Assume that we have six positive real numbers whose sum is 150. Prove that there exist two of them whose difference is less than 10.

Question

I'm trying to answer this questions using contradiction but I don't know if it's right. $$a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = 150$$ Assuming that all the differences of $$a_j - a_i \ge 10$$ then.

$a_1 + a_2 + a_3 + a_4 + a_5 + a_6 - (a_1 + a_2 + a_3 + a_4 +a_5 +a_6) = 0$

$(a_1 - a_2) + (a_2 - a_3) + (a_3 - a_4) + (a_4 - a_5) + (a_5 - a_6 ) + (a_6- a_1) \ge 60 $ (contradiction)

Answer 1

Your argument fails because some of the parenthesized terms must be negative. Based on the symmetry you can renumber things so that $a_1 \gt a_2 \gt a_3 \gt a_4 \gt a_5 \gt a_6$ You know $a_6 \gt 0$ so $a_5 \gt 10$. Then $a_4 \gt 20$. Keep going.

Answer 2

Best to start with the smallest one $x>0 $. Then the others in increasing order are $$x+d_1, x+d_1 + d_2, \ldots,x+ d_1 + d_2+d_3 + d_4 + d_5$$ where $d_i\ge 0$ is the difference between $i+1$ th and $i$th number. Now their sum is $$150 = 6 x + 5 d_1 + 4 d_2 + 3 d_3 + 2 d_4 + d_5$$ so if $d$ is the smallest of the $d_i$ we get $$150 > (5+4+3+2+1) d = 15 d$$ so $$0\le d < 10$$ In general, if $n$ positive numbers have sum $S$ then two of them have difference $\ge 0$ and $< S/\binom{n}{2}$

Answer 3

Building on @fleablood's answer: the minimum number may be arbitrarily small, but still positive. We call it a.

The next five numbers (at their smallest) must be:

$$b = 10 + a$$ $$c = 20 + a$$ $$d = 30 + a$$ $$e = 40 + a$$ $$f = 50 + a$$ Thus, the sum of all the numbers is

$$a + b + c + d + e + f = 150 + 6a$$

Since a is positive (though arbitrarily small) the sum must be greater than 150, which is a contradiction.

Answer 4

The smallest of the six terms must be more than $0$.

If the differences of any two is at least $10$, then the second must be more than $10$

The smallest the third must be more than $20$.

Etc.

So all six must add up to more than $0+10+20+30+40+50=150$.

So they can't add to $150$. They can be arbitrarily close to $150$ but they must be more than $50$.

.....

If the smallest one is $a = \epsilon > 0$ and then the next smallest is $b \ge 10 + \epsilon$ and then the next smallest is $c \ge 10 + b \ge 20 +\epsilon$. Etc.

The sum is $a + b + c + d+e +f \ge \epsilon + (10 + \epsilon) + .... + (50 + \epsilon) = 150 + 6\epsilon > 150$.

Answer 5

I'm assuming we're taking $a_1 \geq a_2 \geq a_3 \geq a_4 \geq a_5 \geq a_6$.

However, we can't simply take $a_j - a_i \geq 10$ because $a_i$ might be bigger than $a_j$ (such as the $a_6 - a_1$ that was used to get the contradiction).

The proper way would be $|a_j - a_i| \geq 10$, but this is somewhat difficult to work with.

Instead, maybe consider proving by contradiction by noting that $$a_i \geq a_{i+1} + 10$$

You should be able to work out that $a_1 + a_2 + a_3 + a_4 + a_5 \geq 5a_6 + 150$.

Answer 6

I think you are not doing it right, because we have to prove the absolute value: $|a_{i}-a_{j}|<10$ is true for all $i,j\in(1;2;3;4;5;6)$ and $i\ne j$.

Assume that $a_1\le a_2\le a_3 \le a_4\le a_5\le a_6$ and we have $a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = 150$, also assume that $a_{j}-a_{i}\ge10$ for all $i,j\in(1;2;3;4;5;6)$ and $i<j$.

• $a_1\ge 1$ and $a_2-a_1\ge 10$ $\Rightarrow a_2\ge11.$

• $a_2\ge 11$ and $a_3-a_2\ge 10$ $\Rightarrow a_3\ge21.$

• $a_3\ge 21$ and $a_4-a_3\ge 10$ $\Rightarrow a_4\ge31.$

• $a_4\ge 31$ and $a_5-a_4\ge 10$ $\Rightarrow a_5\ge41.$

• $a_5\ge 41$ and $a_6-a_5\ge 10$ $\Rightarrow a_6\ge41.$

$\Rightarrow a_1+a_2+a_3+a_4+a_5+a_6\ge 156$, contradiction.

Answer 7

Pigeonhole principle: at least two numbers fall into one of the intervals/brackets $[0,10],[10,20],[20,30]$ and etc...?