+0  
 
+4
55
2
avatar+79 

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

iamdaone  Aug 31, 2018
 #1
avatar+7023 
+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 \(k\in \mathbb Z^*\)

MaxWong  Sep 1, 2018
 #2
avatar+79 
+3

thanks

iamdaone  Sep 1, 2018

14 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.