Question 388635
It suffices to choose three (not necessarily distinct) digits between 0 and 9. I'll do it case by case:


Case 0: a, a, a

{0,0,0}, {1,1,1}, ..., {9,9,9} (9 cases)


Case 1: a, a+1, a+2

{0,1,2}, {1,2,3},...,{7,8,9} (8 cases)


Case 2: a, a+2, a+4

{0,2,4}, {1,3,5}, ..., {5,7,9} (6 cases)


Case 3: a, a+3, a+6

{0,3,6},...,{3,6,9} (4 cases)


Case 4: a, a+4, a+8

{0,4,8}, {1,5,9} (2 cases)


For cases 1 through 4, each set that does not have a zero can produce two different numbers (e.g. {4,5,6} produces 456, 654) and each set that has a zero can only produce one number (to account for leading zeroes). There are 16 sets with no zeroes and 4 sets with a zero, so the number of 3-digit numbers in cases 1-4 is 16*2 + 4 = 36.


For case 0, where all the digits are equal, there are nine set {1,1,1}...{9,9,9}. Each set implies one more number, so nine numbers. Therefore the total number of 3-digit numbers is 36+9 = 45.