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The sum of five positive numbers is $100$. Prove that there are two numbers among them whose difference is at most $10$.

attempt: Let $m,n,p,q,r$ be positive real numbers such that $m \leq n \leq p \leq q \leq r$. Then assume $m > 0, n > 10, p> 20, q > 30 , e > 40.$ Can anyone please suggest something ? Thank you

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  • $\begingroup$ If $m>0,n>10,p>20,q>30,r>40$ then $m+n+p+q+r>0+10+20+30+40=100$. $\endgroup$ – arctic tern Feb 2 '17 at 5:47
  • $\begingroup$ The Pigeonhole Principle $\endgroup$ – Nosrati Feb 2 '17 at 5:48
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    $\begingroup$ @MyGlasses For real numbers? $\endgroup$ – arctic tern Feb 2 '17 at 5:48
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Assume all numbers have a difference that is bigger than $10$. Let these $5$ numbers be $a>b>c>d>e>0$. So we have $$a>b+10>c+20>d+30>e+40>40$$So, $$a>e+40, \quad b>e+30, \quad c>e+20, \quad d>e+10$$ So $$100=a+b+c+d+e>5e+100>100$$ Contradiction.

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Let $x_2,x_2,\dots x_5$ be the numbers and assume by contradiction that $x_{i+1}>10+x_i$ for all $i$.

We have: $x_1>0, x_2> 10,x_3 > 20,x_4> 30,x_5>40\implies x_1+x_2+x_3+x_4+x_5>100$

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