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

#Algebra #Pre-RMO

$</code> ... <code>$ $< / c o d e > . . . < c o d e >$ ."> Easy Math Editor

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Sort by:

Top Newest

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, 10 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.

Siddhesh Naik - 4 years, 10 months ago

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 . - 1 year, 2 months ago

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

Least possible value of

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

Krishna Sharma - 4 years, 10 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.

Pranshu Gaba - 4 years, 10 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.

Siddhesh Naik - 4 years, 10 months ago

Is the answer 24?

Siddhartha Srivastava - 4 years, 10 months ago

Excel in math and science

Master concepts by solving fun, challenging problems.

It's hard to learn from lectures and videos

Learn more effectively through short, interactive explorations.

Used and loved by over 8 million people

Learn from a vibrant community of students and enthusiasts, including olympiad champions, researchers, and professionals.

Pre-RMO 2014/18

Comments

Learn from a vibrant community of students and enthusiasts,
including olympiad champions, researchers, and professionals.