UKMT Special (Problem $1$ )

Show that there are at least $3$ primes that are less than $200$ that satisfy the system of equations:

$p + 2$

$p + 6$

$p + 8$

$p + 12$

and result in primes.

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

$q + 2$

$q + 6$

$q + 8$

$q + 12$

$q + 14$

and result in primes.

[UKMT BMO $2019$ , Q $1$ ]

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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

@Mahdi Raza

Yajat Shamji - 7 months ago

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

Yajat Shamji - 7 months ago

I'm interpreting the question as "Find all primes $p$ 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 $p \ge 2$ , we notice that from Pigeonhole Principle, there must be a number from the first $3$ statements that is divisible by $3$ . One can easily check that the number is composite.

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

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

Elijah L - 7 months ago

There is a solution!

I solved it myself!

Yajat Shamji - 7 months ago

Oops!

I accidentally wrote the wrong system of equations!

Re-solve it, @Elijah L

Yajat Shamji - 7 months ago

Note that from the first set of equations, $p$ must be $0$ , $3$ , or $4 \pmod 7$ for sufficiently large $p$ . Looking at it $\pmod 5$ , the only permissible residues mod $5$ are $0$ and $1$ . From the same logic, the number must also be $2 \pmod 3$ . Checking small cases leaves that $p = 5$ is a solution.

Finding primes that are $3$ or $4 \pmod 7$ and congruent to $1 \pmod 5$ is all that is left for part 1. We can strengthen this condition and say that the last digit of the prime must be $1$ , as if it is $6$ the number is not prime. The next prime is $p = 11$ , and the next is $p = 101$ . Hence, there are a minimum of $3$ primes that satisfy the first set of requirements.

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

Elijah L - 7 months ago

@Elijah L – Correct, but different method.

Yajat Shamji - 7 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}`	$\boxed{123}$

UKMT Special (Problem 111)

Comments

UKMT Special (Problem $1$ )