21check this- http://www.goiit.com/posts/list/algebra-another-one-very-challenging-indeed-76530.htm
24it is the same soln given by BT....i was looking for something diff
anyways thx eragon for link
21
eragon24 _Retired
·2010-01-15 04:50:59
hmm....well i m not giving bt ts...so i din knew wat soln was given der
btw i got anoder
see regular polygon with 1982 sides will be having 1982 vertices....and we know that the n roots of unity lie on a unit circle forming a regular polygon of n sides
so the vertices of regular polygon with 1982 sides will be roots of the eq Z^{1982}-1=0
similary for regular polygon having 2973 sides
Z^{2973}-1=0
so common vertices r common roots of Z^{1982}-1=0 and Z^{2973}-1=0
and those common roots r roots of teh eq Z^{d}-1=0
wer d=gcd(2973,1982)=991
i think this was wat u wer looking for ......
btw did this really come in bt exam......they seem to be raising der level now
24
eureka123
·2010-01-15 08:55:18
ya ...it came in BT..
they have raised their level by miles [6]
62I dont knwo if it requires any complex nubmers.. but the answer is imply
gcd(1982,2973) for the simple reason that the angle at the center suspended by each side is 2pi/N
here we have 2 values of N
and the common vertices will be for each multiple of 2pi/N and 2pi/n which are equal
Hence the answer is simply 991
