Where is my solution to 2018 AIME I Problem 1 wrong?
a can be any integer belonging to [1,100], so I count the no.of unordered pairs of positive integers whose sum is less than or equal to 100.
i.e., the no.of unordered pairs (0,1),(0,2)...(0,100)(1,1),(1,2)..(50,50) which is equal to {(101•102)/2} -1 [-1 for excluding (0,0)] which is equal to 5150 which far exceeds 2600 which is the total number.
Where is my approach wrong?
Thank you.
Note that if we have 0 in the x coordinate then, we will have 101 terms that are possible in the y coordinate (let 0,0 work for now.) Then, for 1 in the x coordinate, have 99 terms that will work for the y coordinate (1-99). Repeat this process and add all the y coordinates possible to get the sum $101+99+...+1$ There are 51 terms in this sequence. Add 1 to every term to get $102+100+...+2$ This sum is equal to $104*51/2$ This is equal to $52*51$. We have to subtract 51 from this expression to get $51^2$ and then subtract 1 from $51^2$ because we cannot have the case (0,0). This means we have $51^2-1$ Which is 2600. Therefore, the answer is 600.