Since f(3) = f(2+1) >= f(2) + f(1) = f(1) and 0 = f(2) = f(1+1) >= 2f(1) so that f(1) = 0; f(3) = 1 then f(4) = f(3+1) = 1 Or 2 f(5) = f(3+2) = 1 Or 2 f(6) = f(3+3) >= 2f(3) = 2 so that f(6) = 2 Or 3 f(9) = f(6+3) = f(6) + f(3) + 0 Or 1 so that f(9) = 3, 4 Or 5 f(12) = f(9+3), f(12) = 4, 5, 6, Or 7 min(f(k)) = [k/3] since f(9999) = 3333 & min(f(9999)) = 3333; min(f(9999)) = f(9999); so that f(9999-3) = min(f(9999-3)) = 3332 ... so that f(3x) = x f(1980) = 1980/3 = 660 f(1982) = f(1980 + 2) = f(1980) + f(2) +0 Or 1= 660 Or 661

Actually I think that we can say specifically that f(1982)=660. By letting $m=n$ we have that $f(2m) \geq 2f(m)$. Suppose that $n<800$ (where 800 is rather arbitary, with $4 \times 800 <3333$). Also suppose for contradiction that $f(3n+2)=n+1$. Then $f(6n+4) \geq 2n+2$ and $f(12n+8) \geq 4n+4$. However $f(12n+8) \leq f(12n+9) = f(3(4n+3)) = 4n+3$ (as $4n+3 < 3333$). Contradiction, as $4n+4 > 4n+3$. Hence for all $n<800$, $f(3n+2) \neq n+1$ and hence it equals $n$. Specifically, this is true for $n=660$ and $f(1982)=660$.

How does $f(3(4n+3)) = 4n+3?$

From Kalimdor's post- Letting $m=3k$ and $n=3$ we have $f(3(k+1)) \geq f(3k)+1$. As $f(3)=1$ and $f(9999)=3333$, we must have $f(3k)=k$ for all $1 \leq k \leq 3333$.

I still don't understand... just because it works for two values, how does it work for the rest?

Suppose (for contradiction) that for some $m$, $1<m<3333$, $f(3m)>m$. Then $f(3(m+1))>m+1$ (by my previous post) and by induction $f(3(m+r))>m+r$. But for some $r$ this gives $f(3 \times 3333)>3333$, which is false. Now suppose that for some such $m$, $f(3m)<m$. By my previous post, $f(3(m-1))<m-1$, and by induction $f(3(m-r))<m-r$. But for $r=m-1$ this gives $f(3)<1$, which is also false. Hence $f(3m)=m$.

Clearly $f(1) \ge 1 \Rightarrow f(m+1) \ge f(m)+f(1) \ge f(m)+1$ so $f(9999) \ge 9999$.Contradiction!So $f(1)=0$.This forces $f(3)=1$.Hence $f(3k+3) \ge f(3k)+f(3)>f(3k)$ so the inequality $f(3)<f(6)<\cdots<f(9999)=3333$ forces $f(3k)=k \forall k \le 3333$.Now $f(3k+2) \ge k+1 \Rightarrow f(6k+4) \ge 2k+2 \Rightarrow f(12k+8) \ge 4k+4 \le f(12k+9)=4k+3$(Note:This is valid for $12k+9 \le 9999$ or $3k+2 \le 2499$).Contradiction!Hence the non-decreasing nature of $f$ gives $f(3k+1)=k$.Hence $f(n)=\lfloor\frac{n}{3}\rfloor \forall 1\le n \le 2499$. So $f(1982)=\lfloor\frac{1982}{3}\rfloor=660$.

First observe that $$3333=f(9999)\geq f(9996)+f(3)\geq f(9993)+2f(3)\geq \cdots\geq 3333f(3)$$Since $f(3)$ is a positive integer, we need $f(3)=1$. Next, observe that \begin{align*} 3333=f(9999)\geq 5f(1980)+33f(3)=5f(1980)+33\quad\Longrightarrow\quad f(1980)\leq 660 \end{align*}On the other hand, $f(1980)\geq 660f(3)=660$, so combine the two inequalities we obtain $f(1980)=660$. Finally, write $$f(1982)=f(1980)+f(2)+0\vee 1=660\vee 661$$Suppose that $f(1982)=661$, then \begin{align*} 3333=f(9999)\geq 5f(1982)+29f(3)=3305+29=3334 \end{align*}a contradiction. Hence we conclude that $f(1982)=660$.

