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 Christian Warjri
3 years, 10 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.

- 3 years, 10 months ago

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.

Staff - 3 years, 10 months ago

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

- 3 years, 10 months ago

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

- 2 years, 7 months ago

$m>11 \\ m = 11 + a ; ~ a \in Q\\ a = \pm n, ~ n \in Q \\ m = x + y, ~x = 11, y = \pm n$

- 3 years, 10 months ago

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.

Staff - 3 years, 10 months ago

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

- 3 years, 10 months ago

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?

Staff - 3 years, 10 months ago

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

- 3 years, 10 months ago

13 = 9 + 4

- 2 years, 9 months ago