Difference between revisions of "2008 iTest Problems/Problem 94"
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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 | + | ==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 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.
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 93 |
Followed by: Problem 95 | |
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