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 discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
When posting on Brilliant:
Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
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
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.
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).
I don't think f(x)=xn 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.
f(x)=xn 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.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
We can write f(999)=(f(3))3⋅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)=⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧13723px=1x=2x=3x=37x=p,p is a prime=2,3,37
This gives f(999)=8×3=24.
Log in to reply
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
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).
Log in to reply
f(x)=xn (n is odd number)
Least possible value of
f(999)=0,when n →−∞
Log in to reply
I don't think f(x)=xn 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
f(x)=xn 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.
Is the answer 24?