Indulul's function

$f(x+1) - f(x) \leq 1$

$f(x+2) - f(x+1) \leq 1$

$f(x+3) - f(x+2) \leq 1$

$f(x+4) - f(x+3) \leq 1$

$f(x+5) - f(x+4) \leq 1$

Add them.

$f(x+5) - f(x) \leq 5$

$\Rightarrow f(x+5) \leq f(x) + 5$

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So, $f(x+5) = f(x) + 5$

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$f(x+5) - f(x) = 5$ and $f(x+1) - f(x) \leq 1$

Hence $f(x+1) - f(x) = 1$

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Now $f(x+1) = f(x) + 1$

$f(x+2) = f(x+1) + 1 = f(x) + 2$

.......

.......

.......

$f(x+2012) = f(x) + 2012$

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$f(2013) = f(1+ 2012) = f(1) + 2012 = 101 + 2012 = 2113$

Last three digits $\boxed{113}$

Well I don't have exact proof but I consider it on a pattern.

• f(x + 5) ≥ f(x) + 5
• f(-4 + 5) ≥ f(-4) + 5
• 101 ≥ f(-4) + 5
• f(-4) ≤ 96

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• f(x + 1) ≤ f(x) + 1
• f(0 + 1) ≤ f(0) + 1
• f(0) ≥ 100

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• f(x + 5) ≥ f(x) + 5
• f(1 + 5) ≥ f(1) + 5
• f(6) ≥ 106

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• f(x +1) ≤ f(x) + 1
• f(1 + 1) ≤ f(1) + 1
• f(2) ≤ 102

Writing them all :

• f(-4) ≤ 96
• f(0) ≥ 100
• f(1) = 101
• f(2) ≤ 102
• f(6) ≥ 106

It just tells that f(x) = 100 + x

Thus : f(2013) = 100 + 2013 - f(2013) = 2113

$\boxed{113}$

First, we evaluate the minimum value of $f(2013)$ by putting $x=2008$ into the first inequality and successively reducing the value of $x$ .

$\begin{aligned} f(2013) &\ge f(2008) + 5 \\ &\ge f(2003) + 5 + 5 \\ & . \\ &. \\ &. \\ &\ge f(3) + 402(5) \\ &\ge f(3) + 2010\end{aligned}$

Now, we find the maximum value of $f(2013)$ by putting $x = 2012$ into the second inequality and successively reducing the value of $x$ .

(\begin{align } f(2013) &\le f(2012) + 1 \ &\le f(2011) + 1 + 1 \ &.\&.\&.\&\le f(1) + 2012(1)\end{align })

This is pretty silly. $$f(x)+5=f(x+5)\le f(x+4) + 1 \le \dots \le f(x) + 5$$ so equality must hold and $$f(x+1) = f(x) + 1$$ for all $x.$ Thus, $f(2013) = f(0) + 2013 = 2113$ so the answer is $\boxed {113}.$ (Ok I know $f(0)$ doesn't actually exist but it doesn't matter and makes the computation more obvious)

whatever be the value of x , the value of f(x) is just 100 more than that , so when x=2013, f(x)= 2013 +100= 2113 and the last three digits are 113..

By logic we get 2nd statement as $f\left( x+5 \right) \le f\left( x \right) +5$ .

But as we know 1st statement is $f\left( x+5 \right) \le f\left( x \right) +5$ .

Therefore $f\left( x+5 \right) =f\left( x \right) +5$ .

When x=1 we get $f\left( 6 \right)=106$

Continuing this process we get general rule as

$f\left( n \right)=100+n$

Therefore

$f\left( 2013 \right)=2113$

Therefore last three digits are 113.

I will prove that $f(x+1)=f(x)+1$ or all $x$ , and from this we easily get that, since $f(1)=101$ , then $f(2013)=f(1+2012)=f(1)+2012=2113$ and so the answer would be $113$ .

