Difference between revisions of "2008 iTest Problems/Problem 94"
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== Solution == | == Solution == | ||
+ | ===Solution 1=== | ||
The largest prime number less than <math>2008</math> is <math>2003</math>; we claim that this is the answer. Indeed, we claim that the <math>6007</math>th term divides <math>2003</math>, where <math>6007</math> is prime (and hence [[relatively prime]] to <math>2003</math>). | The largest prime number less than <math>2008</math> is <math>2003</math>; we claim that this is the answer. Indeed, we claim that the <math>6007</math>th term divides <math>2003</math>, where <math>6007</math> is prime (and hence [[relatively prime]] to <math>2003</math>). | ||
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as desired. | as desired. | ||
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+ | ===Solution 2 (Official Solution)=== | ||
+ | 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 == | ||
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] |
Revision as of 14:37, 7 October 2017
Problem
Find the largest prime number less than that is a divisor of some integer in the infinite
sequence
Solution
Solution 1
The largest prime number less than is
; we claim that this is the answer. Indeed, we claim that the
th term divides
, where
is prime (and hence relatively prime to
).
To do so, we claim that
holds, and since is prime the result follows. Indeed,
, where
denotes the fractional part of a number. So
becomes
By Fermat's Little Theorem, we have , so
. Also,
is equivalent to the remainder when
is divided by
, and by Fermat's Little Theorem again, we have
. Hence, equation
reduces to
as desired.
Solution 2 (Official Solution)
Fermat's Little Theorem tells us that for a prime that is not a divisor of
,
, so
. When
, then
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:
The most general piece we have to work with is the summation
. We look for some way to determine which primes
can be factors of this expression. So long as
, then by Fermat's Little Theorem,
divides
and so
Now we note that
divides evenly into
when
is a divisor of
. So, if there exists such a prime
, then
is a divisor of some term of the given sequence. Dirichlet's Theorem guarantees that within the arithmetic sequence
there are infinitely many primes. ONe of them,
implies that
. For sufficiently large
, this completes our proof that each prime
that is relatively prime to
and
must be a divisor of some term in the given sequence. The largest prime less than
is
, which is our answer.