Combinatorics #6

Suppose $S=a_{1},a_{2},\cdots,a_{15}$ is a set of $15$ distinct positive integers chosen from $2,3,\cdots,2014$ such that every two of them are coprime. Prove that $S$ contains a prime number. (Note: Two positive integers $m,n$ are coprime if their only common factor is $1$.

#Combinatorics

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:

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
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 `$` ... `$` or `\[` ... `\]` 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

Damn my first solution was lost in the preview...had to rewrite.

Note that two numbers are coprime if they have no alike prime factors. If we assume $S$ are all composite, then we can represent each $a_i=p_i*q_i*k$ where all $p_i,q_i$ are distinct primes and $p_i\le q_i$ Hence $2014\ge a_i>p_i^2\Rightarrow p_i\le 43$ . Since $43$ is the 14th prime, but there are $15$ numbers, hence there must exists $1\le r<s\le15$ such that $p_r=p_s$ which is a contradiction.

Xuming Liang - 6 years, 11 months ago

You forgot that $p_i=q_i$ is a possibility.

Daniel Liu - 6 years, 11 months ago

I did specify that $p_i\le q_i$ . Note that the intent for doing all this is to show that the smallest prime factor of each $a_i$ must be under $43$ .

Xuming Liang - 6 years, 11 months ago

You can simply use logic to solve this. Assuming $S$ has no primes, the maximum number of distinct numbers it can contain is $14$ . Logically thinking, for $S$ to contain as many numbers of possible, we would square prime numbers, thus maximising the number of prime factors as the prime factors of a prime squared would be itself only. We are given that the 15th prime is $47$ and $47 \times 47$ gives us $2209$ which is bigger than $2014$ .

(Sorry if some of my reasoning is unclear)

Charlton Teo - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` 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}`	$\boxed{123}$