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Cyclic property question

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by mathewmithun » Sun Oct 21, 2012 10:54 am
I am aware of cyclic principle and their use in finding remainders or units digit. I tried to use this method to find the remainder of (8^36)/6.
the cyclic value of 8 is 4 that is 8^4 gives unit digit 6 and when we multiply another 8 that is 8^5, the unit digit repeats.
So i rewrite 8^36 as 8^(4*9) which is (XXXXX6)^9 as 8^4 gives us a a number with unit digit as 6. So what is the remainder when (xxxxx6)^9 is divided by 6 and I am stuck. Some one pls help me....
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by sanrisenew » Sun Oct 21, 2012 11:12 am
Why r u splitting 36 into 4*9, 8^36= XXXXXXXX6 after the repetition of 9 cycles.
I will approach this problem differently
8^36=( 6+2)^36 and now doing its binomial expansion and looking for remainder all members except 2^36 are divisible by 6
now 2^36=8^12= ( 6+2)^12 repeating the same we will get 2^12
now 2^12 = 8^4 = ( 6 +2)^4 and remainder will be 2^4, now 2^4 is 16 and when 16 is divided by 6 remainder will be 4(ans).
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by mathewmithun » Sun Oct 21, 2012 11:20 am
sanrisenew wrote:Why r u splitting 36 into 4*9, 8^36= XXXXXXXX6 after the repetition of 9 cycles.
I will approach this problem differently
8^36=( 6+2)^36 and now doing its binomial expansion and looking for remainder all members except 2^36 are divisible by 6
now 2^36=8^12= ( 6+2)^12 repeating the same we will get 2^12
now 2^12 = 8^4 = ( 6 +2)^4 and remainder will be 2^4, now 2^4 is 16 and when 16 is divided by 6 remainder will be 4(ans).
Wow, I had studies binomial expansion, but it simply did not cross my mind. Now I need to brush up binomial expansion once again.Thank you sanrisenew for the reply.
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