Difference between revisions of "2008 iTest Problems/Problem 48"
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We have <math>\underbrace{111 \cdots 1}_{n\ \text{ones}} = 10^{n-1} + 10^{n-2} + \cdots + 1 = \frac{10^n - 1}{9}</math>. Since <math>9</math> and <math>97</math> are [[relatively prime]], it suffices for <math>10^{n} - 1 \equiv 0 \pmod{97}</math>, or <math>10^{n} \equiv 1 \pmod{97}</math>. | We have <math>\underbrace{111 \cdots 1}_{n\ \text{ones}} = 10^{n-1} + 10^{n-2} + \cdots + 1 = \frac{10^n - 1}{9}</math>. Since <math>9</math> and <math>97</math> are [[relatively prime]], it suffices for <math>10^{n} - 1 \equiv 0 \pmod{97}</math>, or <math>10^{n} \equiv 1 \pmod{97}</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>. | + | 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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+ | {{2008 iTest box|num-b=47|num-a=49}} | ||
== See also == | == See also == | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] |
Revision as of 19:33, 12 July 2018
Problem
A repunit is a natural number whose digits are all . For instance,
are the four smallest repunits. How many digits are there in the smallest repunit that is
divisible by ?
Solution
We have . Since
and
are relatively prime, it suffices for
, or
.
The smallest integer such that
where
is
, where
is the Carmichael function. For primes
,
, and so the answer is
.
2008 iTest (Problems) | ||
Preceded by: Problem 47 |
Followed by: Problem 49 | |
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