Math Help - Sum of real numbers

1. Sum of real numbers

Several real numbers (not necessarily are different) are chosen. The sum of these numbers is 10. Is it possible that the sum of the squares of these numbers is less than one-millionth? Justify your answer.

Any hints or help would be greatly appreciated. Thanks!

Several real numbers (not necessarily are different) are chosen. The sum of these numbers is 10. Is it possible that the sum of the squares of these numbers is less than one-millionth? Justify your answer.

Any hints or help would be greatly appreciated. Thanks!
Consider the case when the numbers are equal, i.e. all the numbers equal $\frac{10}{n}$, if there are $n$ of them. What is the sum of their squares?

3. Originally Posted by Laurent
Consider the case when the numbers are equal, i.e. all the numbers equal $\frac{10}{n}$, if there are $n$ of them. What is the sum of their squares?
I'm sorry this isn't my type of math but I don't see a pattern in having the numbers equal. If all the numbers are 1, then I try to have all of them be 2, I get two totally different sums of their squares.

I'm sorry this isn't my type of math but I don't see a pattern in having the numbers equal. If all the numbers are 1, then I try to have all of them be 2, I get two totally different sums of their squares.
Since you don't know how many numbers are chosen, you can't just pick a specific number. But Laurent showed you that you can choose all equal numbers, which are is possible by the hypothesis, then compute the sum of their squares. You should follow Laurent's hint.
Square each number, each of them is $\frac{10}{n}$, so its square is $(\frac{10}{n})^2$.
Then add up all of these square $n$ times. See what you get, and check if it is possible that the sum you just computed is less than one-millionth.

5. Originally Posted by namelessguy
Since you don't know how many numbers are chosen, you can't just pick a specific number. But Laurent showed you that you can choose all equal numbers, which are is possible by the hypothesis, then compute the sum of their squares. You should follow Laurent's hint.
Square each number, each of them is $\frac{10}{n}$, so its square is $(\frac{10}{n})^2$.
Then add up all of these square $n$ times. See what you get, and check if it is possible that the sum you just computed is less than one-millionth.
I tried doing what you said trying to get the sum to 10 but I keep getting 1. Does that mean it's impossible?

I tried doing what you said trying to get the sum to 10 but I keep getting 1. Does that mean it's impossible?
I don't understand why you tried to get the sum to 10. Which sum are you mentioning here?
Here is a sketch of what you should do.
$(\frac{10}{n})^2+(\frac{10}{n})^2+...+(\frac{10}{n })^2$ There are $n$ terms.
So this sum is equal to $n\times (\frac{10}{n})^2=\frac{100}{n}$
Now, is it possible that $\frac{100}{n}$ is less than one-millionth?

7. And the point of looking at all numbers equal is that you can show that the more they differ, the larger the sum of squares will be. The case with all numbers equal is the smallest sum of squares. If that sum can't be "less than one-millionth" then the other sums certainly can't be.

8. Originally Posted by namelessguy
I don't understand why you tried to get the sum to 10. Which sum are you mentioning here?
Here is a sketch of what you should do.
$(\frac{10}{n})^2+(\frac{10}{n})^2+...+(\frac{10}{n })^2$ There are $n$ terms.
So this sum is equal to $n\times (\frac{10}{n})^2=\frac{100}{n}$
Now, is it possible that $\frac{100}{n}$ is less than one-millionth?
Ok I kinda get what you're saying here. By using your equation above I can plug in 100,000,000 for n and get exactly one-millionth. The problem I'm having with this is how the sum of these numbers can even equal 10. This is why I tried to get the sum to 10 in my previous post but couldn't.

To everyone that's posted, I will give thanks when I figure this out but I don't see the picture everyone is trying to paint here.

The problem I'm having with this is how the sum of these numbers can even equal 10.
If you have $n$ numbers, all of which equal $\frac{10}n,$ then the sum is

$\underbrace{\frac{10}n+\frac{10}n+\frac{10}n+\cdot s+\frac{10}n}_{n\text{ terms}}=n\cdot\frac{10}n = \frac{10n}n = 10\text.$

For example, suppose 3 numbers are to be chosen. Then you choose them all to be $\frac{10}3$ and you get a sum of $\frac{10}3+\frac{10}3+\frac{10}3=\frac{30}3=10\tex t.$ But this works for any $n,$ not just $n=3\text.$

10. Originally Posted by Reckoner
If you have $n$ numbers, all of which equal $\frac{10}n,$ then the sum is

$\underbrace{\frac{10}n+\frac{10}n+\frac{10}n+\cdot s+\frac{10}n}_{n\text{ terms}}=n\cdot\frac{10}n = \frac{10n}n = 10\text.$

For example, suppose 3 numbers are to be chosen. Then you choose them all to be $\frac{10}3$ and you get a sum of $\frac{10}3+\frac{10}3+\frac{10}3=\frac{30}3=10\tex t.$ But this works for any $n,$ not just $n=3\text.$
Haha wow I feel like an idiot. Thank you very much, squaring 10/n through me off but now I understand how it can equal 10. But was my understanding correct, that I can make n a really large number to get the desired less than one-millionth?