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).
I wonder if this is a number-based-on-the-year problem that was adapted from a different year where there is a definite answer. For 2025 it doesn't quite work.
Based on the problem
f(1)=1
f(2)≤2 (rule 2)
f(3)≤3 (rule 2)
f(4)≤4 (rule 2)
f(5)≤5 (rule 2)
however, f(6)≤6 (rule 2) but also f(6)≥6 (rule 1) so
f(6)=6
f(7)≤6 (rule 2) but also f(7)≥f(2)+5 (rule 1) but since f(2) can be as small as we want, so can f(7).
similar with f(8), f(9), f(10)
f(11) is constrained by f(6) so f(11)=11
We know for sure that f(5n+1)=5n+1, but there is not enough information for other values.
f(2021)=2021
but the best we can do is f(2025)≤2025.
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Posted by Jer
on 2025-03-18 13:08:58 |
Unsolvable
