let the 3 terms be: a,a*r,a*r^2
then their sum is:
a+a*r+a*r^2=a*(1+r+r^2)=111
r^2+r+1=111/a=v
r^2+r+1v=0
r=(1+sqrt(14(1v)))/2
r=(1+sqrt(4v3))/2
now for each of a,a*r,a*r^2 to be integers, we need r to be rational.
thus for r to be real
4v3>=0
v>=3/4
111/a>=3/4
0<a<=148
for each of these a's, we get 2 possible values for r. And for each of
these pairs of (a,r) we get a possible sequence. Below are all the
pairs of (a,r) with their corresponding sequences. I have already
eliminated the (a,r)'s where r is not rational.
(1,10) : {1,10,100}
(1,11) : {1,11,121}
(27,4/3) : {27,36,48}
(27,7/3) : {27,63,147}
(37,1) : {37,37,37}
(37,2) : {37,74,148}
(48,3/4) : {48,36,27}
(48,7/4) : {48,84,147}
(100,1/10) : {100,10,1}
(100,11/10) : {100,110,121}
(111,0) : {111,0,0}
(111,1) : {111,111,111}
(121,1/11) : {121,11,1}
(121,10/11) : {121,110,100}
(147,3/7) : {147,63,27}
(147,4/7) : {147,84,48}
(148,1/2) : {148,74,37}
now we can further eliminate sequences which either contain zero or
a duplicate value. Which gives us:
(1,10) : {1,10,100}
(1,11) : {1,11,121}
(27,4/3) : {27,36,48}
(27,7/3) : {27,63,147}
(37,2) : {37,74,148}
(48,3/4) : {48,36,27}
(48,7/4) : {48,84,147}
(100,1/10) : {100,10,1}
(100,11/10) : {100,110,121}
(121,1/11) : {121,11,1}
(121,10/11) : {121,110,100}
(147,3/7) : {147,63,27}
(147,4/7) : {147,84,48}
(148,1/2) : {148,74,37}
this list can be further simplified using the fact that if you have
(a,r) (a,b,c) as a sequence, then
(c,1/r) (c,b,a) is also a sequence
so by requiring a<c we can narrow it down to the following solutions:
(1,10) : {1,10,100}
(1,11) : {1,11,121}
(27,4/3) : {27,36,48}
(27,7/3) : {27,63,147}
(37,2) : {37,74,148}
(48,7/4) : {48,84,147}
(100,11/10) : {100,110,121}

Posted by Daniel
on 20130808 10:55:54 