Remainders.

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Remainders.

by vineeshp » Thu May 12, 2011 6:33 am
If n and m are positive integers, what is the remainder when 3^(4n + 2 + m) is divided by 10 ?

(1) n = 2 (2) m = 1

OA Later.
Vineesh,
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by manpsingh87 » Thu May 12, 2011 6:55 am
vineeshp wrote:If n and m are positive integers, what is the remainder when 3^(4n + 2 + m) is divided by 10 ?

(1) n = 2 (2) m = 1

OA Later.
here we are interested in knowing the last digit of the expression 3^(4n + 2 + m); because when we divide any number, remainder is equal to the last digit of the expression, for example when we divide 49/10 remainder is 9,similarly when we divide 99/10 remainder is still 9 because last digit is 9;

1) n=2;
3^(4*2 + 2 + m); 3^(8 + 2 + m);3^(10 + m); since we don't know about m, therefore here different answers are possible depending upon the different values of m,for example when m=1; remainder would be7, and when m=2; remainder would be 1; hence 1 alone is not sufficient to answer the question,

2)m=1;
3^(4n + 2 + 1); 3^(4n + 3); also observe following;
3^1=3;
3^2=9;
3^3=27;
3^4=81;
---------
3^5=243;
here last digit start repeating itself after every fourth power of three, i.e. its last digit repeats after every 4 cycles,

now when (4n +3) is divided by 4; remainder will always be 3; hence last digit will be 7..!!!
as 2 alone is sufficient to answer the question hence answer should be B
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by vineeshp » Thu May 12, 2011 7:03 am
Of course OA is B. :)
Vineesh,
Just telling you what I know and think. I am not the expert. :)