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Difference between revisions of "2009 IMO Problems/Problem 1"

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''Author: Ross Atkins, Australia''
 
''Author: Ross Atkins, Australia''
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--[[User:Bugi|Bugi]] 10:23, 23 July 2009 (UTC)Bugi
 
  
 
== Solution ==
 
== Solution ==
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Note <math>q\mid a_2-1</math> implies <math>(q,a_2)=1</math> and hence <math>q\mid a_3-1</math>. Similarly one has <math>q\mid a_i-1</math> for all <math>i</math>'s, in particular, <math>p\mid a_1</math> and <math>q\mid a_1-1</math> force <math>(p,q)=1</math>. Now <math>(p,a_1-1)=1</math> gives <math>p\mid a_k</math>, similarly one has <math>p\mid a_i</math> for all <math>i</math>'s, that is <math>a_i</math>'s satisfy <math>p\mid a_i</math> and <math>q\mid a_i-1</math>, but there should be at most one such integer satisfies them within the range of <math>1,2,\ldots,n</math> for <math>n=pq</math> and <math>(p,q)=1</math>. A contradiction!!!
 
Note <math>q\mid a_2-1</math> implies <math>(q,a_2)=1</math> and hence <math>q\mid a_3-1</math>. Similarly one has <math>q\mid a_i-1</math> for all <math>i</math>'s, in particular, <math>p\mid a_1</math> and <math>q\mid a_1-1</math> force <math>(p,q)=1</math>. Now <math>(p,a_1-1)=1</math> gives <math>p\mid a_k</math>, similarly one has <math>p\mid a_i</math> for all <math>i</math>'s, that is <math>a_i</math>'s satisfy <math>p\mid a_i</math> and <math>q\mid a_i-1</math>, but there should be at most one such integer satisfies them within the range of <math>1,2,\ldots,n</math> for <math>n=pq</math> and <math>(p,q)=1</math>. A contradiction!!!
  
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''Solution by ychjae''
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== Solution 2 ==
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Let <math>n = p_1^{b_1}p_2^{b_2} \cdots p_s^{b_s}</math>. Then after toying around with the <math>p_i^{b_i}</math> and what they divide, we have that <math>p_i^{b_i} \nmid a_k</math>, and so in particular, <math>n \nmid a_k</math>.
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Assume by way of contradiction that <math>n \mid a_k(a_1 - 1)</math>. Then <math>n \mid a_1 - 1</math>. Now we shift our view towards the <math>a_i(a_{i + 1} - 1)</math>. Here each <math>p_i^{b_i}</math> divides <math>a_i(a_{i + 1} - 1) \implies a_ia_{i + 1} \equiv a_i \pmod{p_i^{b_i}}</math>. Hence we have the chain of equivalences <math>a_1a_2 \equiv a_1 \pmod{p_i^{b_i}}, a_2a_3 \equiv a_2 \pmod{p_i^{b_i}}, \dots, a_{k - 1}a_k \equiv a_{k - 1} \pmod {p_i^{b_i}}</math>. Now we also have that <math>p_i^{b_i} \mid n \mid a_1 - 1</math>. Thus <math>a_1 \equiv 1 \pmod{p_i^{b_i}}</math>. Now plugging this value of <math>a_1</math> modulo <math>p_i^{b_i}</math>, we obtain that <math>a_1 \equiv a_2 \equiv a_3 \equiv \cdots a_k \equiv 1 \pmod{p_i^{b_i}}</math>. Hence this chain of congruences is also true for <math>n</math> as <math>p_i</math> was arbitrary. However as all the <math>a_i \in \{1, 2, \dots, n\}</math> we have that not all the <math>a_i</math> are distinct, and so this is a contradiction.
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==See Also==
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{{IMO box|year=2009|before=First Problem|num-a=2}}

Latest revision as of 00:16, 19 November 2023

Problem

Let $n$ be a positive integer and let $a_1,\ldots,a_k (k\ge2)$ be distinct integers in the set $\{1,\ldots,n\}$ such that $n$ divides $a_i(a_{i+1}-1)$ for $i=1,\ldots,k-1$. Prove that $n$ doesn't divide $a_k(a_1-1)$.

Author: Ross Atkins, Australia

Solution

Let $n=pq$ such that $p\mid a_1$ and $q\mid a_2-1$. Suppose $n$ divides $a_k(a_1-1)$. Note $q\mid a_2-1$ implies $(q,a_2)=1$ and hence $q\mid a_3-1$. Similarly one has $q\mid a_i-1$ for all $i$'s, in particular, $p\mid a_1$ and $q\mid a_1-1$ force $(p,q)=1$. Now $(p,a_1-1)=1$ gives $p\mid a_k$, similarly one has $p\mid a_i$ for all $i$'s, that is $a_i$'s satisfy $p\mid a_i$ and $q\mid a_i-1$, but there should be at most one such integer satisfies them within the range of $1,2,\ldots,n$ for $n=pq$ and $(p,q)=1$. A contradiction!!!

Solution 2

Let $n = p_1^{b_1}p_2^{b_2} \cdots p_s^{b_s}$. Then after toying around with the $p_i^{b_i}$ and what they divide, we have that $p_i^{b_i} \nmid a_k$, and so in particular, $n \nmid a_k$.

Assume by way of contradiction that $n \mid a_k(a_1 - 1)$. Then $n \mid a_1 - 1$. Now we shift our view towards the $a_i(a_{i + 1} - 1)$. Here each $p_i^{b_i}$ divides $a_i(a_{i + 1} - 1) \implies a_ia_{i + 1} \equiv a_i \pmod{p_i^{b_i}}$. Hence we have the chain of equivalences $a_1a_2 \equiv a_1 \pmod{p_i^{b_i}}, a_2a_3 \equiv a_2 \pmod{p_i^{b_i}}, \dots, a_{k - 1}a_k \equiv a_{k - 1} \pmod {p_i^{b_i}}$. Now we also have that $p_i^{b_i} \mid n \mid a_1 - 1$. Thus $a_1 \equiv 1 \pmod{p_i^{b_i}}$. Now plugging this value of $a_1$ modulo $p_i^{b_i}$, we obtain that $a_1 \equiv a_2 \equiv a_3 \equiv \cdots a_k \equiv 1 \pmod{p_i^{b_i}}$. Hence this chain of congruences is also true for $n$ as $p_i$ was arbitrary. However as all the $a_i \in \{1, 2, \dots, n\}$ we have that not all the $a_i$ are distinct, and so this is a contradiction.

See Also

2009 IMO (Problems) • Resources
Preceded by
First Problem 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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