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If N = 4^31 + 1, what is the remainder when N is divided by 3?

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If N = 4^31 + 1, what is the remainder when N is divided by 3?  [#permalink]

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New post 26 Apr 2016, 23:05
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

83% (00:40) correct 17% (01:02) wrong based on 169 sessions

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Re: If N = 4^31 + 1, what is the remainder when N is divided by 3?  [#permalink]

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New post 27 Apr 2016, 01:21
4
1
Bunuel wrote:
If N = 4^31 + 1, what is the remainder when N is divided by 3?

A. 0
B. 1
C. 2
D. 3
E. 4


Hi,

two methods--

1) JUST looking at CHOICES--
a) remainder of 0 = remainder of 3..
we cannot have two answers A and D..
we can eliminate A and D..
b) remainder of 1 = remainder of 4 or 4 cannot be remainder so it also means 4-3 =1
again two answers B and E..
c) ONLY C is left
ans C

2) Binomial expression..
\(4^{31} = (3+1)^{31}\), this will give us remainder 1 when div by 3..
so \(4^{31} + 1\) will give \(1+1 = 2\) as remainder..
ans C
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Re: If N = 4^31 + 1, what is the remainder when N is divided by 3?  [#permalink]

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New post 27 Apr 2016, 02:14
Bunuel wrote:
If N = 4^31 + 1, what is the remainder when N is divided by 3?

A. 0
B. 1
C. 2
D. 3
E. 4


4^31 + 1 = (3+1)^31 + 1
On expanding, (3+1)^31 will have powers of 3 in all elements apart from 1
Therefore (3+1)^31 will leave a remainder 1

Hence (3+1)^31 + 1 will leave a remainder 2

COrrect Option: C
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Re: If N = 4^31 + 1, what is the remainder when N is divided by 3?  [#permalink]

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New post 27 Apr 2016, 07:52
1
2
Bunuel wrote:
If N = 4^31 + 1, what is the remainder when N is divided by 3?

A. 0
B. 1
C. 2
D. 3
E. 4


\(4^1\) / 3 Will have remainder 1
\(4^2\) / 3 Will have remainder 1
\(4^3\) / 3 Will have remainder 1
\(4^4\) / 3 Will have remainder 1

Thus 4^ { Any number } when divided by 3 will have the same remainder 1

So,

\(4^31\) / 3 ; Will have remainder 1

1/3 ; Will have remainder 1

Finally (1 + 1 ) / 3 will have remainder as 2

Thus answer will be (C)
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Re: If N = 4^31 + 1, what is the remainder when N is divided by 3?  [#permalink]

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New post 28 Dec 2016, 22:48
I have solved it with different strategy so sharing that too-

4^1=4
4^2=16
4^3=64
4^4=256
.
.
4^31 = ..4

so taking only the unit place in consideration 4+1=5/3 = 2

I have recently started the preparation so please share your thoughts. Also correct me if this approach is not right. Thanks :)
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Re: If N = 4^31 + 1, what is the remainder when N is divided by 3?  [#permalink]

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New post 17 Dec 2018, 00:13
ashishsuman wrote:
I have solved it with different strategy so sharing that too-

4^1=4
4^2=16
4^3=64
4^4=256
.
.
4^31 = ..4

so taking only the unit place in consideration 4+1=5/3 = 2

I have recently started the preparation so please share your thoughts. Also correct me if this approach is not right. Thanks :)


Hi,

I don't think this is the right approach because divisibility test of 3 says that *Sum* of all digits should be divisible by 3, and not just the last digit. Your approach would be correct if we were checking divisibility by 2 where we only need to focus on last digit.

Hence, we need to write 4^31 as (3 + 1)^31 and then find the answer using binomial expansion, or by alternative approaches that others have mentioned.

Hope this makes sense!

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Re: If N = 4^31 + 1, what is the remainder when N is divided by 3? &nbs [#permalink] 17 Dec 2018, 00:13
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If N = 4^31 + 1, what is the remainder when N is divided by 3?

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