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

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

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TopNewestWe 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 · 2 years, 7 months ago

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– Siddhesh Naik · 2 years, 7 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.Log in to reply

Is the answer 24? – Siddhartha Srivastava · 2 years, 7 months ago

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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 · 2 years, 7 months ago

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– Pranshu Gaba · 2 years, 7 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.Log in to reply

– Siddhesh Naik · 2 years, 7 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.Log in to reply