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+2+3+4+...+nS=n+(n−1)+(n−2)+(n−3)+...+1 | Add these sums together. |
| 2S=(n+1)+(n+1)+(n+1)+...+(n+1)⏟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=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=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=n2+n | Move everything to one side. | ||
| n2+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=−1±√12−4(1)(−1332)2(1) | Now, simplify | ||
| n=−1±√53292 | The radicand happens to be a perfect square. | ||
| n=−1±732 | 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+2+3+4+...+nS=n+(n−1)+(n−2)+(n−3)+...+1 | Add these sums together. |
| 2S=(n+1)+(n+1)+(n+1)+...+(n+1)⏟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=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=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=n2+n | Move everything to one side. | ||
| n2+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=−1±√12−4(1)(−1332)2(1) | Now, simplify | ||
| n=−1±√53292 | The radicand happens to be a perfect square. | ||
| n=−1±732 | 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.
