Difference between revisions of "2002 USAMO Problems/Problem 5"
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== Problem == | == Problem == | ||
− | Let <math>a, b </math> be integers greater than 2. Prove that there exists a positive integer <math>k </math> and a finite sequence <math>n_1, n_2, \ldots, n_k </math> of positive integers such that <math>n_1 = a</math>, <math>n_k = b </math>, and <math> | + | Let <math>a, b </math> be integers greater than 2. Prove that there exists a positive integer <math>k </math> and a finite sequence <math>n_1, n_2, \ldots, n_k </math> of positive integers such that <math>n_1 = a</math>, <math>n_k = b </math>, and <math>n_in_{i+1} </math> is divisible by <math>n_i + n_{i+1} </math> for each <math>i </math> (<math> 1 \le i \le k </math>). |
== Solutions == | == Solutions == | ||
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[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:39, 28 March 2015
Contents
Problem
Let be integers greater than 2. Prove that there exists a positive integer and a finite sequence of positive integers such that , , and is divisible by for each ().
Solutions
We may say that two integers and are connected (and write ) if there exists such a sequence of integers as described in the problem. For reference, we note that is an equivalence relation: it is reflexive (), symmetric ( implies ), and transitive ( implies ).
Solution 1
Note that for any divisor of some , , so . It follows that in fact , for any nonnegative integer , and
for any positive integer .
For all integers , there exists some integer such that . Let
.
We have
.
But we also have
.
Hence , so , as desired.
Solution 2
We note that for any integer , , for
It follows that for ,
.
Thus all integers greater than 2 are connected.
Solution 3
We note that if and only if
.
Therefore for any divisor of , .
Now, for ,
.
Also,
.
This means that all positive multiples of 4 are connected.
Furthermore, for ,
,
which implies that every even number greater than 2 is connected to a multiple of 4.
Finally, for ,
.
Since is even, this means that all integers greater than or equal to 2 are connected to some even number.
Together, these imply that all integers greater than 2 are connected.
Solution 4
We note that if is a divisor of , then .
We say a positive integer is safe if for all integers , . Note that the product of two safe numbers is also a safe number. Define () to be the smallest divisor of that is greater than 2. We will prove that 2 is safe, by strong induction on . For the case , we have . If , we note that , since must be less than all the divisors of . Thus the inductive hypothesis gives us . Furthermore, we have and , both from our initial note. Thus .
We will now prove that every prime is safe, by strong induction. We have already proven the base case . Now, for odd , is the product of odd primes less than , so is safe. Then we have
.
Thus the induction is complete, and all primes are safe.
We have now shown that all integers greater than 1 are safe. Specifically, and are safe, and .
Solution 5
Similarly to the first solution, we observe that for any integer and any positive integers ,
.
We also note that if , then for any positive integer , , for implies .
It is sufficient to prove that for any , . If is composite, then there exist such that , and
.
If, on the other hand, is prime, then we use strong induction. For the base case, , we note . Now, assuming that and that this holds for all primes less than (we know it holds for all composites), it is sufficient to show that , since is composite and therefore . But since and are both even, it suffices to show that . But this is true, since either composite, or a prime less than , and it is greater than 3, since . Thus the induction is complete.
See also
2002 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.