21Since the number is divisible by 5 and 9, it must also be divisible by 45. We also know it to be a 3-digit number. Ignore the fact that we could list out all such numbers - there should be about 20 of them.
ABC is our number. 100A + 10B + C = N, and N = 45M
Because of the progression, we have B = A + d, C = A + 2d, so that A + C = 2B
100A + 10B + (2B-A) = N
99A + 12B = 45M
Now, we are down to two variables A and B.
33A + 4B = 15M
33 and 15 are multiples of 3, so that B must be a multiple of 3. B = 3F
33A + 12F = 15M
11A + 4F = 5M
Since 3F is a digit, F = 0, 1, 2, or 3.
F = 0, B = 0.
11A = 5M
But then A = 5, so M = 11. This makes the original number 45*11 = 495. No good.
F = 1.
11A + 4 = 5M
A = 1, M = 3 is one way. A = 6, M = 14 is another. Note that 6-1 = 5 and 14 - 3 = 11. Thus, once you find one solution, you found them all.
3 * 45 = 135, which is a solution. 14 * 45 = 630, which is good.
F = 2.
11A + 8 = 5M
A = 2, M = 6 (6 * 45 = 270, which is not good.) or A = 7, M = 17 (765, which is good)
F = 3.
11A + 12 = 5M
A = 3, M = 9, N = 405, which is not good.
A = 8, M = 20-> 900, which is not good.
Thus, 135, 630, 765
