If N is a positive integer what is the remainder when 2^N

Asked: If N is a positive interger what is the remainder when \(2^N\) is divided by 10?

1. N is divisible by 2
N = 2k
2^N = 2^{2k}
The remainder when \(2^N\) is divided by 10 = {4,6}
NOT SUFFICIENT

2. N is divisible by 4
N = 4k
2^N = 2^{4k}
The remainder when \(2^N\) is divided by 10 = 6
SUFFICIENT

IMO B

The question says that If N is a positive integer, so it cannot be 0.

Missed to note this, if like Bunuel pointed out Positive integer wasnt mentioned i would have gotten this question wrong .. Kudos to you for thinking it..

If N is a positive integer what is the remainder when \(2^N\) is divided by 10?
This question reduces to what is unit digit of \(2^N\)

1. N is divisible by 2
if N = 2, unit digit of \(2^N\)= 4, if n = 4, unit digit of \(2^N\) = 6. So \(NOT SUFFICIENT.\)
2. N is divisible by 4
If N = 4n, unit digit of \(2^N\) = 6. hence \(SUFFICIENT.\)

Answer B

Calculating unit digit:
a^(4n+r) has the same unit digit as \(a^r\), if exponent is multiple of 4, find unit digit of \(a^4\)
Principle to calculate unit digit.
1) divide the exponent by 4, and find remainder.
2) unit digit of expression will be the unit digit of base raised to power of remainder.
If remainder is 0. Raise to power of 4.

THIS METHOD IS APPLICABLE FOR ALL DIGITS, no need to remember cyclicity for different digits.

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