If k is a positive integer, What is the remainder when 2^k is divided

General approach is correct, though the red parts are not.

The last digit of 2^k repeats in pattern of 4 (cyclicity is 4):
2^1=2 --> last digit is 2;
2^2=4 --> last digit is 4;
2^3=8 --> last digit is 8;
2^4=16 --> last digit is 6;
2^5=32 --> last digit is 2 again;

Now, when k itself is a multiple of 4 (when there is no remainder upon division k by cyclicity number), then the last digit will be the last digit of 2^4 (4th in pattern), so 6 not 1 (taking 2^0) as you've written.

If k is a positive integer, what is the remainder when 2^k is divided by 10?

Notice that all we need to know to answer the question is the last digit of 2^k.

(1) k is divisible by 10 --> different multiples of 10 yield different remainders upon division by 4 (for example 10/4 yields 2 and 20/4 yields 0), thus we can not get the single numerical value of the last digit of 2^k. Not sufficient.

(2) k is divisible by 4 --> as discussed, when k is a multiple of 4, the last digit of 2^k equals to the last digit of 2^4, which is 6. Integer ending with 6 yields remainder of 6 upon division by 10. Sufficient.

Answer: B.

Hope it's clear.
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Understand. Thank u bunnel.
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2^k divided by 10. The cycliicity of 2 when divided by 10 is 4.

1 - k is divisible by 10 - the number can be 10 (2) or 20(0) - Not Sufficient
2 - k is divisible by 4 - Sufficient.

Ans. B

To add some clarity for myself and viewers:

Since the last digit in 2^k repeats in cycles of 4, you will ALWAYS know the last digit (and remainder) if k is a multiple of 4.

Therefore 2^4, 2^8,2^12. 2_16, etc.... will always have a last digit of 6.

If k is a multiple of 10, you know if k = 10, the last digit will be 4, and if k=20 the last digit will be 6, k=30 the last digit will be 4, etc... in repeating pattern. However without knowing the exact value of k you won't know the remainder.

remainder by 10 means units digit.

1) k is div by 10
k = 10 ; 2^10 ends in 4
k = 20 ; 2^20 ends in 6
insufficient.

2) k is div by 4
2^(4k) always ends in 6
sufficient.

B.
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here the trick is to realise that the cylicity of 2 => Four
hence statement 2 is sufficient and the remainder will be always => 6
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\(2^1= unit \ digit \ 2\)

\(2^2=unit \ digit\ 4\)

\(2^3=unit \ digit\ 8\)

\(2^4=unit \ digit\ 6\)

\(2^5=unit \ digit\ 2\)

The cyclicity is \(4\)

(1) If \(k=10\), \(\frac{10}{4}\)= remainder 2; \(\frac{2^{10}}{10}\) remainder will 4

If \(k =20, \frac{20}{5}\)= No Remainder; as there is no remainder upon division k by cyclicity number, then the last digit will be the last digit of 2^4 (4th in the pattern), so 6 is remainder when \(\frac{2^4}{10}.\)

Two different answers 4 and 6, So \(Insufficient. \)

(2) K is a multiple of 4 (the cyclicity). So, as there is no remainder upon division k by cyclicity number, then the last digit will be the last digit of 2^4 (4th in the pattern), so 6 is remainder when \(\frac{2^4}{10}.\) or \(\frac{2^8}{10}.\) \(Sufficient \)

The answer is \(B\)
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