How many 4-digit numbers have the last digit equal to the sum of the first two digits?

The first two can add to 1,2,3,4,5,6,7,8,9

First I am just looking at combinations for the first 2 digits.

01 to 09 9 ways ** This line was wrong, there is only 9 10,20,30,40,50,60,70,80,90**

11 to 18 8+7 ways 11,12,13,14,15,16,17,18, 21,31,41,51,61,71,81

22 to 27 6+5 ways 22,23,24,25,26,27, 32,42,52,62,72,

33 to 36 4+3 ways 33,34,35,36, 43,53,63,

44 to 45 2+1ways 44,45, 54

**9**+15+11+7+3 = **45**

45 combinations for the first two digits

10 combinations for the 3rd digit

The last digit is preset

**45***10*1 = **450** possible numbers

OK Mine is fixed now, i just had a stupid error. We seem to be in agreement that the answer is 450.

Thanks guest and Rom