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That's my attempt.

We know that among $11$ integers by pigeonhole principle at least $3$ leave the same remainder after division with $5$. We take those $3$. Of them at least $2$ leave the same remainder after division by $2$. But now, since no numbers of them are divisible by $20$, they leave the same remainder if they are again divided by $2$. Choose them and we have that $20 \mid a-b$.

Is this right? If this is wrong I would appreciate only hints because I would like to finish it myself.

Bill Dubuque
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Arthr
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    You need divisibility by $4$, not just $2$ – lulu Nov 02 '24 at 10:05
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    More broadly, you are trying to prove a stronger statement, one without the $a+b$ option. But maybe you need that option. – lulu Nov 02 '24 at 10:06
  • If you want a hint: Suppose you are trying to build a counterexample. note that selecting a remainder on division by $20$, say $4$ you are effectively killing off two options for future choices, you are removing both $4$ and $16$. What does that tell you? (Note that if your selection has remainder $10$ then you are only removing $10$ as a remainder, since $20-10=10$). – lulu Nov 02 '24 at 10:29
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Nov 02 '24 at 18:04
  • It seems to me that we need only 10 numbers, not 11. Am I mistaken? – MJD Nov 02 '24 at 18:15
  • Hint By pigeonhole: two collide in the $10$ nonzero signed remainder buckets $,\pm1,,\pm2,\ldots,\pm9,10\ \ $ – Bill Dubuque Nov 02 '24 at 20:51
  • @MJD There are $10$ (not $9)$ holes (possible signed remainders), if you present the the argument uniformly, see my comments above and also on Aditya's answer. Btw, I am confident that this question is a duplicate, besides being an SV missing context. – Bill Dubuque Nov 02 '24 at 21:40
  • Generally we can $\rm\color{#c00}{pair}$ elements via any $\rm\color{#c00}{reflection}$ (involution) i.e. a nontrivial map $f$ with $f^2(x) = x\ $ (here modular $\rm\color{#c00}{negation}$ $,f(x) \equiv -x),,$ e.g. see Wilson's theorem and Gauss's grade school summation trick. $\ \ $ – Bill Dubuque Nov 02 '24 at 22:03
  • @BillDubuque I don't understand whether your answer says that 10 is sufficient or 11 is required. If the latter, what 10-set witnesses it? – MJD Nov 03 '24 at 02:26
  • @MJD As said, there are $10$ nonzero signed remainder buckets, so we need at least $11$ integers for a collision. – Bill Dubuque Nov 03 '24 at 03:21
  • Then what is a set of 10 integers that fails to satisfy the condition that there are $a,b$ with $a+b$ or $a-b$ a multiple of 20? – MJD Nov 03 '24 at 03:37
  • @MJD I may be misunderstanding the question, but what about {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}? (I assume it is intended that a and b must be distinct.) – dfan Nov 03 '24 at 15:11
  • (Uh, yeah, of course they must be distinct or the $a-b$ clause would make it trivial.) – dfan Nov 03 '24 at 15:17
  • I was thinking $10+10$ is a multiple of $20$, but you must be correct. Thanks for pointing out what I had missed. – MJD Nov 03 '24 at 16:22

1 Answers1

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Use pigeon-hole principle. Each number is of the form $20q+r$, where $q$ and $r$ are whole numbers. If you have $11$ numbers not divisible by $20$, then you have $11$ numbers $20q_i+r_i$, and $r_i$ is non-zero for all $i$.

If $r_i=r_j$ for any pair $(i,j)$, we are done, otherwise we have $r_i$ taking $11$ different values. For each value that $r_i$ takes for some $i$, if we have $r_j=20-r_i$, for some j, we are done.

We have $10$ holes of capacity $2$: $(1,19),(2,18).....,(9,11),(10,10)$. But we have $11$ pigeons. So there will exist some hole with $2$ pigeons sitting in it. Therefore if all $r_i$ are unique, then there exists some pair $i,j$ such that $r_i+r_j$ is divisible by $20$. Hence we are done.

Aditya Mishra
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  • i.e. $11$ into $10$ nonzero signed remainder buckets $,\pm1,,\pm2,\ldots,\pm9,10,$ so two $,a,b,,$ are in same bucket $,\pm r,$ so $,{a,b}\equiv_{20} \color{#0a0}{{r,-r}}$ or $,\color{#c00}{{r}\ {\rm or}\ ,{-r}},$ so $,20\mid \color{#0a0}{a+b},$ or $,20\mid\color{#c00}{a-b}.\ \ $ – Bill Dubuque Nov 02 '24 at 20:45