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05 Mar 2025, 04:33
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Try to break down 4 such that one term is a multiple of 6, we can write 4 as => 6 - 2
Now we need to find remainder of \(\frac{(6 - 2)^{96} }{ 6}\)
Using binomial theorem: \((6 - 2)^{96}\) = \({96_C_0} * 6^{96} * (-2)^0\) + \({96_C_1} * 6^{95} * (-2)^1\) + \({96_C_2} * 6^{94} * (-2)^2\) + .................... + \({96_C_{95}} * 6^{1} * (-2)^{95}\) + \({96_C_{96}} * 6^{0} * (-2)^{96}\)
When above is divided by 6, we are left with \(\frac{(-2)^{96} }{ 6}\) = \(\frac{2^{96} }{ 6}\)
Now, using cyclicity of 2 divided by 6 =>
\(\frac{2^1}{6}\) = remainder 2
\(\frac{2^2}{6}\) = remainder 4
\(\frac{2^3}{6}\) = remainder 2
\(\frac{2^4}{6}\) = remainder 4
We can see that when even powers of 2 are divided by 6, we get a remainder 4. Therefore, \(\frac{2^{96} }{ 6}\) = remainder 4
Using binomial theorem is just an additional step here, a more efficient way to do this problem is writing \(4^{96}\) as \(2^{192}\) and checking cyclicity to find remainder.
