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Difference between revisions of "2006 iTest Problems/Problem 34"

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{{iTest box|year=2006|num-b=33|num-a=35|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}
 
{{iTest box|year=2006|num-b=33|num-a=35|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}
  
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[[Category:iTest]] [[Category:Number Theory Problem]]
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[[Category:iTest]] [[Category:Number Theory Problems]]

Revision as of 15:10, 31 July 2020

For each positive integer $n$ let $S_n$ denote the set of positive integers $k$ such that $n^k-1$ is divisible by $2006$. Define the function $P(n)$ by the rule \[P(n):=\begin{cases}\min(s)_{s\in S_n}&\text{if }S_n\neq\emptyset,\\0&\text{otherwise}.\end{cases}\] Let $d$ be the least upper bound of $\{P(1),P(2),P(3),\ldots\}$ and let $m$ be the number of integers $i$ such that $1\leq i\leq 2006$ and $P(i) = d$. Compute the value of $d+m$.

Solution

We find that the prime factorization of $2006$ is $2*17*59$.

Then we can compute $d = \lambda(2006)$ (where $\lambda$ is the Carmichael function) by Carmichael's theorem: it is $\text{lcm}(\lambda(2), \lambda(17), \lambda(59)) = \text{lcm}(1, 16, 58) = 2^4 * 29 = 464$.

As for solving $P(i) = d$, we must have $i$ odd (otherwise it would not be coprime to $2$), and we must also have $i$ be a primitive root modulo $17$ as well as a primitive root modulo $59$. There are $\phi(\phi(17)) = \phi(16) = 8$ primitive roots modulo $17$ (where $\phi$ is the Euler totient function) and $\phi(\phi(59)) = \phi(58) = (2-1)*(29-1) = 28$ primitive roots modulo $59$. Then we have $m = 1 * 8 * 28 = 224$ by the Chinese remainder theorem, so our answer is $d + m = 464 + 224 = 688$ and we are done.

See also

2006 iTest (Problems, Answer Key)
Preceded by:
Problem 33 		Followed by:
Problem 35
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 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10
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