In this cryptarithm, each letter above the line represents a digit differing by 1 from the digit represented by the same letter below the line. For example, if A=3 above the line, then A=2 or A=4 below the line. All occurrences of a letter on the same side of a line represent the same digit. It is also known that there are total of five digits in the solution.

 ADABA
+CACBA
------
 DBABC

Ones Column: A+A = 2A, so the ones digit of 2A (which is C below the line) is even. Since the same letters above and below the line are off by one, this means that C must be odd, above the line (1).

Tens column: Let’s look at the column of Bs. We don’t know if A is 5 or greater, so we don’t know if a one is carried. Let’s look at all the cases possible:

For A<5 it turns out that B above the line can only be 1 or 9.

For A>4 it turns out that B above the line can only be 0 or 8 (2).

Hundreds column: Again, we don’t know if a one is carried from the tens column, but let’s think about each case.

For B<5: (remember, C above the line must be odd (by 1), and not equal to A) In this case, C must be 9 or 1, and A can be anything (except that it can’t be the same as C). But look at the TenThousands column. If C is 9 (and assuming we aren’t using leading zeros) then a 1 will be carried in the solution, but there aren’t any more digits. So C must be 1 in this case (3).

For B>4: (remember, C above the line must be odd (by 1), and not equal to A) This doesn’t work at all. If C could be even, it would work when C=8, but since it must be odd, we can’t have a case where B +B carries a one.

So C=1.

So, C=1 means that B must be <5 (by 3), and we already found B can only be 0, 1, 8, or 9, (by 2) so this leaves us with only 0 or 1. We already said that C=1, so B=0.

B=0 means that A>4 (by 2).

Also, B below the line will have to be 1, since a 1 is carried from A+A.

Let’s look at the ones column again. A can only be 5, 6, 7, 8, or 9, and the C below the line can only be 0 or 2. So A can only be 5 or 6.

Let’s see what we have so far.

ADA0A + 1A10A =  D1A1C

If A=5, then we would have this:

5D505 + 15105 = D1610

In this case, D above the line must be 6. This is because no positive integer can make D+5=1, and in order for D+5=11 D must be 6. Then a one is carried, and we have 1+5+1=7. This works.

Just to check, if A = 6 we would have this:

6D606 + 16106 = D1712

D would have to be 6. This is because no positive integer can make D+6=1, an in order for D+6=11 D must be 5. Then a one is carried and we have 1+6+1=8. This does not work.

So the solution is 56505 + 15105 = 71610 and only 5 digits were used the whole time!

Posted by nikki on 2004-08-11 14:44:02