We show that $f(n) = [n/3]$ for $n <= 9999$, where [ ] denotes the integral part. We show first that $f(3) = 1$. $f(1)$ must be $0$, otherwise $f(2) - f(1) - f(1)$ would be negative. Hence $f(3) = f(2) + f(1) + 0$ or $1$ = $0$ or $1$. But we are told $f(3) > 0$, so $f(3) = 1$. It follows by induction that $f(3n) >= n$. For $f(3n+3) = f(3) + f(3n)$ + $0$ or $1 = f(3n) + 1$ or $2$. Moreover if we ever get $f(3n) > n$, then the same argument shows that $f(3m) > m$ for all $m > n$. But $f(3.3333) = 3333$, so $f(3n) = n$ for all $n <= 3333$. Now $f(3n+1) = f(3n) + f(1) + 0$ or $1$ = $n$ or $n + 1$. But $3n+1 = f(9n+3) >= f(6n+2) + f(3n+1) >= 3f(3n+1)$, so $f(3n+1) < n+1$. Hence $f(3n+1) = n.$ Similarly, $4f(3n+2) = n$. In particular $f(1982) = 660$.

Proof: Lemma 1: $f(mn)\geq nf(m)$ Let, $P(m,m)$ be assertion. $$f(2m) \geq 2f(m)$$Similarly,we can induct to get $f(nm)\geq nf(m)$. $$f(3m) \geq f(2m)+f(m) \geq 3f(m)\implies f(nm)\geq \underbrace{f(m(n-1))+f(m)\geq .....}_{\text{n times}}\geq nf(m)$$Lemma proved. Then we see that, $$f(9999)\geq 3333f(3) \implies f(3)=1:f(3)>0$$Then, $$f(9999) \geq 5f(1980)+33f(3) \implies 3300 \geq 5f(1980) \implies f(1980) \leq 660$$$$f(1980) \geq 660f(3) \implies f(1980)=660$$Then we can easily get,by assertion $P(1980,2)$ $$f(1982)=660 \space or \space f(1980)=661$$Hence, $\boxed{f(1980)=660,661}$.And, we are done. $\blacksquare$ (Also, this post looks alike the previous one, but I found it genuinely myself.)

Hey everyone, I am facing some problems while solving this problem. Kindly check my approach : We are given that $f:\mathbb{N}\to\mathbb{N}_0$ $P(1,1)\rightarrow -2f(1)=0 or 1$ Due to the codomain of $f$, $f(1)=0$ $P(m,1)\rightarrow f(m+1)-f(m)=0 or 1$ If $f(m+1)=f(m)$ then, for $m=2$, $f(3)=0$ but this contradicts $f(3)>0$. Thus, $f(m+1)=f(m)+1$ Then, by induction, $f(n)=n-2\forall n>1$. But this contradicts $f(9999)=3333$. So how do we resolve this problem, as I exhausted all possibilities?

First note that $f(4) = 0 or 1$ but also $f(4) = f(3) + f(1) or f(3) + f(1) + 1 $, since $f(3) > 0$, it follows that $f(4) = 1$ and $f(1) = 0$\par Now we will prove that $f(3k) = k$ for all integers $k$,ergo $f(m+n)-f(m)-f(n)=0$ if $3\mid m+n$,notice that this holds with $f(9999) =f(3333\cdot3) = 3333$\par Suppose this statement holds until some $g < 9999$, then we would have that $f(3g)=g+1$, thus summing one to all multiples to 3 greater than $g$ and $f(9999) = 3334$, a contradiction\par \begin{align*} f(1982) &= f(1982) \\ f(1984) &= f(1982) or f(1982) + 1 \\ f(1986) &= f(1984) or f(1984) + 1 = 662 \\ f(1983) &= 661 = f(1982) + 1 or f(1982). \end{align*}Thus $\boxed{f(1982) = 661 or 660}$

