1roots of α5 - 1 = 0 are 1, α, α2, α3, α4
sum of roots = 1 + α + α2 + α3 + α4 = 0
α + α2 + α3 + α4 = -1
log√3 | α + α2 + α3 + α4 - 2α | = log√3 | -3α | = log√3 3
(since |α| = 1 as α is fifth root of unity)
= 2
if α≠1
α5=1
then value of log√3 mod(1+α+α2+α3-2α)
i know we hav to use concept of nth root of unity....but not getting
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9 Answers
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1yeah i was also getting 2....but answer given was 3......may be some printing mistake must be der...
thnx btw[1]
1i m not able to post this q in a sperate thread.....so posting in this only
If m is the number of five-element subsets that can be chosen from the set of the first 14 natural numbers so that atleast two of the five numbers are consecutive. find the last digit of m
this is a doubt
62total 5 digit subsets = 14C5
If there are no consecutives,
let the numbers be 1+a, 1+a+2+k, 1+a+2+k+2+l, 1+a+2+k+2+l+2+m, 1+a+2+k+2+l+2+m+2+n
we want 1+a+2+k+2+l+2+m+2+n<=14
a+k+l+m+n<=14-9
a+k+l+m+n+b=5
so you have to partition 5 into 6 numbers
hence 10C5
now you have to find the last digit of 14C5-10C5
1a dumb question--""so you have to partition 5 into 6 numbers "" din get that[2]
62a+k+l+m+n+b=5
this means that you have to distribute 5 into 6 parts.... a, k, l, m, n and b
so you have 5 dividers and 5 numbers to arrange in a row..
we have solved a lot of such problems... try to see the old posts..
1oh yeah got it now....so silly of me
btw thanx ...... and u r very much helpful......keep helping me like that[1]