Does there exist an positive integer $n$, so that for any positive integer $m<1002$, there exists an integer $k$ so that \[\displaystyle \frac{m}{1002} < \frac{k}{n} < \frac {m+1}{1003}\] holds? If $n$ does not exist, prove it; if $n$ exists, determine the minimum value of it. I know this problem was easy, but it still appeared on our TST, and so I posted it here.
Problem
Source: Taiwan 1st TST 2005, 1st independent study, problem 2
Tags: algebra solved, algebra
ZetaX
12.08.2005 16:51
Indeed setting $m=1002$ will give a contradiction. But if $m$ has to be smaller than $1002$ it's as good as the same problem we had on our TST's: http://www.mathlinks.ro/Forum/viewtopic.php?t=39365
k2c901_1
12.08.2005 17:18
Actually, there was the assumption that $m<1002$. I seem to have omitted it in my original post. Sorry.
pywolf
24.08.2005 00:00
i think n does exist: since m/1002 and m/1003 are both rational numbers, there exist: 1/2{(m/1002)+[(m+1)/1003]} at the midpoint. and assume the smallest common denominator of 1002 & 1003 is 1002*1003 = 1005006. therefore n= 2010012. but i m not sure about the min value for n. then for all integer m < 1002 there exist an integer k = 2005m+1002
pywolf
25.08.2005 03:20
for the minimum value of n, i think this might work. since the distance between m/1002 and (m+1)/1003 is: (m+1)/1003 - m/1002 == (1002-m) / (1003*1002). and by given m < 1002, therefore (1002-m) / (1003*1002) always > 0 and 1002 = 2*3*167, 1003=17*59, so if we can find the factors for 1002-m. i chose m = 3*167 then 1002-m= 2*3*167 - (3*167) = (3*167)*(2-1) = 3*167 1003*1002=17*59*2*3*167 ==> there we can cancell 3*167 n = 17*59*2 = 2006 ... i think this is the minimum value of n.