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 Question from Lisa, a parent: Jill and Bill are two members of the Hill family. Jill is 8 years old and Bill is 9 years old. What will Bill's age be when the total of the digits in both their ages is once again 17?

Hi Lisa.

Let's use letters as variables for each digit:
Jj = Jill's age. The capital J is the tens digit and the small j is the ones digit.
Bb = Bill's age.

Now recall that Bill is always one year older than Jill (we assume that they have the same birthday!)

When Bill is 10, 20, 30, ... then Jill is always 9, 19, 29, ... so

Most of the time, b = j +1 and B = J. But b is 0 (ie. Bill is 10, 20, 30...) when j = 9 and at that time B = J + 1 (ie, Bill is 30 and Jill is 29).

So we have really two cases: when b is zero and when b is not zero.

Case 1: b is zero.
In this case, we know j = 9. We also know B + b + J + j = 17 (that's the situation described in the question). We can substitute in 0 for b and 9 for j, so we get B + 0 + J + 9 = 17. This simplifies to B + J = 8. But remember that B = J + 1, so we replace B with (J + 1) in the former equation: (J + 1) + J = 8. Thus, 2J + 1 = 8, which reduces to J = 3.5. That's not permitted, since J must be a whole number. Therefore, there are no solutions for case 1 and so b must be non-zero if there is a solution at all.

Case 2: b is non-zero.
Remember B + b + J + j = 17. What do we know that we can substitute in?
b = j + 1 and B = J. So (J) + (j + 1) + J + j = 17. This reduces to 2J + 2j = 16 which means J + j = 8. There are lots of solutions for this! Just pick a value of J larger than zero and smaller than nine. That
lets you find the corresponding value of j, giving Jill's age. Bill is one year older than that.

So if J = B, then b = j + 1. If J < B, then J = B -1 and j = 0 and b = 0.

Hope this helps,
Stephen La Rocque.

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.