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Difference between revisions of "1978 IMO Problems/Problem 1"

(Created page with "==Problem== <math>m</math> and <math>n</math> are positive integers with <math>m < n</math>. The last three decimal digits of <math>1978^m</math> are the same as the last thre...")
 
 
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==Problem==
 
==Problem==
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<math>m</math> and <math>n</math> are positive integers with <math>m < n</math>. The last three decimal digits of <math>1978^m</math> are the same as the last three decimal digits of <math>1978^n</math>. Find <math>m</math> and <math>n</math> such that <math>m + n</math> has the least possible value.
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Let <math> m</math> and <math> n</math> be positive integers such that <math> 1 \le m < n</math>. In their decimal representations, the last three digits of <math> 1978^m</math> are equal, respectively, to the last three digits of <math> 1978^n</math>. Find <math> m</math> and <math> n</math> such that <math> m + n</math> has its least value.
  
 
==Solution==
 
==Solution==
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{{Solution}}
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We have <math>1978^m\equiv 1978^n\pmod {1000}</math>, or <math>978^m-978^n=1000k</math> for some positive integer <math>k</math> (if it is not positive just do <math>978^n-978^m=-1000k</math>). Hence <math>978^n\mid 1000k</math>. So dividing through by <math>978^n</math> we get <math>978^{m-n}-1=\frac{1000k}{978^n}</math>. Observe that <math>2\nmid LHS</math>, so <math>2\nmid RHS</math>. So since <math>2|| 978^n</math>, clearly the minimum possible value of <math>n</math> is <math>3</math> (and then <math>489^n\mid k</math>). We will show later that if <math>n</math> is minimal then <math>m</math> is minimal. We have <math>978^{m-3}-1\equiv 0\pmod {125}\Leftrightarrow 103^{m-3}\equiv 1\pmod {125}</math>. Hence, <math>m-3\mid \varphi(125)\Rightarrow m-3\mid 100</math>. Checking by hand we find that only <math>m-3=100</math> works (this also shows that minimality of <math>m</math> depends on <math>n</math>, as claimed above). So <math>m=103</math>. Consequently, <math>m+n=106</math> with <math>\boxed{(m,n)=(103,3)}</math>.
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The above solution was posted and copyrighted by cobbler and Andreas. The original thread for this problem can be found here: [https://aops.com/community/p393524] and [https://aops.com/community/p3332509]
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==Video Solution==
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https://www.youtube.com/watch?v=SRl4Wnd60os
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== See Also == {{IMO box|year=1978|before=First question|num-a=2}}

Latest revision as of 23:21, 20 October 2021

Problem

Let $m$ and $n$ be positive integers such that $1 \le m < n$. In their decimal representations, the last three digits of $1978^m$ are equal, respectively, to the last three digits of $1978^n$. Find $m$ and $n$ such that $m + n$ has its least value.

Solution

We have $1978^m\equiv 1978^n\pmod {1000}$, or $978^m-978^n=1000k$ for some positive integer $k$ (if it is not positive just do $978^n-978^m=-1000k$). Hence $978^n\mid 1000k$. So dividing through by $978^n$ we get $978^{m-n}-1=\frac{1000k}{978^n}$. Observe that $2\nmid LHS$, so $2\nmid RHS$. So since $2|| 978^n$, clearly the minimum possible value of $n$ is $3$ (and then $489^n\mid k$). We will show later that if $n$ is minimal then $m$ is minimal. We have $978^{m-3}-1\equiv 0\pmod {125}\Leftrightarrow 103^{m-3}\equiv 1\pmod {125}$. Hence, $m-3\mid \varphi(125)\Rightarrow m-3\mid 100$. Checking by hand we find that only $m-3=100$ works (this also shows that minimality of $m$ depends on $n$, as claimed above). So $m=103$. Consequently, $m+n=106$ with $\boxed{(m,n)=(103,3)}$.

The above solution was posted and copyrighted by cobbler and Andreas. The original thread for this problem can be found here: [1] and [2]

Video Solution

https://www.youtube.com/watch?v=SRl4Wnd60os

See Also

1978 IMO (Problems) • Resources
Preceded by
First question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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