+0

# Some Number Theory

0
286
1

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

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.
 $$n=\frac{-1+73}{2}$$ $$n=\frac{-1-73}{2}$$

 $$n=\frac{72}{2}=36$$ $$n=\frac{-74}{2}=-37$$

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.

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.
 $$n=\frac{-1+73}{2}$$ $$n=\frac{-1-73}{2}$$

 $$n=\frac{72}{2}=36$$ $$n=\frac{-74}{2}=-37$$

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.

TheXSquaredFactor Dec 3, 2017