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> | ||
== See also == | == See also == | ||
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] |
Revision as of 13:37, 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.