Let f be a function defined on all real numbers such that for all x, we have f(x + 5) ≥ f(x) + 5, and f(x + 1) ≤ f(x) + 1. If f(1) = 1, find f(2025).
Iterate f(x+1) <= f(x) + 1 onto itself four more times to create the compound inequality f(x+5) <= f(x+4) + 1 <= f(x+3) + 2 <= f(x+2) + 3 <= f(x+1) + 4 <= f(x) + 5.
Then the extreme left and right side form f(x+5) <= f(x) + 5. This together with the given f(x+5) >= f(x) + 5 means we must have f(x+5) = f(x) + 5.
Now substitute this result back into the first term of the compound inequality to get f(x) + 5 <= f(x+4) + 1 <= f(x+3) + 2 <= f(x+2) + 3 <= f(x+1) + 4 <= f(x) + 5.
The extreme left and right sides are equal so then all the intermediate terms must also be equal to that value. In particular f(x+1) + 4 = f(x) + 5. Subtract 4 from each side to leave f(x+1) = f(x) + 1.
Given f(1)=1 then apply this relation 2024 times to get f(2025) = 2025.
Solution
