sandman13
PKN
pattern of hundreds digit of \((201)^n\):-
2-4-6-8-0
Cyclicity:- 5
PKN, how do you know the pattern is 2-4-6-8-0?
I solved this using binomial expansion.
If you think of 201^201 = (200+1)^201 = 201C0*200^201 + 201C1*200^200 + ... +
201C200*200^1 + 201C201*200^0
^the highlighted term equal 201*200 = 40
200
^the highlighted part is the hundreds digit. All other terms will have higher powers and won't contribute to the hundreds term.
Answer: CHi
sandman13,
If you observe carefully, We have each of the digit of 201 , lies between 0 and 2 inclusive. On successive powers of 201 from 1 to infinity, the tens digit of 201 decides the hundreds digit(1*any number=that number.no carry forward)). Here the tens digit is zero, hence 0*any number=0 with no carry forward.
Hundreds digit of \(201^1\)=2
Hundreds digit of \(201^2\)= 2+Hundreds digit of \(201^1=2+2=4\)
Hundreds digit of \(201^3\)= 2+Hundreds digit of \(201^2=2+4=6\)
Hundreds digit of \(201^4\)= 2+Hundreds digit of \(201^3=2+6=8\)
Hundreds digit of \(201^5\)= 2+Hundreds digit of \(201^4=2+8=0\)
Hundreds digit of \(201^6\)= 2+Hundreds digit of \(201^5=2+0=2\)
Hundreds digit of \(201^7\)= 2+Hundreds digit of \(201^6=2+2=4\)
This was my approach, it mayn't be a standard one.