»  Solutions to puzzles in my VDARE.com monthly Diary

—————————

¶  Solution.

Solving cryptarithmetic puzzles — "cryptarithms" for short — rarely depends on flashes of inspiration. You just have to beat the darn things to death.

¶ Preamble:   I need some language to deal with this one.

• Since both the digit zero and the uppercase letter "O" play roles in the puzzle, to avoid confusion I'll write zero as "Ø."
  
  
• I'll refer to the five columns of the sum as, from left to right, "column 1," "column 2," …
  
  
• Since it's addition we are doing, there may be numbers "carried" from a column to the column at its immediate left. I'll refer to these carries with lower-case "c" as c₁, c₂, c₃, and c₄. There is of course no carry digit under column 5.
  
  How big can these carry numbers be? Well, since every column has the form 3X+2Y, with X and Y different single digits, the biggest possible value for 3X+2Y is when X=9, Y=8, giving 43. So each carry number is either Ø, 1, 2, 3, or 4.

OK, let the beating commence!

¶ Column 1:   I'll work the columns, beginning with column 1, the leftmost.

Column 1 adds up to 3N+2T+c₁. There are only five digits in WHOLE, so this sum must be a one-digit number; and since there are no leading zeroes, neither N nor T can be Ø.

If c₁ is 4, 3N+2T can be at most 5. This is not possible with different N and T.

If c₁ is 3, 3N+2T can be at most 6; also not possible.

So c₁ is Ø, 1, or 2.

This gives just six possibilities for (N, T, c₁, W).

(1, 3, Ø, 9)
(2, 1, Ø, 8)
(1, 2, Ø, 7)
(2, 1, 1, 9)
(1, 2, 1, 8)
(1, 2, 2, 9)

¶ Column 2:   Shifting attention to column 2, the first thing to be said about it actually relates to column 1. It is: c₁ can't be Ø.

Why can't c₁ be Ø? Because column 2 tells us that 3I+2H+c₂ gives us digit H in the total. That's only possible if there's a carry digit — if, in other words, 3I+2H+c₂ = 1Ø+H or 2Ø+H or 3Ø+H or 4Ø+H. So c₁ can't be Ø.

That trims down the possibilities for (N, T, c₁, W) to

(2, 1, 1, 9)
(1, 2, 1, 8)
(1, 2, 2, 9)

OK, now I can concentrate on column 2, right? Well, yes; but first, glancing rightwards to column 3, I see that c₂ can't be bigger than 2.

Why not? Because column 3 adds to 3N+2I+c₃. We know that N is at most 2. Even with the biggest possible values of I and c₃ that only adds to 28.

Column 2 sums to 3I+2H+c₂, and the total is some number whose "units" digit is H. A moment ago I expressed that fact like this: 3I+2H+c₂ = 1Ø+H or 2Ø+H or 3Ø+H or 4Ø+H. To simplify, I can just subtract an H from both sides of the equals sign in each case, giving me

3I+H+c₂ = 1Ø, 2Ø, 3Ø, or 4Ø.

I can forget about the 40. 3I+2H+c₂=40 with c₂=Ø, 1, or 2 would mean 3I+2H=40 or 39 or 38. There are no solutions for that with single-digit I and H.

Given that 1 and 2 are spoken for (they are N and T in some order) and neither c₁ nor c₂ can be bigger than 2, how many possibilities are there? I make it ten. Expressed as (I, H, c₁, c₂, 3I+2H+c₂), they are:

(4, 8, 2, 2, 28)
(0, 9, 1, 1, 19)
(3, 0, 1, 1, 10)
(4, 7, 2, 1, 27)
(5, 4, 2, 1, 24)
(0, 8, 1, 2, 18)
(6, 0, 2, 2, 20)
(4, 6, 2, 2, 26)
(3, 9, 2, 2, 29)
(5, 3, 2, 2, 23)

Sure enough: scrutinizing the last number in each of those brackets, I see that its "tens" digit is c₁ and its "units" digit is H.

¶ Column 3:   Onwards to column 3.

First let's get a handle on c₃, the carry digit at the bottom of column 3. It is of course the "tens" digit from adding up 3T+2R+c₄. Since T is at most 2 and R at most 9 and c₄ at most 4, 3T+2R is at most 28, so c₃ is at most 2.

