If n is a positive integer, what is the remainder when 3^(4n+2) + 1 is divided by 10?
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Number is [3^(4n+2)]+1 = [(3^2)^(2n+1) ] + 1 = 9^(2n+1) + 1frizo wrote:If n is a positive integer, what is the remainder when 3^(4n+2) + 1 is divided by 10?
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When 9 is divided by 10, we can take the remainder as -1 (instead of 9)
Since 9 is raised to (2n+1) which is always odd...the remainder when 9^(an odd number) is divided by 10 will always be -1 i.e 9.
Therefore 9^(2n+1) + 1 is exactly divisible by 10.
Hence the remainder must be 0.
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Another way of looking at this is...when a number N is divided by 10, the remainder is nothing but the unit digit of that number N
Also the unit's digit of 3^(4n+2) will be 9 irrespective of value of n.
So the given number i.e 3^(4n+2) + 1 ends with 0 and hence the remainder when it is divided by 10 will be 0
