# 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.

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)

• First line should equal 150. I can't edit it myself, because the edit has to be at least six characters long... – infinitezero Apr 25 '18 at 9:58
• If the difference of every pair of numbers is greater than ten then a number minus itself must also be greater than ten, and therefore zero is greater than ten. But we could attempt a fix by saying the difference of every pair of unequal numbers must be greater than ten. That doesn't work either, because it implies that x - y and y - x are both not negative which is only true when x and y are equal. Now do you see why your proof attempt fails? – Eric Lippert Apr 25 '18 at 13:43
• Just to add to your problem, the precondition is not specified correctly, there certainly could exist j=i and so it's already contradition of precondition.. With these preconditions come into the contradiction you cannot be sure if ai>aj or ai<aj.. – Polostor Apr 25 '18 at 13:50
• Is your question what is wrong with your proof, or what a correct way to prove it is? Or neither: are you looking for a correct proof using contradiction? – NH. Apr 25 '18 at 15:12
• Sum of (10,12,18,30,40,50) = 150. 12-10 = 2. 18-12 = 6. Why do you need the additional assumption? I feel like I'm missing something. – CrossRoads Apr 25 '18 at 19:11

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.

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 $< S/\binom{n}{2}$

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.

• Welcome to MSE. Please use MathJax. – José Carlos Santos Apr 25 '18 at 10:18
• "the minimum number has to be infinitesimally small". Admittedly I made an error in assuming integers and thus i should stand corrected. However this correction isn't quite correct. There is no minimum number and no actual number is "infinitesimally small". And, of course, it doesn't have to be infinitesimally small (even if it could be, which it can't). A proper statement would be "$a$ may be arbitrarily small, but it must be more than $0$". – fleablood Apr 25 '18 at 15:08
• Hi flea blood, yes you’re right - I’ll edit my answer to reflect this. – Joseph Mulligan Apr 25 '18 at 21:02
• Instead of a, I'd suggest using the common variable ε which is the usual stand-in for an arbitrarily small positive value. – Sandy Chapman Apr 25 '18 at 22:30
• @fleablood "no actual number is 'infinitesimally small'"... Well, not necessarily, though certainly no real number is.... – Reinstate Monica Apr 26 '18 at 0:50

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

• The problem statement declares 'six positive real numbers', not integer numbers, so the smallest one may also be $1/2$ or $0.000001$ – CiaPan Apr 25 '18 at 7:05
• Ah, okay, ... that actually makes the question better. But it doesn't change the the argument. – fleablood Apr 25 '18 at 14:53

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 $a_1 + a_2 + a_3 + a_4 + a_5 \geq 5a_6 + 150$.

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.

• Associative and commutative rules allow the numbers to be sorted in increasing order and still have the same sum. So just work each term at least ten greater than the one before. Then fleablood's answer above is right on and is actually a more powerful generalization because it extends to real numbers. – richard1941 May 1 '18 at 23:00

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