Regional Olympiad Problem

Prove that mm is an integer bigger than 11 then there always exist 2 different composite integers xx and yy such that m=x+ym=x+y.

Note by Christian Warjri
3 years ago

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If we take m0(mod4)m \equiv 0 \pmod4, we can take x=8x=8 and y as the required multiple of 4. Note that 16 has to written as 10+6.

If we take m1(mod4)m \equiv 1 \pmod4, we can take x=9x=9 and y as the required multiple of 4.

If we take m2(mod4)m \equiv 2 \pmod4, we can take x=10x=10 and y as the required multiple of 4.

If we take m3(mod4)m \equiv 3 \pmod4, we can take x=15x=15 and y as the required multiple of 4.

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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 - 3 years ago

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Oh right! 15 can be written as 6+96+9.

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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 - 1 year, 10 months ago

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m>11m=11+a; aQa=±n, nQm=x+y, x=11,y=±nm>11 \\ m = 11 + a ; ~ a \in Q\\ a = \pm n, ~ n \in Q \\ m = x + y, ~x = 11, y = \pm n

Viki Zeta - 3 years 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 m = 13 , 13=11+2 13 = 11 + 2 , 2 isn't composite either.

Calvin Lin Staff - 3 years ago

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

Christian Warjri - 3 years ago

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@Christian Warjri 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 - 3 years ago

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

Viki Zeta - 3 years ago

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@Viki Zeta 13 = 9 + 4

Edwin Gray - 1 year, 11 months ago

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