Now, suppose for some x, $f(x+1) ≠ f(x)+1$ . Then, by the second inequality, $f(x+1) < f(x)+1$ . Therefore

$f(x) + 5 \le f(x+5) \le f(x+4)+1 \le ... \le f(x+1)+4 < f(x)+5$

a contradiction!

We're done. Answer: $\boxed{113}$

f(1+5)>=f(1)+5

f(1+1)<=f(1)+1

f(2+1)<=f(2)+1

f(3+1)<=f(3)+1

f(4+1)<=f(4)+1

f(5+1)<=f(5)+1

==>f(6)<=f(1)+5>=f(1)+5

==>f(6)=f(1)+5

==>f(x)=f(1)+x-1

f (1+1) < or = 101+1 >>> f (2) < or = 102

f (1+5) > or = 101+5 >>> f (6) > or = 106 ..................................................................(i)

f (2+1) < or = 102+1 >>> f (3) < or = 103

f (3+1) < or = 103+1 >>> f (4) < or = 104

f (4+1) < or = 104+1 >>> f (5) < or = 105

f (5+1) < or = 105+1 >>> f (6) < or = 106............................................................(ii) But, from (i) f (6) > or = 106

So that, f(6)= 106 so, f(x)= 100+x

so, f(2013)= 100+2013=2113

see that $f(x+1)\leq f(x)+1$ then now we know that $f(x+5)\leq f(x+4)+1\leq...\leq f(x)+5$ but we know that $f(x+5)\geq f(x)+5$ so f(x+5)=f(x) + 5 and this condition can holds if and only if f(x+1)=f(x)+1 so f(2013) = f(1) + 2012 = 101 + 2012 = 2113 so the last three digits of f(2013) is 113

The following info is given:

$\begin{aligned} f(1)=101\\ f(x+5)\ge f(x)+5\\ f(x+1)\le f(x)+1 \end{aligned}$

From that, we conclude:

$\begin{aligned} f(2013)=f(2008+5)\ge f(2008) + 5\\ f(2013)=f(2012+1)\le f(2012) + 1 \end{aligned}$

And since $f(2012)\le f(2011) +1 \le (f(2010) + 1) +1 \le \cdots \le f(2008) + 5$ ,

$\begin{aligned} f(2013) \ge f(2008) + 5 \\ f(2013) \le f(2008) +5 \end{aligned}$

Because we have to satisfy both of the statements, we find:

$f(2013) = f(2008) + 5$

Basically, we see that $f(A) + B = f(A+B)$ . Therefore:

$\begin{aligned} f(2013)=f(1)+2012=101+2012=\underline{\mathbf{2113}} \end{aligned}$

We end up with $\displaystyle\mathbf{\underline{2113}}$ , the last 3 digits of which are $\mathbf{\underline{\boxed{113}}}$ .

P.S.: How do you align stuff to the center? begin{align} doesn't apparently center text on solutions.

Let's see the pattern below:

$f(x)+5<=f(x+5)<=f(x+4)+1<=f(x+3)+2<=f(x+2)+3<=f(x+1)+4<=f(x)+5$

Hence, equality holds for all relations above. Thus, $f(x+1)=f(x)+1$ . And since $f(1)=101$ , then $f(2)=102$ , $f(3)=103$ , and so on till $f(2013)=2113$

• $f(x+5)\le f(x+4)+1\le f(x+3)+2\le f(x+2)+3\le f(x+1)+4\le f(x)+5$
• $f(x+5)\ge f(x)+5$
• $\therefore f(x)+5\ge f(x+5)\ge f(x)+5$
• $\therefore f(x)+5=f(x+5)\hspace{50pt}\mbox{(Squeeze theorem)}$

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• $\because f(x+1)\le f(x)+1$
• $\therefore f(x+1)-f(x)\le1$

