Prove that \(m\) is an integer bigger than 11 then there always exist 2 different composite integers \(x\) and \(y\) such that \(m=x+y\).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestIf we take \(m \equiv 0 \pmod4\), we can take \(x=8\) and y as the required multiple of 4. Note that 16 has to written as 10+6.

If we take \(m \equiv 1 \pmod4\), we can take \(x=9\) and y as the required multiple of 4.

If we take \(m \equiv 2 \pmod4\), we can take \(x=10\) and y as the required multiple of 4.

If we take \(m \equiv 3 \pmod4\), we can take \(x=15\) and y as the required multiple of 4. – Svatejas Shivakumar · 5 months, 3 weeks ago

Log in to reply

Depending on the intent of the question, you might be missing one small case.

Note: Most people do not consider 0 a composite number. – Calvin Lin Staff · 5 months, 3 weeks ago

Log in to reply

– Svatejas Shivakumar · 5 months, 3 weeks ago

Oh right! 15 can be written as \(6+9\).Log in to reply

\(m>11 \\ m = 11 + a ; ~ a \in Q\\ a = \pm n, ~ n \in Q \\ m = x + y, ~x = 11, y = \pm n \) – Vicky Vignesh · 5 months, 3 weeks ago

Log in to reply

Log in to reply

– S M · 5 months, 3 weeks ago

Ahh, is this sum that tough? I am doomed.Log in to reply

It's actually a pretty easy problem, just give it a try. What have you tried? Where did you get stuck? – Calvin Lin Staff · 5 months, 3 weeks ago

Log in to reply

– Vicky Vignesh · 5 months, 3 weeks ago

oh pretty sure i dont notice the condition it needs to be a composite numberLog in to reply