How many consecutive whole numbers, starting with 1, must be added to get a some of 666?
+2446 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. | ||
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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.
+2446 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. | ||
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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.
