How many consecutive whole numbers, starting with 1, must be added to get a some of 666?

Guest Dec 3, 2017

#1**+2 **

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. | ||

| |||

| |||

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.

TheXSquaredFactor
Dec 3, 2017

#1**+2 **

Best Answer

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. | ||

| |||

| |||

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.

TheXSquaredFactor
Dec 3, 2017