2006 iTest Problems/Problem 34

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.

2006 iTest (Problems, Answer Key)
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2006 iTest Problems/Problem 34

Solution

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