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MBA.Aspirant
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winniethepooh
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Frankenstein
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Hi,
3^64 = 3.3^63 = 3.(3^3)^21 = 3.(27)^21
=3.(28-1)^21
(28-1)^21 when expanded(binomial expansion) gives 28^21 + 21C1.(28^20)(-1)+...+28C27.(28).(-1)^20+(-1)^21
So, every term of this sum is a multiple of 28 except the last term i.e. (-1)^21 which is (-1)
So, 3(28-1)^21 = 3(28*p - 1) = 3.28p - 3 = 28(3p-1)+25
So, remainder is 25
3^64 = 3.3^63 = 3.(3^3)^21 = 3.(27)^21
=3.(28-1)^21
(28-1)^21 when expanded(binomial expansion) gives 28^21 + 21C1.(28^20)(-1)+...+28C27.(28).(-1)^20+(-1)^21
So, every term of this sum is a multiple of 28 except the last term i.e. (-1)^21 which is (-1)
So, 3(28-1)^21 = 3(28*p - 1) = 3.28p - 3 = 28(3p-1)+25
So, remainder is 25
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
