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555-605 Level|   Remainders|      
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If k is a positive integer, What is the remainder when 2^k is divided Updated on: 20 Oct 2014, 08:35
Originally posted by MHIKER on 23 Jan 2012, 21:55.
Last edited by Bunuel on 20 Oct 2014, 08:35, edited 2 times in total.
Edited the question.
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68% (01:35) correct 32% (01:47) wrong based on 2203 sessions

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

(1) k is divisible by 10
(2) k is divisible by 4

My approach is as follows:
Show Spoiler
(1) k could be 10, 20, 30...
case i. if k = 10, 2^10, the cyclicity of 2 is 4 (10/4 = reminder 2) so 2^2 is divided by 10 and reminder is 4
case ii. if k = 20, 2^20, the cyclicity of 2 is 4 (20/4 = 5, 5/4 = reminder 1) so 2^1 is divided by 10 and reminder is 2
Insufficient.

(2) k = 4, 8, 12
2^4, the cyclicity of 2 is 4 (4/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
2^8, the cyclicity of 2 is 4 (8/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
Sufficient.

Ans. B

Please help whether the above approach can be applied in the problem?
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B
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If k is a positive integer, What is the remainder when 2^k is divided 24 Jan 2012, 02:05
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Baten80
If k is a positive integer, What is the remainder when 2^k is divided by 10?
1) k is dividable by 10
2) k is dividable by 4

My approach is as follows:
(1) k could be 10, 20, 30...
case i. if k = 10, 2^10, the cyclicity of 2 is 4 (10/4 = reminder 2) so 2^2 is divided by 10 and reminder is 4
case ii. if k = 20, 2^20, the cyclicity of 2 is 4 (20/4 = 5, 5/4 = reminder 1) so 2^1 is divided by 10 and reminder is 2
Insufficient.

(2) k = 4, 8, 12
2^4, the cyclicity of 2 is 4 (4/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
2^8, the cyclicity of 2 is 4 (8/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
Sufficient.

Ans. B

Please help whether the above approach can be applied in the problem?
Show more

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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If k is a positive integer, What is the remainder when 2^k is divided 25 Jan 2012, 09:57
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Understand. Thank u bunnel.
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If k is a positive integer, What is the remainder when 2^k is divided 06 Jun 2013, 06:10
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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If k is a positive integer, What is the remainder when 2^k is divided 20 Oct 2014, 08:27
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If k is a positive integer, What is the remainder when 2^k is divided by 10?

(1) k is divisible by 10
(2) k is divisible by 4

Show more

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
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If k is a positive integer, What is the remainder when 2^k is divided 20 Oct 2014, 09:31
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Bunuel
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If k is a positive integer, What is the remainder when 2^k is divided by 10?
1) k is dividable by 10
2) k is dividable by 4

My approach is as follows:
(1) k could be 10, 20, 30...
case i. if k = 10, 2^10, the cyclicity of 2 is 4 (10/4 = reminder 2) so 2^2 is divided by 10 and reminder is 4
case ii. if k = 20, 2^20, the cyclicity of 2 is 4 (20/4 = 5, 5/4 = reminder 1) so 2^1 is divided by 10 and reminder is 2
Insufficient.

(2) k = 4, 8, 12
2^4, the cyclicity of 2 is 4 (4/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
2^8, the cyclicity of 2 is 4 (8/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
Sufficient.

Ans. B

Please help whether the above approach can be applied in the problem?

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.
Show more

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.
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If k is a positive integer, What is the remainder when 2^k is divided 02 Nov 2014, 02:22
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If k is a positive integer, What is the remainder when 2^k is divided by 10?

(1) k is divisible by 10
(2) k is divisible by 4

My approach is as follows:
Show Spoiler
(1) k could be 10, 20, 30...
case i. if k = 10, 2^10, the cyclicity of 2 is 4 (10/4 = reminder 2) so 2^2 is divided by 10 and reminder is 4
case ii. if k = 20, 2^20, the cyclicity of 2 is 4 (20/4 = 5, 5/4 = reminder 1) so 2^1 is divided by 10 and reminder is 2
Insufficient.

(2) k = 4, 8, 12
2^4, the cyclicity of 2 is 4 (4/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
2^8, the cyclicity of 2 is 4 (8/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
Sufficient.

Ans. B

Please help whether the above approach can be applied in the problem?
Show more

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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If k is a positive integer, What is the remainder when 2^k is divided 22 Mar 2016, 02:47
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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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If k is a positive integer, What is the remainder when 2^k is divided 07 Dec 2020, 14:13
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MHIKER
If k is a positive integer, What is the remainder when 2^k is divided by 10?

(1) k is divisible by 10

(2) k is divisible by 4
Show more

\(2^1= unit \ digit \ 2\)

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

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

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

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

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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If k is a positive integer, What is the remainder when 2^k is divided 28 May 2025, 03:46
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is this approach ok Bunuel

k is divisible by 10 so
2^10 = 1024
so remainder 4
2^ 20 = 1024*1024 = units = 6 so remainder 6

hence in suff.

(2)
k is divisible by 4 always units is 6, hence remainder always 6
MHIKER
If k is a positive integer, What is the remainder when 2^k is divided by 10?

(1) k is divisible by 10
(2) k is divisible by 4

My approach is as follows:
Show Spoiler
(1) k could be 10, 20, 30...
case i. if k = 10, 2^10, the cyclicity of 2 is 4 (10/4 = reminder 2) so 2^2 is divided by 10 and reminder is 4
case ii. if k = 20, 2^20, the cyclicity of 2 is 4 (20/4 = 5, 5/4 = reminder 1) so 2^1 is divided by 10 and reminder is 2
Insufficient.

(2) k = 4, 8, 12
2^4, the cyclicity of 2 is 4 (4/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
2^8, the cyclicity of 2 is 4 (8/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
Sufficient.

Ans. B

Please help whether the above approach can be applied in the problem?
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