Math.Is.Beautiful wrote: Hey everyone, I am facing some problems while solving this problem. Kindly check my approach : We are given that $f:\mathbb{N}\to\mathbb{N}_0$ $P(1,1)\rightarrow -2f(1)=0 or 1$ Due to the codomain of $f$, $f(1)=0$ $P(m,1)\rightarrow f(m+1)-f(m)=0 or 1$ If $f(m+1)=f(m)$ then, for $m=2$, $f(3)=0$ but this contradicts $f(3)>0$. Thus, $f(m+1)=f(m)+1$ Then, by induction, $f(n)=n-2\forall n>1$. But this contradicts $f(9999)=3333$. So how do we resolve this problem, as I exhausted all possibilities? “If $f(m+1)=f(m)$ then, for $m=2$, $f(3)=0$ but this contradicts $f(3)>0$. Thus, $f(m+1)=f(m)+1$.” That makes no sense. Since when $m=2$, we have $f(m+1)=f(m)+1$, while $m=3$, we have $f(m+1)=f(m)$.

The answer is $660$ only (which means some of the above solutions are wrong), which can be achieved by $f(x) \equiv \lfloor \frac{x}{3} \rfloor$. Let $P(m,n)$ denote the assertion that $$f(m+n)-f(m)-f(n) \in \{0,1\}.$$By $P(1,1)$, we have $f(1)=0$. Assume FTSOC that $f(3) \geq 2$. By $P(2,3)$, we have $P(5) \geq 2$, by $P(5,3)$, we have $P(8) \geq 4$, and so on. We eventually get that $P(9998) \geq 6664$, and by $P(9998,1)$, we get $P(9999) \geq 6664$, a contradiction. Thus, $f(3)=1$. By repeating $P(m,3)$, we get $f(x) \geq \lfloor \frac{x}{3} \rfloor$ for all positive integers $x$. Thus, $f(1982) \geq 660$. Assume FTSOC that $f(1982) \geq 661$. Then, we use $P(1982,1982)$ to get $f(3964) \geq 1322$, $P(3964,1982)$ to get $f(5946) \geq 1983$, and $P(4053,5946)$ to get $f(9999) \geq 3334$, a contradiction. Therefore, $f(1982)=660$.

This is easy.

There are 3 solutions that have got 2 values, but it seems their reasoning is faulty.

We claim that $f(1982) = \boxed{660}$. The function $f(x)= \left\lfloor \frac{x}{3}\right\rfloor$ fits. Let $P(m,n)$ denote the assertion that \[f(m+n) - f(m) - f(n)\in \{0,1\}\] Notice that $f(m+2) - f(m)\in \{0,1\}$. Claim: $f(1) = 0$ and $f(3) = 1$. Proof: $P(1,1)$ gives that $0 = f(2)\in \{2f(1), 2f(1) + 1\}$. Here we can see that $f(1) = 0$. We have \[f(m+6) - f(m) = (f(m+6) - f(m+4)) + (f(m+4) - f(m+2)) + (f(m+2) - f(m))\in \{0,1,2,3\}\] We also have \[f(m+6)- f(m) = (f(m+6) - f(m+3)) + (f(m+3) - f(m))\in \{2f(3), 2f(3) +1, 2f(3) + 2\}\] If $f(3)>1$, then $2f(3)>3$, contradiction. So $f(3) = 1$. $\square$ $P(m,1): f(m+1) - f(m)\in \{0,1\}$. $P(m,3): f(m+3) - f(m)\in \{1,2\}$. Note that $f(m+3) \ge f(m) + 1$, so $f(3m)\ge m$ for each $m$ because $f(3)= 1$. Claim: If $f(3x)>x$, then $f(3(x+1))>(x+1)$. Proof: If $f(3x)>x$, we have $f(3(x+1)) \ge f(3x) + 1 > x+1$, as desired. $\square$ If $f(3x)>x$ for some $x$, then $x>3333$, so $f(3x) = x\forall x\le 3333$. This gives $f(1980) = 660$, so $f(1982)\in \{660,661\}$. Claim: $f(1982) = 660$. Proof: Suppose $f(1982) = 661$. $P(1982,1982)$ gives $f(1982\cdot 2 ) \ge 1322$. $P(1982\cdot 2, 1982): 1982 = f(1982\cdot 3) \ge f(1982\cdot 2) + f(1982) \ge 1983$, contradiction. $\square$ Therefore $f(1982) = 660$.

