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Difference between revisions of "2008 iTest Problems/Problem 48"

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The smallest integer <math>m</math> such that <math>a^m \equiv 1 \pmod{p}</math> where <math>\text{gcd}\,(a,p)=1</math> is <math>m = \lambda (p)</math>, where <math>\lambda (p)</math> is the [[Carmichael function]]. For primes <math>p</math>, <math>\lambda(p) = p-1</math>, and so the answer is <math>n = \boxed{96}</math>.
 
The smallest integer <math>m</math> such that <math>a^m \equiv 1 \pmod{p}</math> where <math>\text{gcd}\,(a,p)=1</math> is <math>m = \lambda (p)</math>, where <math>\lambda (p)</math> is the [[Carmichael function]]. For primes <math>p</math>, <math>\lambda(p) = p-1</math>, and so the answer is <math>n = \boxed{96}</math>.
  
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== See also ==
 
{{2008 iTest box|num-b=47|num-a=49}}
 
{{2008 iTest box|num-b=47|num-a=49}}
  
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== See also ==
 
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Latest revision as of 16:14, 14 July 2018

Problem

A repunit is a natural number whose digits are all $1$. For instance,

\[1,11,111,1111, \ldots\]

are the four smallest repunits. How many digits are there in the smallest repunit that is divisible by $97$?

Solution

We have $\underbrace{111 \cdots 1}_{n\ \text{ones}} = 10^{n-1} + 10^{n-2} + \cdots + 1 = \frac{10^n - 1}{9}$. Since $9$ and $97$ are relatively prime, it suffices for $10^{n} - 1 \equiv 0 \pmod{97}$, or $10^{n} \equiv 1 \pmod{97}$.

The smallest integer $m$ such that $a^m \equiv 1 \pmod{p}$ where $\text{gcd}\,(a,p)=1$ is $m = \lambda (p)$, where $\lambda (p)$ is the Carmichael function. For primes $p$, $\lambda(p) = p-1$, and so the answer is $n = \boxed{96}$.

See also

2008 iTest (Problems)
Preceded by:
Problem 47 		Followed by:
Problem 49
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