UKMT Special (Problem 11)

Show that there are at least 33 primes that are less than 200200 that satisfy the system of equations:

p+2p + 2

p+6p + 6

p+8p + 8

p+12p + 12

and result in primes.

Also show that there is only one one prime that satisfy the system of equations:

q+2q + 2

q+6q + 6

q+8q + 8

q+12q + 12

q+14q + 14

and result in primes.

[UKMT BMO 20192019, Q11]

Note by Yajat Shamji
2 weeks, 6 days ago

No vote yet
1 vote

  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.

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 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

@Mahdi Raza

Yajat Shamji - 2 weeks, 6 days ago

Log in to reply

You have until next Saturday, 3:003:00pm!

Yajat Shamji - 2 weeks, 6 days ago

Log in to reply

I'm interpreting the question as "Find all primes pp such that the list of expressions above are all prime". However, there are no such primes that even satisfy the first set of requirements! For p2p \ge 2, we notice that from Pigeonhole Principle, there must be a number from the first 33 statements that is divisible by 33. One can easily check that the number is composite.

The only number that actually satisfies this restriction is p=1p = 1, and even still 11 is not a prime.

Since the weaker condition cannot be satisfied, neither can the stronger condition involving qq.

Elijah L - 2 weeks, 5 days ago

Log in to reply

There is a solution!

I solved it myself!

Yajat Shamji - 2 weeks, 4 days ago

Log in to reply

Oops!

I accidentally wrote the wrong system of equations!

Re-solve it, @Elijah L

Yajat Shamji - 2 weeks, 4 days ago

Log in to reply

Note that from the first set of equations, pp must be 00, 33, or 4(mod7)4 \pmod 7 for sufficiently large pp. Looking at it (mod5)\pmod 5, the only permissible residues mod 55 are 00 and 11. From the same logic, the number must also be 2(mod3)2 \pmod 3. Checking small cases leaves that p=5p = 5 is a solution.

Finding primes that are 33 or 4(mod7)4 \pmod 7 and congruent to 1(mod5)1 \pmod 5 is all that is left for part 1. We can strengthen this condition and say that the last digit of the prime must be 11, as if it is 66 the number is not prime. The next prime is p=11p = 11, and the next is p=101p = 101. Hence, there are a minimum of 33 primes that satisfy the first set of requirements.

For the part of the question involving qq, note that 55 is the only number that satisfies the requirements. This is because the only permissible residue mod 55 is 0, and 55 is the only prime that is congruent to 0(mod5)0 \pmod 5.

Elijah L - 2 weeks, 4 days ago

Log in to reply

@Elijah L Correct, but different method.

Yajat Shamji - 2 weeks, 3 days ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...