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
You can limit the number of formulas that you need to memorize by knowing and understanding where the formulas come from.
There are still a lot that it is better just to memorize of course but I know a lot less formulas than most other people of my level.
Most times this does not go against me.
For instance, the distance and midpoint formulas look aweful but if you understand where they come from you do not need to know the formulas at all!
The midpoint formula is just (the average of xs, the average of ys) which is totally logical.
And the distance formula is just pythagoras's theorem.