Cryptarithm: Half + Fifth + Tenth + Tenth + Tenth = Whole

Difficulty:
★★★★☆

Problem:

Figure out which digit each letter represents. Every letter is a different digit and there are no leading zeros.

    H A L F
  F I F T H
  T E N T H
  T E N T H
+ T E N T H
------------
  W H O L E



Solution:

Since carry5 is 0, we can immediately notice that T should be small. In fact, T can only be either 1 or 2 -- If T is 3 or higher, carry5 would be greater than 0.

Let's first consider the case where T is 1. The second column from the right indicates that the first digit of L + 1 + 1 + 1 + 1 + carry1 is L. This means carry1 needs to be 6, which would be impossible even if H and F were assigned the highest digits (8 + 9 + 9 + 9 < 60). Hence we can eliminate the possibility that T is 1 and deduce that T must be 2. In addition, since the first digit of L + 2 + 2 + 2 + 2 + carry1 is L, we can conclude that carry1 is 2. 

    H A L F
  F I F 2 H
  2 E N 2 H
  2 E N 2 H
+ 2 E N 2 H
------------
  W H O L E

Let's again think about the fact that carry5 is 0. That means F needs to be 3 or less. Since 2 is already taken, F is either 1 or 3.

Let's pretend F was 3. Then W would be 9, and carry4 would have to be 0. Thus, E is either 0 or 1 (since 2 is taken) -- anything higher would mean carry4 is greater than 0. Now look at the last column. Given that carry1 is 2, we can deduce that F + 4H = 3 + 4H = 20 + E. Thus, 4H - E = 17. No matter whether E is 0 or 1, there is no H that fits the equation. So now we can conclude that F cannot be 3 -- it must be 1.

    H A L 1
  1 I 1 2 H
  2 E N 2 H
  2 E N 2 H
+ 2 E N 2 H
------------
  W H O L E

Since carry1 is 2, H must be either 5, 6, or 7. Anything other than those would result in carry1 being too small or too large. Let's consider all three cases.

• H is 5. From the last column, E must be 1. However, since 1 is already taken, this cannot be the case. So we can eliminate this case.
• H is 6. From the last column, E must be 5.
• H is 7. From the last column, E must be 9.

So now we have two options: (1) H = 6, E = 5 and (2) H = 7, E = 9. We can show that the second option is impossible. Suppose it is true. Since 7 and 9 are both taken, W must now be 8, meaning that carry4 must be 1. However, since E is 9, carry4 needs to be at least 2. This contradiction allows us to eliminate option 2 and conclude that H is 6 and E is 5.

    6 A L 1

  1 I 1 2 6
  2 5 N 2 6
  2 5 N 2 6
+ 2 5 N 2 6
------------
  W 6 O L 5

Since carry4 is at least 2, we can also figure out that W has to be 9. And as a result, carry4 has to be 2 (it cannot be greater).

    6 A L 1

  1 I 1 2 6
  2 5 N 2 6
  2 5 N 2 6
+ 2 5 N 2 6
------------
  9 6 O L 5

The fact that carry4 is 2 only leaves us few options for I. Note that carry has to be 4 or less; even if all digits in a column were 9, the carry would be 4. If I was 0, carry3 would have to be 5 -- fail. 1, 2, 5, and 6 are taken. If I was 7, carry3 would have to be 8 -- fail. If I was 8, carry3 would have to be 7 -- fail. 9 is taken. That leaves us with only 3 and 4 as options for I.

Let's pretend I is 3. Then carry3 would be 2. Excluding the digits that are already taken, N can only be 0, 4, 7, or 8. It cannot be 0 because then carry3 would be less than 2. From the middle column we know that A + 1 + 3N = 20 + O, which can be simplified to A + 3N - O = 19. Now let's consider the options: N = 4, N = 7, and N = 8.

• N = 4. This means A + 12 - O = 19, simplified to A - O = 7. The only option for A and O that satisfy this equation is A = 7 and O = 0. Thus, carry2 would have to be 2, which is impossible regardless of the value of L. So we can conclude that N is not 4.
• N = 7. This means A + 21 - O = 19, simplified to A - O = -2. There are no pair of digits that aren't taken that satisfy this equation. Thus, N is not 7.
• N = 8. This means A + 24 - O = 19, simplified to A - O = -5. Again there are no pair of digits that aren't taken that satisfy this equation (keep in mind that we're assuming that N is 8, so it can't be the case that A is 3 and O is 8). So N cannot be 8.

We've exhausted our options, which leads us to conclude that I cannot be 3. Thus, I must be 4.

    6 A L 1

  1 4 1 2 6
  2 5 N 2 6
  2 5 N 2 6
+ 2 5 N 2 6
------------
  9 6 O L 5

As a result, we also learn that carry3 is 1. Given our available digits, N must be either 0 or 3. However, 0 is too small for N, because then A would have to be 8 in order to make carry3 1, in which case O must be 0 (remember that carry2 is at most 1), which would be taken. So N must be 3.

    6 A L 1

  1 4 1 2 6
  2 5 3 2 6
  2 5 3 2 6
+ 2 5 3 2 6
------------
  9 6 O L 5

We only have 0, 7, and 8 available. Since carry2 is at most 1, we know that O is greater than A by 1. The only combination that works is A = 7, O = 8, L = 0.

    6 7 0 1

  1 4 1 2 6
  2 5 3 2 6
  2 5 3 2 6
+ 2 5 3 2 6
------------
  9 6 8 0 5