 25 + 65 = J5 + 90 = 1F5, so F < J, thus F = 3 and J = 4:

  1. 35 + 25 + 45 = 105
  2. 15 + 75 + 25 = 115
  3. 45 + 25 + 65 = 135
  4. 35 + 45 + 15 = 95
  5. 65 + 15 + 85 = 165
  6. 75 + 35 + 45 = 155
  7. 25 + 65 + 15 = 105
  8. 85 + 75 + 95 = 255

By my reckoning, we do not need lines 6 or 7, but that’s okay. Having them there makes the problem a little easier, works as a nice cross-check, and is a distraction for people who notice (as I did) that there are eight different numbers on the left, so if they’re variables, it’s possible to solve for them.

Now that we have a map, we can return to the last line: 60 + 45 + 15 becomes 25 + 73 + 93 = 191. But for a truly complete answer, we need to map that back to 313.

As I said, what truly appeals to me about this problem is that it reinforces that, when we break the conventional rules in order to construct a problem, it’s not always apparent¬†which conventional rules we’ve broken. In this case, the addition and the equality are perfectly well-behaved, it’s the numerals that are mucked up.

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