How many different ways are there to write 2004 as a sum of one or more positive integers which are all "aproximately equal" to each other? Two numbers are called aproximately equal if their difference is at most 1. The order of terms does not matter: two ways which only differ in the order of terms are not considered different.
Problem
Source: Tournament of towns, Junior B-Level paper, Fall 2004
Tags: modular arithmetic, number theory unsolved, number theory
paganinio
25.12.2004 16:56
"aproximately equal" to each other? then
it must be like n, n+1, n, n+1, ...
so that n*m+(n+1)*m=2004
m(2n+1)=2004 (*)
list all possible 'm's, you will got the result.
then we look at how many 'm's there are.
2004=2*2*3*167
m could be 1, 2, 3, 167, 4, 6, 2*167, 3*167, 12, 2*2*167, 6*167, 2004
then put them into (*) one by one, see if n and n+1 are both positive integers.
paganinio
25.12.2004 17:06
paganinio wrote: "aproximately equal" to each other? then
it must be like n, n+1, n, n+1, ...
so that n*m+(n+1)*m=2004
m(2n+1)=2004 (*)
list all possible 'm's, you will got the result.
then we look at how many 'm's there are.
2004=2*2*3*167
m could be 1, 2, 3, 167, 4, 6, 2*167, 3*167, 12, 2*2*167, 6*167, 2004
then put them into (*) one by one, see if n and n+1 are both positive integers.
here's a serious mistake
2n+1 is an odd number
so (*) should be n*m+(n+1)*(m+1)=2004
therefore 2mn+m+n=2003
then find integer solutions for this equation.
Xixas
26.12.2004 20:41
Why do you assume, that there are m terms equal to n in the sum and m+1 (or m in your first post) terms equal to n+1? Can't there be m terms equal to n and m+k terms equal to n+1 for some integer k, and then nm + (n+1)(m+k)=2004 ? Sorry, if I missed something trivial.
TripleM
27.12.2004 04:08
Lets say there are K terms, with a being equal to n+1 and the rest equal to n. We can assume there must be at least one term equal to n (otherwise we count n+n+n.. again as (n+1)+(n+1) + ..) Our sum is $nK+a=2004$. So for any given K, $0\le a<K$, but also $a\equiv 2004 \pmod{K}$, so there is exactly one value for a. Thus, since K can be from 1 to 2004, there are 2004 solutions.