Happy 2016!

Algebra Level 1

$1+2+3+4+\cdots + n = 2016$ Solve for $n.$

Hint : $2016 = 2^5 \cdot 3^2 \cdot 7.$

4 solutions

Rishabh Jain
Jan 1, 2016

$\frac{(n)(n+1)}{2}=\frac{4032}{2}=\frac{2^6\times3^2\times7}{2}$ $\Rightarrow (n)(n+1)=63\times 64\Rightarrow$ $\color{magenta}{n=63}\\$

simple answer! I took the hard way of using the sum of an arithmetic progression! haha

Justin Ruaya - 5 years, 5 months ago

Rishbash did the same thing, he just used a special case with the first term as 1.

Ryoha Mitsuya - 5 years, 5 months ago

Pablo Ruiz
Jan 4, 2016

$\frac{n(n+1)}{2}$ = $2016$ $\Rightarrow$ $n(n+1)$ =4032 $\Rightarrow$ $n^2+n-4032=0$ $\Rightarrow$ $(n+64)(n-63)=0$ $\Rightarrow$ $n$ = $63, -64$

We are looking for a positive value

$\boxed{n=63}$

Bettie Wallace
Aug 20, 2016

2016=n/2(1+n), 4032=n+n^2, n^2+n-4032=0, n=63 or -32, Therefore n=63.

Youssef Mohammed
Jan 5, 2016

Actually i didn't use the hint as 1+2+3+........+n = 2016 So, n(n+1)/2 = 2016 N^2 + n - 4032 = 0 Then , (x+64)(x-63)=0 X = -64 , 63 Solved!!!! :))