Some Number Theory

The way I would approach this problem is to find a formula for the sum of consecutive integers 1,2,3,4 ..., n. 

 

Let S = sum of consecutive whole numbers starting with 1

 

\(S=1+2+3+4+...+n \)

 

Now, let's reverse the sum. You will see where this is headed in a moment.

 

\(S= n+(n-1)+(n-2)+(n-3)+...+1\)

 

Both of the sums above are the same. Now, let's add them together.

 

\(S= 1+\hspace{1cm}2\hspace{2mm}+\hspace{1cm}3\hspace{2mm}+\hspace{9mm}4\hspace{2mm}+...+n\\ S=n+(n-1)+(n-2)+(n-3)+...+1\)	Add these sums together.
\(2S=\underbrace{(n+1)+(n+1)+(n+1)+...+(n+1)}\\ \hspace{3cm}\text{n occurences of n+1}\)	Knowing that there are n-terms present, we can simplify this slightly.
\(2S=n(n+1)\)	Divide by 2 on both sides to figure out the sum of consecutive whole numbers up to n.
\(S=\frac{n(n+1)}{2}\)

 

Now that we have generated a formula (a well-known one at that), we can proceed with this problem. I have proved this sum. 

 

We know that the sum must be equal to 666, so let's plug that in and solve.

 

\(666=\frac{n(n+1)}{2}\)	Let's eliminate the fraction immediately by multiplying by 2 on both sides.
\(1332=n(n+1)\)	Expand the right hand side of the equation.
\(1332=n^2+n\)	Move everything to one side.
\(n^2+n-1332=0\)	I will tell you beforehand that this quadratic is factorable, but it is pretty difficult to factor, so let's try using the quadratic formula instead.
\(n=\frac{-1\pm\sqrt{1^2-4(1)(-1332)}}{2(1)}\)	Now, simplify
\(n=\frac{-1\pm\sqrt{5329}}{2}\)	The radicand happens to be a perfect square.
\(n=\frac{-1\pm73}{2}\)	Now, solve for both values of n.
\(n=\frac{-1+73}{2}\) \(n=\frac{-1-73}{2}\)
\(n=\frac{72}{2}=36\) \(n=\frac{-74}{2}=-37\)

 

We reject the negative answer because we only care about whole-number solutions, and -37 is not a whole number. 

 

What this means is that \(666=1+2+3+4...+36\), so one must add consecutively 36 positive whole numbers, starting with 1, to get a sum of 666.