Difference between revisions of "2008 iTest Problems/Problem 48"
(solution) |
Rockmanex3 (talk | contribs) m |
||
| (One intermediate revision by the same user not shown) | |||
| Line 10: | Line 10: | ||
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>. |
== See also == | == See also == | ||
| + | {{2008 iTest box|num-b=47|num-a=49}} | ||
| + | |||
[[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 
. For instance,
![\[1,11,111,1111, \ldots\]](http://latex.artofproblemsolving.com/6/3/8/6386858630e6455adad0971f9ba1465e78468343.png)
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 
.
See also
| 2008 iTest (Problems) | ||
| Preceded by: Problem 47 |
Followed by: Problem 49 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100 | ||
