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Math Expert V
Joined: 02 Sep 2009
Posts: 65290
If N = 4^31 + 1, what is the remainder when N is divided by 3?  [#permalink]

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Difficulty:   15% (low)

Question Stats: 78% (01:05) correct 22% (01:19) wrong based on 232 sessions

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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

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Math Expert V
Joined: 02 Aug 2009
Posts: 8757
Re: If N = 4^31 + 1, what is the remainder when N is divided by 3?  [#permalink]

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6
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

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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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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2
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

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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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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 _________________
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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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!

Posted from my mobile device
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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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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

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

N-1mod3 = 4^31mod3 = 1^31mod3 = 1
Nmod3 = 1+1 = 2

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

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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

Hope the method i have used is correct:

Given : N= 4^31 +1
Using cyclicity 4 is 2. That means the unit digit will be 4.

Thus , 4+1 = 5
Which when divided by 3 gives us the remainder 2.

Ans. C.
Math Expert V
Joined: 02 Aug 2009
Posts: 8757
Re: If N = 4^31 + 1, what is the remainder when N is divided by 3?  [#permalink]

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minirana wrote:
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

Hope the method i have used is correct:

Given : N= 4^31 +1
Using cyclicity 4 is 2. That means the unit digit will be 4.

Thus , 4+1 = 5
Which when divided by 3 gives us the remainder 2.

Ans. C.

No, you cannot use units digit for finding remainder when divided by 3.
Units digit is helpful in finding remainders when divided by 2, 5, and 10.
Last two digit when divided by 2^2 or 4
Sum of digits when divided by 3 and 9. Example 4753=4+7+5+3=19 so remainder when divided by 3 or 9 is 1 as 18=18+1.
Sum of alternate digits when divided by 11.
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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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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*4^(4*7)+2]
[4*4^(4*7)*16]
when divided by 3, remainder will multiply
1*1*1=1
and when 1 divided by 3 remainder is 1
so 1+1=2
C:) Re: If N = 4^31 + 1, what is the remainder when N is divided by 3?   [#permalink] 31 May 2020, 20:46

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