Waste less time on Facebook — follow Brilliant.
×

Easy Peasy Prime Proof Problem

Given that \(p\) and \(q\) are primes such that \(q - p = 2\), prove that there doesn't exist a non-negative integer \(k\) such that

\[pq + 10^k\]

is also prime.


Hint:

If you want this hint, it is that you may prove the problem using divisibility rules.

Note by Sharky Kesa
3 years ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. 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"
MathAppears 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

Sort by:

Top Newest

I hope I can make myself expressible.

\(p\) and \(q\) are twin primes, that is, they differ by \(2\). Let \(r\) be the only positive integer between \(p\) and \(q\). Then \(prq\), being a product of \(3\) consecutive integers, is divisible by \(3\). Since \(p\) and \(q\) are primes, \(3\) divides \(r\), and hence, \(p=3m+2\) and \(q=3n+1\), where \(m\) and \(n\) are positive integers.

Their product, thus, is of the form \(3l+2\), where \(l\) is a positive integer. Lastly, \(10^k=(9+1)^k\) is of the form \(3o+1\), where \(o\) is a positive integer.

Adding them together, \(pq+10^k\) is of the form \(3l+2+3o+1\), which is of the form \(3z\), where \(z\) is a positive integer.

This is divisible by \(3\), thus this cannot be a prime number.

Satvik Golechha - 3 years ago

Log in to reply

nice one, but I think that \(p=3m-1\) and \(q=3m+1\) since you say they are twins

Trevor Arashiro - 3 years ago

Log in to reply

Note that this doesn't hold for \( p = 3, q = 5 \).

This case has to be dealt with separately. (obvious fix).

The error in the initial logic is making the wrong assumption that \(3 \mid pqr \rightarrow 3 \mid r \).

Calvin Lin Staff - 3 years ago

Log in to reply

The first part can also be proved using the fact that all primes \( \geq 5 \) are of the form \( 6k + 1 \) or \( 6k + 5 \), where \( k \) is a positive integer. Anyway, nice solution :)

Tan Li Xuan - 3 years ago

Log in to reply

Yeah! Thanks, I just wanted to use simple divisibility rules, the hint Sharky gave.

Satvik Golechha - 3 years ago

Log in to reply

Good one!!! @Satvik Golechha and @Sharky Kesa

Shivamani Patil - 3 years ago

Log in to reply

Thanks! Gave RMO?

Satvik Golechha - 3 years ago

Log in to reply

@Satvik Golechha No.Did you give?

Shivamani Patil - 3 years ago

Log in to reply

@Shivamani Patil Yeah!

Satvik Golechha - 3 years ago

Log in to reply

Big hint: Consider the equation modulo 3.

Jordi Bosch - 3 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...