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Difference between revisions of "2022 SSMO Speed Round Problems/Problem 6"

(Created page with "==Problem== Find the smallest odd prime that does not divide <math>2^{75!} - 1</math>. ==Solution== Let this odd prime be <math>p</math>. Note that <math>2^{75!} - 1</math>...")
 
 
(2 intermediate revisions by the same user not shown)
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==Problem==
 
==Problem==
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Find the smallest odd prime that does not divide <math>2^{75!} - 1</math>.
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At the beginning of day <math>1</math>, there is a single bacterium in a petri dish. During each day, each bacterium in the petri dish divides into <math>a>1</math> new bacteria, and <math>b\ge 1</math> bacteria are added to the petri dish (these bacteria do not divide on the day they were added). For example, at the end of day <math>1</math>, there are <math>a+b</math> bacteria in the petri dish. If, at the end of day <math>4</math>, the number of bacteria in the petri dish is a multiple of <math>48</math>, find the minimum possible value of <math>a+b</math>.
  
 
==Solution==
 
==Solution==
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Let this odd prime be <math>p</math>.
 
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Note that <math>2^{75!} - 1</math> is divisible by <math>p</math> if
 
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<cmath>
 
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    2^{75!} \equiv 1 \pmod{p}
 
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</cmath>
 
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or <math>p - 1 \mid 75!</math>.
 
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As such, <math>p</math> is the smallest prime of the form <math>2q + 1</math> where
 
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<math>q > 75</math> is also prime.
 
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This is called a \textit{safe} prime in literature and checking
 
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that <math>\boxed{167}</math> is the first such <math>p</math>.
 

Latest revision as of 19:14, 2 May 2025

Problem

At the beginning of day $1$, there is a single bacterium in a petri dish. During each day, each bacterium in the petri dish divides into $a>1$ new bacteria, and $b\ge 1$ bacteria are added to the petri dish (these bacteria do not divide on the day they were added). For example, at the end of day $1$, there are $a+b$ bacteria in the petri dish. If, at the end of day $4$, the number of bacteria in the petri dish is a multiple of $48$, find the minimum possible value of $a+b$.

Solution

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