The answer is $f(1982)=660$. We let $P(m,n)$ denote the assertion of our FE. Claim: $f(3)=1$ and $f(3n)=n$ for all integers $n\leq 3333$. Proof: Using induction one can show that for any $n$, $f(3n)=nf(3)+k$ for some $k\in \{0,1,2,...,n-1\}$. Plugging $n=3333$ into this gives $3333=f(9999)=3333f(3)+k$. Since $f(3)>0$, $f(3)=1$ must hold, but then $k=0$ and $f(3n)=nf(3)=n$ is guaranteed for all $n\leq 3333$. Claim: $f(1982)=660$. Proof: Note that $f(1980)=f(3\times 660)=660$, and so $f(1980)+f(2)=660$. From $P(1980, 2): f(1982)=660$ or $661$. Assume FTSOC $f(1982)=661$, then $1982=f(1982\times 3)\geq 3f(1982)=3\times 661=1983$. Which is the contradiction. Thus $f(1982)=660$.

Note that \[f(9999)\ge f(3)\cdot 3333\ge 3333\]with equality case $f(3k)=k$ for all positive integers $k\le 3333$. We know that \[1982-f(1982)\cdot 3=f(1982\cdot 3)-f(1982)\cdot 3\in\{0,1,2\}.\]Taking mod $3$, it must be $2$, so $f(1982)=\boxed{660}$.

Can anyone please help me to identify my mistake in the following idea . Since f(2) = 0 , it's easy to show that f(1)=0 and f(3)=1 . P(m,n) : f(m+n) - f(m) - f(n) = 0,1 P(m,1) : f(m+1) - f(m) = (0,1)+ f(1) = 0,1 ....(i) P(m,3) : f(m+3) - f(m) = (0,1)+ f(3) = 1,2 ....(ii) P(m+1,2) : f(m+3) - f(m+1) = (0,1)+ f(2) = 0,1 ..(iii) By adding (i) and (iii) and comparing that with (ii) we conclude that Either, f(m+3) - f(m+1) = 1 or f(m+1)- f(m) = 1 or both . Which gives us , f(x+2)=f(x)+1 or f(x+1)=f(x)+1 but in this case the answers don't match.

Let $P(m, n)$ denote the given assertion. $P(1, 1)$ gives $f(1)=0$. $P(m, 1)$ gives $f(m+1)-f(m)\in\{0,1\}$. $P(m, 2)$ gives $f(m+2)-f(m)\in \{0,1\}$. $P(2, 1)$ and $f(3)>0$ gives $f(3)=1$. $P(m, 3)$ gives $f(m+3)-f(m)\in \{1, 2\}$. Notice that $$f(9999)=(f(9999)-f(9996))+\dots+(f(6)-f(3))+f(3)\ge 1\cdot 3332+1=3333$$with equality only when $f(m+3)-f(m)=1$ for all $3|m$. This means that $f(3k)=k$ for integers $k>0$. Since $f(1980)=660$, $f(1982)=660$ or $f(1982)=661$. If $f(1982)=661$, then $P(1982, 1982)$ gives $f(1982\cdot2)\ge 1322$ but $P(3964, 1982)$ gives $f(1982\cdot3)\ge 1983$ which is false since $f(1982\cdot3)=1982$. Therefore, $f(1982)=\boxed{660}$. Plugging $f(x)=\lfloor \frac{x}{3}\rfloor$ shows that this works.

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