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Find all integers such that n(2n+1)(7n+1) is divisible by 12.

 Aug 31, 2018
 #1
avatar+7499 
+1

Notice: \(\dfrac{n(2n+1)(7n+1)}{6} = (n+1)^2+(n+2)^2+\dots+(2n)^2\)

So the question is equivalent to: Find all integers n such that \((n+1)^2+(n+2)^2+\dots+(2n)^2\) is an even number.

 

But if x^2 is an odd number, then x is an odd number, also if x^2 is an even number, then x is an even number(why?)

So the question is just asking: Find all integers n such that \((n+1)+(n+2)+\dots+(2n)\) is an even number.

Next, we will find the formula for \((n+1)+(n+2)+\dots+(2n)\).

\(\quad(n+1)+(n+2)+\dots+(2n)\\ =n^2+(1+2+3+\dots+n)\\ =n^2+\dfrac{1}{2}(n)(n+1) \\ = \dfrac{n}{2}(3n+1)\)

 

So, finally, the question is equivalent to: Find all integers n such that n(3n+1) is divisble by 4.

For some reason, we arrived at a simpler question. Let's try to solve it now.

n = 4k for k = 1,2,3,4,... is a trivial solution, because if n is divisible by 4, then n(3n+1) is also divisible by 4.

n = 4k+1 for k = 0,1,2,3,4,... is another set of solution. Let's substitute n = 4k+1 into n(3n+1) to see why.

n(3n+1) = (4k+1)(12k+3+1) = (4k+1)(12k+4) = 4(4k+1)(3k+1)

Therefore, when n = 4k+1, for k = 0,1,2,3,4,..., it is also a solution.

 

Final solution: n = 4k for \(k\in \mathbb Z^+\), or n = 4k+1 for \(k\in \mathbb Z^*\)

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 Sep 1, 2018
 #2
avatar+95 
+4

thanks

 Sep 1, 2018

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