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

(Solution: Added official solution (which is less computational and more general than Solution 1))
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Fermat's Little Theorem tells us that for a prime <math>p</math> that is not a divisor of <math>2008</math>, <math>2008^p\equiv 2008\pmod p</math>, so <math>p\mid (2008^p - 2008)</math>.  When <math>p>2008</math>, then <cmath>\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*}</cmath> 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: <cmath>\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.</cmath> The most general piece we have to work with is the summation <math>\sum_{k=0}^{p-2}2008^k</math>.  We look for some way to determine which primes <math>q</math> can be factors of this expression.  So long as <math>\gcd(q,2007\cdot 2008) = 1</math>, then by Fermat's Little Theorem, <math>q</math> divides <cmath>2008^{q-1} - 1 = (2008-1)(2008^{q-2} + 2008^{q-3} + \cdots + 2008 + 1),</cmath> and so <cmath>q\mid (2008^{q-2} + 2008^{q-3} + \cdots + 2008 + 1).</cmath> Now we note that <math>(2008^{q-2} + 2008^{q-3} + \cdots + 2008 + 1)</math> divides evenly into <math>\sum_{k=0}^{p-2}2008^k</math> when <math>q-1</math> is a divisor of <math>p-1</math>.  So, if there exists such a prime <math>p</math>, then <math>q</math> is a divisor of some term of the given sequence.  Dirichlet's Theorem guarantees that within the arithmetic sequence <cmath>q,\quad 2q-1,\quad 3q-2,\quad 4q-3,\quad\ldots,</cmath> there are infinitely many primes.  ONe of them, <math>mq-m+1=p</math> implies that <math>p-1 = m(q-1)</math>.  For sufficiently large <math>p</math>, this completes our proof that each prime <math>q</math> that is relatively prime to <math>2007</math> and <math>2008</math> must be a divisor of some term in the given sequence.  The largest prime less than <math>2008</math> is <math>2003</math>, which is our answer.
 
Fermat's Little Theorem tells us that for a prime <math>p</math> that is not a divisor of <math>2008</math>, <math>2008^p\equiv 2008\pmod p</math>, so <math>p\mid (2008^p - 2008)</math>.  When <math>p>2008</math>, then <cmath>\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*}</cmath> 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: <cmath>\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.</cmath> The most general piece we have to work with is the summation <math>\sum_{k=0}^{p-2}2008^k</math>.  We look for some way to determine which primes <math>q</math> can be factors of this expression.  So long as <math>\gcd(q,2007\cdot 2008) = 1</math>, then by Fermat's Little Theorem, <math>q</math> divides <cmath>2008^{q-1} - 1 = (2008-1)(2008^{q-2} + 2008^{q-3} + \cdots + 2008 + 1),</cmath> and so <cmath>q\mid (2008^{q-2} + 2008^{q-3} + \cdots + 2008 + 1).</cmath> Now we note that <math>(2008^{q-2} + 2008^{q-3} + \cdots + 2008 + 1)</math> divides evenly into <math>\sum_{k=0}^{p-2}2008^k</math> when <math>q-1</math> is a divisor of <math>p-1</math>.  So, if there exists such a prime <math>p</math>, then <math>q</math> is a divisor of some term of the given sequence.  Dirichlet's Theorem guarantees that within the arithmetic sequence <cmath>q,\quad 2q-1,\quad 3q-2,\quad 4q-3,\quad\ldots,</cmath> there are infinitely many primes.  ONe of them, <math>mq-m+1=p</math> implies that <math>p-1 = m(q-1)</math>.  For sufficiently large <math>p</math>, this completes our proof that each prime <math>q</math> that is relatively prime to <math>2007</math> and <math>2008</math> must be a divisor of some term in the given sequence.  The largest prime less than <math>2008</math> is <math>2003</math>, which is our answer.
  
== See also ==
+
==See Also==
 +
{{2008 iTest box|num-b=93|num-a=95}}
  
 
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]

Latest revision as of 21:21, 22 November 2018

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
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