I'm now going to enlarge that list of ten possibilities I just produced. Into each bracket I'll inset N in the first position and the total sum of column 3, including carry digit, in the last position. So now it is a list of (N, I, H, c₁, c₂, 3I+2H+c₂, 3N+2I+c₃). Unfortunately, since N can be either 1 or 2, the list is not only broader, it's now twice as long.

(1, 4, 8, 2, 2, 28, 11+c₃)
(1, 0, 9, 1, 1, 19, 3+c₃)
(1, 3, 0, 1, 1, 10, 9+c₃)
(1, 4, 7, 2, 1, 27, 11+c₃)
(1, 5, 4, 2, 1, 24, 13+c₃)
(1, 0, 8, 1, 2, 18, 3+c₃
(1, 6, 0, 2, 2, 20, 15+c₃)
(1, 4, 6, 2, 2, 26, 11+c₃)
(1, 3, 9, 2, 2, 29, 9+c₃)
(1, 5, 3, 2, 2, 23, 13+c₃)
(2, 4, 8, 2, 2, 28, 14+c₃)
(2, 0, 9, 1, 1, 19, 6+c₃)
(2, 3, 0, 1, 1, 10, 12+c₃)
(2, 4, 7, 2, 1, 27, 14+c₃)
(2, 5, 4, 2, 1, 24, 16+c₃)
(2, 0, 8, 1, 2, 18, 6+c₃
(2, 6, 0, 2, 2, 20, 18+c₃)
(2, 4, 6, 2, 2, 26, 14+c₃)
(2, 3, 9, 2, 2, 29, 12+c₃)
(2, 5, 3, 2, 2, 23, 16+c₃)

Whoa, this is getting out of control. Let's not panic, though. We can cull those twenty possibilities.

The "tens" digit of the column 3 total is, by definition, c₂, so it must match the fourth number in each bracket, which is c₂ by definition.

That culls the twenty possibilities for (N, I, H, c₁, c₂, 3I+2H+c₂, 3N+2I+c₃) down to just five.

(1, 4, 7, 2, 1, 27, 11+c₃)
(1, 5, 4, 2, 1, 24, 13+c₃)
(2, 4, 7, 2, 1, 27, 14+c₃)
(2, 5, 4, 2, 1, 24, 16+c₃)
(2, 6, Ø, 2, 2, 2Ø, 18+c₃)

Note that in every one of these cases, c₁ is 2. Looking back to my researches on column 1 I see that (N, T, c₁, W) must then be (1, 2, 2, 9). So N=1, T=2, W=9, and only the first two items on my list are possible: (N, I, H, c₁, c₂, 3I+2H+c₂, 3N+2I+c₃) is either (1, 4, 7, 2, 1, 27, 11+c₃) or (1, 5, 4, 2, 1, 24, 13+c₃)

We can just try these two possibilities in turn.

¶ Dénouement:   Trial & error.

The first of those two possibilities leaves digits Ø, 3, 5, 6, and 8 still unaccounted for; the second leaves Ø, 3, 6, 7, and 8.

At this point I just rewrote the two sums as they appear in those two cases, and ran trial & error on the unaccounted-for digits. The first possibility makes the sum look this:

1 4 1 2 7
1 4 1 2 7
1 4 1 2 7
2 7 4 R D
2 7 4 R D
—————
9 7 O L E

with digits Ø, 3, 5, 6, and 8 still unaccounted for

D=Ø or 5 gives E=1; D=3 or 8 gives E=7. Since 1 and 7 are already accounted for, D must be 6, making E=3. That carries 3, so column 4 is 9+2R.

With 6 and 3 now also spoken for, R is 0, 5, or 8, giving 9+2R equal to 9, 19, or 25 and so L=9 or 5. Since W=9, R must be 8 and L=5, carrying 2 to column 3. But that makes column 3 add to 13, implying O=3, and we already have E=3 …

So the first of our possibilities can't be made to work. We must rest our hope in the second. Here the sum looks like this:

1 5 1 2 4
1 5 1 2 4
1 5 1 2 4
2 4 5 R D
2 4 5 R D
—————
9 4 O L E

with digits Ø, 3, 6, 7, 8 still unaccounted for

Two minutes trial & error delivers the solution:

1 5 1 2 4
1 5 1 2 4
1 5 1 2 4
2 4 5 Ø 7
2 4 5 Ø 7
—————
9 4 3 8 6