Difference between revisions of "Talk:2012 USAMO Problems/Problem 3"

Revision as of 16:14, 3 May 2012

The answer is the set of all integers that are at least 3.

For composite n where there are two primes p_1 and p_2 such that n/2<p_1<p_2<n, here's your construction:

Pick maximal integers j_1 and j_2 such that $((p_1)^(j_1))((p_2)^(j_2))$ divides i.

Pick a minimal positive integer s such that $(n(n+1)/2)+(s-1)(p_1)$ is 0 mod p_2. (You know it exists since p_1 and p_2 are relatively prime.)

Pick an integer t such that $(n(n+1)/2)+(s-1)(p_1)+(t-1)(p_2)=0$ . (It exists because of how we defined s. It also must be negative.)

Then $a_i=(s^(j_1))(t^(j_2))$ .

For n=4:

$a_i=(-1)^(j_1+j_2)$ , where $(2^j_1)(3^j_2)$ divides i.

For n=6:

$a_i=(2^j_1)(-5)^j_2$ , where $(3^j_1)(5^j_2)$ divides i.

For n=10:

$a_i=(2^j_1)(-9)^j_2$ , where $(5^j_1)(7^j_2)$ divides i.

[I don't know LaTeX, so someone else can input it.]

--Mage24365 09:00, 25 April 2012 (EDT)

@@ Line 3: / Line 3: @@
 For composite n where there are two primes p_1 and p_2 such that n/2<p_1<p_2<n, here's your construction:
-Pick maximal integers j_1 and j_2 such that ((p_1)^(j_1))((p_2)^(j_2)) divides i.
+Pick maximal integers j_1 and j_2 such that<math>((p_1)^(j_1))((p_2)^(j_2))</math> divides i.
-Pick a minimal positive integer s such that (n(n+1)/2)+(s-1)(p_1) is 0 mod p_2. (You know it exists since p_1 and p_2 are relatively prime.)
+Pick a minimal positive integer s such that <math> (n(n+1)/2)+(s-1)(p_1)</math> is 0 mod p_2. (You know it exists since p_1 and p_2 are relatively prime.)
-Pick an integer t such that (n(n+1)/2)+(s-1)(p_1)+(t-1)(p_2)=0. (It exists because of how we defined s. It also must be negative.)
+Pick an integer t such that<math> (n(n+1)/2)+(s-1)(p_1)+(t-1)(p_2)=0</math>. (It exists because of how we defined s. It also must be negative.)
-Then a_i=(s^(j_1))(t^(j_2)).
+Then <math>a_i=(s^(j_1))(t^(j_2))</math>.
 For n=4:
-a_i=(-1)^(j_1+j_2), where (2^j_1)(3^j_2) divides i.
+<math>a_i=(-1)^(j_1+j_2)</math>, where<math> (2^j_1)(3^j_2) </math>divides i.
 For n=6:
-a_i=(2^j_1)(-5)^j_2, where (3^j_1)(5^j_2) divides i.
+<math>a_i=(2^j_1)(-5)^j_2</math>, where <math>(3^j_1)(5^j_2)</math>divides i.
 For n=10:
-a_i=(2^j_1)(-9)^j_2, where (5^j_1)(7^j_2) divides i.
+<math>a_i=(2^j_1)(-9)^j_2</math>, where <math>(5^j_1)(7^j_2)</math>divides i.
 [I don't know LaTeX, so someone else can input it.]
 --[[User:Mage24365|Mage24365]] 09:00, 25 April 2012 (EDT)