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# Understand the problem

Let N be the set of natural numbers. Suppose $$f : N \to N$$ is a function satisfying the following conditions:

(a) f(mn) = f(m)f(n)

(b) f(m) < f(n) if m < n

(c) f(2) = 2

What is the value of $$\displaystyle {\sum_{k=1}^{20} f(k)}$$

##### Source of the problem

PRMO (Pre Regional Math Olympiad) 2012, India. Problem 16

##### Topic
Functional Equation
3 out of 10
##### Suggested Book
Functional Equations by Venkatchala

Do you really need a hint? Try it first!

The first property is known as multiplicative property. The second one is known as monotonic.

Strategy 0

Find some values of f(x) for some x. This may yield a pattern.

__________

Notice that f(2) = 2 is given.

Can you find out f(1) (using the multiplicative property)?

Consider the following:

$$f(2) = f(1 \times 2 ) = f(1) \times f(2)$$

Hence we have $$2 = f(1) \times 2$$

Thus f(1) = 1

Can you find out f(4), f(8) and f(16) in a similar manner?

Clearly the following strategy works : $$f(4) = f(2\times 2) = f(2) \times f(20 = 2 \times 2 = 4$$

Similarly f(8) = 8 and f(16) = 16.

Can you find out f(3) and generalize a strategy to find f(n)?

We use the monotonic property.

f(2) < f(3) < f(4)

But that means 2 < f(3) < 4. There is exactly one integer between 2 and 4. That is 3. Hence f(3) = 3.

Now use induction to show f(n) = n for any natural number n. Here is a summary of the argument.

• Assume f(k) = k for all integers less than n> 4
• If k+1 is composite then f(k+1) = k+1 (write k+1 as a product of factors r and s both greater than 1. By inductive hypothesis, we know f(r) = r and f(s) = s, therefore f(k+1) = f(rs) = f(r)f(s) = rs = k+1
• If k+1 is prime then k+2 must be composite (as n > 4 there are no consecutive prime numbers). and k+2 can be written as product of numbers r and s greater than 1 and less than k+1. Thus using the above strategy f(k+2) = k+2
• Finally f(k+1) = k+1 by monotonic property.

Since f(n) = n, f(1) + … + f(20) = 1+ … + 20 = 210

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