Determine the smallest positive integer X, such that X is not expressible in the form (2^{P} – 2^{Q})/(2^{R} – 2^{S}), where each of P, Q, R and S is a positive integer.
answer is 11
first I provide (p,q,r,s) values for 1 through 10
1: (1,0,1,0)
2: (2,1,1,0)
3: (2,0,1,0)
4: (3,2,1,0)
5: (4,0,2,0)
6: (3,1,1,0)
7: (3,0,1,0)
8: (4,3,1,0)
9: (6,0,3,0)
10: (5,1,2,0)
Now I shall prove that no positive integers p,q,r,s will give 11
now obviously p>q and r>s so let
p=q+a
r=s+b
then we have
(2^p2^q)/(2^r2^s)=
2^q*(2^a1)/[2^s*(2^b1)]
now 2^a1 is odd so 2^s must divide 2^q for this to be an integer and thus s<=q and we have
2^(qs)*(2^a1)/(2^b1)=11
2^(qs)*(2^a1)=11*(2^b1)
now the right hand side is odd thus qs=0 and q=s and we are left with
2^a1=11*2^b11
2^a=11*2^b10
2^a=2*(11*2^(b1)5)
2^(a1)=11*2^(b1)5
if b1=0 then we have
2^(a1)=115=6
2^(a1)=6
which has no integer solution for a
now if on the other hand we have b1>0 then the right hand side is odd and the only way the left hand side will also be odd is if a1=0 and thus that gives us
1=11*2^(b1)5
6=11*2^(b1)
6/11=2^(b1)
which obviously has no integer solution and thus there can not be any (p,q,r,s) that gives (2^p2^q)/(2^r2^s)=11
and thus 11 is the smallest such x
Edited on November 8, 2009, 6:54 am

Posted by Daniel
on 20091107 17:18:52 