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

Aug 31, 2018

#1
+7499
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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
+95
+4

thanks

Sep 1, 2018