A division-related Problem

Number Theory Level 2

Given $a$ and $10a+1$ are both prime numbers larger than 3. Is the statement below always true ? (Bonus: Why not or Why?)

$5a+1$ is always divisible by 6.

Paradox. Always True Always False Not always True. Not always False.

1 solution

Tin Le
Jun 9, 2019

We know that the product of 3 consecutive positive numbers is always divisible by 3. Therefore $10a(10a+1)(10a+2)$ is always divisible by 3.

$a$ is a prime number, so it's not divisible by 3. 10 is not divisible by 3 as well, so $10a$ is never divisible by 3. (or $10a$ is coprime to 3)

$10a+1$ is also a prime number, so it's not divisible by 3. (Or $10a+1$ is coprime to 3)

If a product of 3 numbers is divisible by a number $x$ , but 2 of them are coprime to $x$ , then another number must be divisible by $x$ .

Therefore, $10a + 2$ = $2(5a+1)$ is divisible by 3. But 2 is coprime to 3, so $5a+1$ is divisible by 3.

$a$ is a prime number larger than 3, so $a$ is always odd. Therefore, $5a+1$ is always even.

2 and 3 are coprime integers, therefore $5a+1$ always divisible by 6.

Nice proof! But the answer options are a bit confusing (and not mutually exclusive). Isn't the statement also "not always false"?

A better set of options would be "always true", "always false", "sometimes true and sometimes false".

Chris Lewis - 2 years ago