MathLabs

Problem 1

The function f(n)f(n) is defined on the positive integers and takes non-negative integer values. f(2)=0f(2)=0, f(3)>0f(3)>0, f(9999)=3333f(9999)=3333, and for all m,nm,n: f(m+n)−f(m)−f(n)=0 or 1.f(m+n)-f(m)-f(n)=0 \text{ or } 1. Determine f(1982)f(1982).
Step 6 of 7: Tripling superadditivity rules out 661661
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In plain words

The same squeeze that nailed down f(3k)f(3k) for small kk reappears here to forbid the larger of the two remaining candidates, by comparing it against a tripled index whose value we already know exactly.

f(3n)=f(n+n+n)≥f(n)+f(2n)≥f(n)+f(n)+f(n)=3f(n)f(3n)=f(n+n+n)\ge f(n)+f(2n)\ge f(n)+f(n)+f(n)=3f(n)
Detailed analysis

Applying superadditivity twice gives f(3n)≥3f(n)f(3n)\ge3f(n) for every nn. Take n=1982n=1982: if f(1982)=661f(1982)=661, this would force f(5946)=f(3⋅1982)≥3⋅661=1983f(5946)=f(3\cdot1982)\ge3\cdot661=1983. But 1982≤33331982\le3333, so Step 3 gives the exact value f(5946)=f(3⋅1982)=1982f(5946)=f(3\cdot1982)=1982, contradicting 1982≥19831982\ge1983.