$\begin{array}{cl} &f(x+1)-f(x)\\ =&f(x-4)+5-f(x)\\ =&-[f(x)-f(x-4)]+5\\ =&-[f(x)-f(x-1)+f(x-1)-f(x-2)+f(x-2)-f(x-3)+f(x-3)-f(x-4)]+5\\ \ge&-(1+1+1+1)+5\\ =&1\end{array}$

• $\therefore 1\le f(x+1)-f(x)\le 1$
• $\therefore f(x+1)-f(x)=1\hspace{50pt}\mbox{(Squeeze theorem)}$
• $\therefore f(x+1)=f(x)+1$

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• f(2)=f(1)+1=102
• f(3)=f(2)+2=103
• f(n)=100+n
• f(2013)=2113

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Therefore, the last 3 digits of f(2013) is $\boxed{113}$ .

We will prove by induction that $f(n) = 100 + n$ .

Base case:

We see that, as the problem said, $f(1) = 101 = 100 +1$ . Therefore, the base case is valid.

Then let´s assume that this is true for every integer from $1$ to $n$ , and we will try to prove the case $n + 1$ .

For this, we first see that $f(n + 1) \leq f(n) + 1 \leq (100 + n) + 1 \leq 100 + ( n + 1 )$ . $\Rightarrow f(n + 1) \leq 100 + (n + 1) \ldots (i)$

Then, working a little bit with the conditions of the problem: $[f(n + 1) = f((n - 4) + 5)] \geq [f(n - 4) + 5 = (100 + n - 4) + 5 = 100 + (n + 1)]$ $\Rightarrow f(n+1) \geq 100 + (n + 1) \ldots (ii)$

Therefore, thanks to $(i)$ and $(ii)$ : $100 + (n + 1) \geq f(n + 1) \geq 100 + (n + 1)$ $\Rightarrow f(n + 1) = 100 + (n + 1)$

Therefore, it has been proven that $f(n) = 100 + n$ . $\Rightarrow f(2013) = 100 + 2013 = 2113$

Hence, the last three digits of $f(2013)$ are $\boxed{113}$ .

With exclusive-inductive method we get f(x) >= 100 + x and f(x) =< 100 + x, so f(x) = x + 100 ---> f(2013) = 2013 + 100 = 2113. So, the ANSWER is 113.

In this solution, the symbol < means "less than or equal to".

From f(x+1) < f(x) +1, by putting x = x+4, we have f(x+5) < f(x+4) + 1. Since f(x) + 5 < f(x+5), f(x) + 5 < f(x+4) + 1, i.e., f(x) + 4 < f(x+4).

Continuing this process, we have f(x) + 3 < f(x+3) and f(x) + 2 < f(x+2). At last, f(x) + 1 < f(x+1).

With the condition f(x+1) < f(x) + 1, we conclude that f(x+1) = f(x) +1.

So, f(2013) = 2113. Hence, the answer.

This solution is NOT good. It's flawed, dry, raw, tasteless. This is NOT the MasterMath quality!!!

- Gordon Ramsay

I'm sorry that the solution is not good enough. I need anyone to fix this solution, or do this again. T_T

You can see that 2 functions $f(x+1)\leq f(x)+1$ and $f(x+5)\geq f(x)+5$ will give you the exact value (no ranges, just a number) for all numbers $5n+1, n \in I^{+}$ .

Ex: $f(1) = 101, f(6) = f(1) + 5 = 106, f(11) = 11, .....$

We can (nope nope nope nope nope nopidee nopidee nope) say that.....

$f(n) = n + 100, \forall n \in I^{+}$

Therefore, $f(2013) = 2013+100 = 2113.$ Answer $\boxed{113}$ . T_T

First; We suppose that there is an integer x such that f(x+1)<f(x)+1 So f(x)+5>f(x+1)+4>f(x+2)+3>...>f(x+5) So f(x+1)-f(x)=1 : (sum_{i=1}^2012) f(i+1)-f(i)=2012 So f(2013)=2012+101=2113