+586 How many 4-digit numbers have the last digit equal to the sum of the first two digits?
Beginning with 1000, there are 10+9+8+7+6+5+4+3+2+1 = 55 such numbers.
Each additional 1000 [2000, 3000, 4000.....etc] has exactly the same number. Hence:
55 x 9 = 495 four-digit numbers.
The above answer is accurate for the LAST 3 digits from the RIGHT. I think the question says the LAST digit from the RIGHT = Sum FIRST TWO digits from the LEFT. If that is accurate, then the number would be:
1000s = 90, 2000s = 80, 3000s = 70, 4000s = 60, 5000s = 50, 6000s = 40, 7000s = 30, 8000s = 20, 9000s =10.
So, the total =90+80+70+60+50+40+30+20+10 = 450 such numbers.
+118608 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 
