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2008 iTest Problems/Problem 94

Revision as of 13:37, 7 October 2017 by Djmathman (talk | contribs) (→‎Solution: Added official solution (which is less computational and more general than Solution 1))

Problem

Find the largest prime number less than $2008$ that is a divisor of some integer in the infinite sequence

\[\left\lfloor \frac{2008}{1} \right\rfloor, \left\lfloor \frac{2008^2}{2} \right\rfloor, \left\lfloor \frac{2008^3}{3}\right\rfloor, \left\lfloor \frac{2008^4}{4} \right\rfloor, \cdots\]

Solution

Solution 1

The largest prime number less than $2008$ is $2003$; we claim that this is the answer. Indeed, we claim that the $6007$th term divides $2003$, where $6007$ is prime (and hence relatively prime to $2003$).

To do so, we claim that

\begin{align*} f(6007) \equiv 6007\left\lfloor \frac{2008^{6007}}{6007} \right\rfloor \equiv 0 \pmod{2003} \tag{1} \end{align*}

holds, and since $6007$ is prime the result follows. Indeed, $\left\lfloor \frac{2008^{6007}}{6007} \right\rfloor = \frac{2008^{6007}}{6007} - \left\{\frac{2008^{6007}}{6007}\right\}$, where $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of a number. So $(1)$ becomes

\begin{align*} f(6007) \equiv 2008^{6007} - 6007\left\{\frac{2008^{6007}}{6007}\right\} \pmod{2003} \tag{2} \end{align*}

By Fermat's Little Theorem, we have $2008^{2002} \equiv 1 \pmod{2003}$, so $2008^{6007} \equiv 2008^{2002} \cdot 2008 \equiv 2008 \pmod{2003}$. Also, $6007\left\{\frac{2008^{6007}}{6007}\right\}$ is equivalent to the remainder when $2008^{6007}$ is divided by $6007$, and by Fermat's Little Theorem again, we have $2008^{6007} \equiv 2008 \pmod{6007}$. Hence, equation $(2)$ reduces to

\begin{align*}f(6007)\equiv 2008 - 2008 \equiv 0 \pmod{2003} \end{align*}

as desired.

Solution 2 (Official Solution)

Fermat's Little Theorem tells us that for a prime $p$ that is not a divisor of $2008$, $2008^p\equiv 2008\pmod p$, so $p\mid (2008^p - 2008)$. When $p>2008$, then \begin{align*}\left\lfloor\frac{2008^p}p\right\rfloor &= \left\lfloor\frac{2008^p - 2008}p\right\rfloor \\ &= \frac{2008^p - 2008}p.\end{align*} Now that we have an expression to work with that doesn't involve the floor function, we begin to manipulate it in order to make it useful: \[\frac{2008^p - 2008}p = \dfrac{2008}p(2008 - 1)\sum_{k=0}^{p-2}2008^k = \left(\frac{2007\cdot 2008}p\right)\sum_{k=0}^{p-2}2008^k.\] The most general piece we have to work with is the summation $\sum_{k=0}^{p-2}2008^k$. We look for some way to determine which primes $q$ can be factors of this expression. So long as $\gcd(q,2007\cdot 2008) = 1$, then by Fermat's Little Theorem, $q$ divides \[2008^{q-1} - 1 = (2008-1)(2008^{q-2} + 2008^{q-3} + \cdots + 2008 + 1),\] and so \[q\mid (2008^{q-2} + 2008^{q-3} + \cdots + 2008 + 1).\] Now we note that $(2008^{q-2} + 2008^{q-3} + \cdots + 2008 + 1)$ divides evenly into $\sum_{k=0}^{p-2}2008^k$ when $q-1$ is a divisor of $p-1$. So, if there exists such a prime $p$, then $q$ is a divisor of some term of the given sequence. Dirichlet's Theorem guarantees that within the arithmetic sequence \[q,\quad 2q-1,\quad 3q-2,\quad 4q-3,\quad\ldots,\] there are infinitely many primes. ONe of them, $mq-m+1=p$ implies that $p-1 = m(q-1)$. For sufficiently large $p$, this completes our proof that each prime $q$ that is relatively prime to $2007$ and $2008$ must be a divisor of some term in the given sequence. The largest prime less than $2008$ is $2003$, which is our answer.

See also

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