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What is the remainder when
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03 Jan 2020, 06:37
03 Jan 2020, 06:37
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Question Stats:
77% (00:40) correct
22% (00:42) wrong based on 18 sessions
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What is the remainder when \(3^{35}\) is divided by 5 ?
A. 0
B. 1
C. 2
D. 3
E. 4
Kudos for the right answer and explanation
A. 0
B. 1
C. 2
D. 3
E. 4
Kudos for the right answer and explanation
Re: What is the remainder when
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03 Jan 2020, 06:53
03 Jan 2020, 06:53
1
For remainders, we need to check the cyclicity or find a number that leaves a remainder 1 when divided by the divisor.
\(3^{1}\) = 3
\(3^{2}\) = 9
\(3^{3}\) = 27
\(3^{4}\) = 81
Now \(3^{4}\) divided by 5 leaves a remainder of 1.
So we are given \(3^{32} \times 3^3\) divided by 5
\(3^{32}\) will leave a remainder of 1 so we need to check the remainder of \(3^{3}\)
\(3^{3}\) = 27
27 divided by 5 leaves a remainder of 2.
OA, C
\(3^{1}\) = 3
\(3^{2}\) = 9
\(3^{3}\) = 27
\(3^{4}\) = 81
Now \(3^{4}\) divided by 5 leaves a remainder of 1.
So we are given \(3^{32} \times 3^3\) divided by 5
\(3^{32}\) will leave a remainder of 1 so we need to check the remainder of \(3^{3}\)
\(3^{3}\) = 27
27 divided by 5 leaves a remainder of 2.
OA, C
Carcass wrote:
What is the remainder when \(3^{35}\) is divided by 5 ?
A. 0
B. 1
C. 2
D. 3
E. 4
Kudos for the right answer and explanation
A. 0
B. 1
C. 2
D. 3
E. 4
Kudos for the right answer and explanation
Re: What is the remainder when
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05 Feb 2020, 09:12
05 Feb 2020, 09:12
1
1
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One more way to solve
calculate the remainder for each 3^1,3^2,3^3,3^4,3^5 when divided by 5
For 3^1 remainder = 3
For 3^2 remainder = 4
For 3^3 remainder = 2
For 3^4 remainder = 1
For 3^5 remainder = 3
we can see that the series of the remainder is repeating after 3^4
so, we can conclude that the remainders of the upcoming numbers will be in this same order i.e 3,4,2,1
3^5 will have a remainder = 3
3^6 will have a remainder of 4
3^7 will have a remainder of 2
...and so on
thus we can conclude that after every 4 places we will get 1 as the remainder
After this, we select a number less than 35(cause we are asked for the value of 3^35) which is divisible by 4(cause we know that for every 4 places we will get 1 as a remainder)
such number is 32
from the above conclusion, we can say that the remainder of 3^32 will be 1
Now, we continue our series of the remainder(3,4,2,1) for 3 more places(cause we need to find the value of 35 and 35 is after 3 places from 32)so we get 1....3,4,2
thus we can conclude that the remainder of 3^35 will be 2
I know this is a lengthy description but once you understand the concept it will be damn faster!
calculate the remainder for each 3^1,3^2,3^3,3^4,3^5 when divided by 5
For 3^1 remainder = 3
For 3^2 remainder = 4
For 3^3 remainder = 2
For 3^4 remainder = 1
For 3^5 remainder = 3
we can see that the series of the remainder is repeating after 3^4
so, we can conclude that the remainders of the upcoming numbers will be in this same order i.e 3,4,2,1
3^5 will have a remainder = 3
3^6 will have a remainder of 4
3^7 will have a remainder of 2
...and so on
thus we can conclude that after every 4 places we will get 1 as the remainder
After this, we select a number less than 35(cause we are asked for the value of 3^35) which is divisible by 4(cause we know that for every 4 places we will get 1 as a remainder)
such number is 32
from the above conclusion, we can say that the remainder of 3^32 will be 1
Now, we continue our series of the remainder(3,4,2,1) for 3 more places(cause we need to find the value of 35 and 35 is after 3 places from 32)so we get 1....3,4,2
thus we can conclude that the remainder of 3^35 will be 2
I know this is a lengthy description but once you understand the concept it will be damn faster!
