Yesterday I solved this problem.
You can find a nice analysis of the solution here.
A Modified Version:
You can find a nice analysis of the solution here.
A Modified Version:
Find the multiplication of all the divisors of (2004)x modulo 29
Let D be the total number of divisors of a number, N = p1^a1 * p2^a2 .... pn^an
ReplyDeleteThen D = (a1+1) (a2+1) .... (an+1)
Multiplication of all divisors,
M = p1 ^ (a1D/2) * p2 ^ (a2D/2) * .... * pn ^ (anD/2)
We can use this formula to solve the problem.
nice! :)
ReplyDeleteHere is another version of the formula for M:
if D is even: (N^x is not a square number)
M = N^(D/2)
if D is odd: (N^x is a square number)
M = (N^(D/2))*(sq) [ here, sq = sqrt(N^x) ]
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ReplyDeletegreat :)
ReplyDelete