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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
1 year, 2 months 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 - 1 year, 2 months ago

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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 - 1 year, 2 months ago

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Oh right! 15 can be written as \(6+9\).

Brilliant Member - 1 year, 2 months ago

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Will you please help me out in solving a question as There is a right angle isosceles triangle ABC 90 degree angle at B and AC is hypotenuse. There are two points D, E in between AC such that AD:DE:EC=3:5:4 then prove that angle DBE=45degree

Ayush Kumar - 6 days, 3 hours 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 - 1 year, 2 months ago

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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 - 1 year, 2 months ago

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Ahh, is this sum that tough? I am doomed.

S M - 1 year, 2 months 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 - 1 year, 2 months ago

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@Calvin Lin oh pretty sure i dont notice the condition it needs to be a composite number

Vicky Vignesh - 1 year, 2 months ago

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@Vicky Vignesh 13 = 9 + 4

Edwin Gray - 1 month, 1 week ago

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