# Difference between revisions of "2008 iTest Problems/Problem 94"

## 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

 2008 iTest (Problems) Preceded by:Problem 93 Followed by:Problem 95 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
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