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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, 7 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 \(f(999) = 8 \times 3 = \boxed{24}\). Pranshu Gaba · 2 years, 7 months ago

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@Pranshu Gaba 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 · 2 years, 7 months ago

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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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@Krishna Sharma 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 · 2 years, 7 months ago

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@Pranshu Gaba \(f(x)=x^n\) 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 · 2 years, 7 months ago

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