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
4 years 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 \(f(999) = 8 \times 3 = \boxed{24}\).

Pranshu Gaba - 4 years ago

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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.

Siddhesh Naik - 3 years, 11 months ago

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It is clearly justified, isn't it. For 1,2,3,37, it is visible that it is one -one. For all numbers other than 2, 3, 37 it is one-one because if the number is prime then it maps the prime number to itself hence one-one, and if it is composite then it can be written as a product of its primes which will yield product of one-one functions which is one-one itself.(as the function gives unique primes).

Geeta . - 4 months ago

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@Geeta . thanks Geeta

Bhaskar Pandey - 2 months ago

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Is the answer 24?

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

Least possible value of

\(f(999) = 0, \text{when n }\to -\infty\)

Krishna Sharma - 4 years ago

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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.

Pranshu Gaba - 4 years ago

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\(f(x)=x^n\) 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.

Siddhesh Naik - 3 years, 11 months ago

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