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Bunuel
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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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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ashishsuman
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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Bunuel
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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Bunuel
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.
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minirana
Bunuel
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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Bunuel
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:)
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