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 \(f(999)\)?

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## Comments

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

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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.

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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).

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Is the answer 24?

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

Least possible value of

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

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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.

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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.

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