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Regional Olympiad Problem

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

Note by S M
11 months, 3 weeks ago

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If 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. Brilliant Member · 11 months, 2 weeks ago

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@Brilliant Member Great!

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 · 11 months, 2 weeks ago

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@Calvin Lin Oh right! 15 can be written as \(6+9\). Brilliant Member · 11 months, 2 weeks ago

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

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@Vicky Vignesh Can you explain what you are trying to do? I have several concerns:

  1. 11 isn't composite.
  2. For \( m = 13 \), \( 13 = 11 + 2 \), 2 isn't composite either.
Calvin Lin Staff · 11 months, 3 weeks ago

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@Calvin Lin Ahh, is this sum that tough? I am doomed. S M · 11 months, 3 weeks ago

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@S M Nope, I am pointing out that the solution makes no sense, and does not answer the problem at all.

It's actually a pretty easy problem, just give it a try. What have you tried? Where did you get stuck? Calvin Lin Staff · 11 months, 3 weeks ago

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@Calvin Lin oh pretty sure i dont notice the condition it needs to be a composite number Vicky Vignesh · 11 months, 3 weeks ago

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