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# Pre-RMO 2014/18

Let $$f$$ be a one-to-one function from the set of natural numbers to itself such that $$f(mn) = f(m)f(n)$$ for all natural numbers $$m$$ and $$n$$. What is the least possible value of $$f(999)$$?

This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
2 years, 5 months ago

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We can write $$f(999) = (f(3))^3 \cdot f(37)$$

To minimize this expression we can define the function $$f$$ as follows:

If $$x$$ is composite, it must be prime factorized, as $$f(999)$$ is factorized above. Then:

$$f(x) = \begin{cases} 1 & x = 1 \\ 37 & x = 2 \\ 2 & x = 3 \\ 3 & x = 37 \\ p & x = p, p \text{ is a prime} \neq 2, 3, 37\end{cases}$$

This gives $$f(999) = 8 \times 3 = \boxed{24}$$. · 2 years, 5 months ago

Good. But you need to justify that this function is one-one. This is not required in the exam, but still as a matter of learning, one should leave no loop hole. · 2 years, 5 months ago

Is the answer 24? · 2 years, 5 months ago

$$f(x) = x^{n}$$ (n is odd number)

Least possible value of

$$f(999) = 0, \text{when n }\to -\infty$$ · 2 years, 5 months ago

I don't think $$f(x) = x^n$$ is a one-to-one function, as not all natural numbers are included in the range of $$f(x)$$. Also, if $$n$$ is negative, $$f(x)$$ will be a fraction, but we want it to be a natural number. · 2 years, 5 months ago

$$f(x)=x^n$$ is indeed a one-one function if $$n$$ is odd, but it is not mapped on set of Natural numbers if $$n$$ is negative, so this function does not satisfy the given conditions. · 2 years, 5 months ago