Jinglander wrote:This is from memory so it may be non exact. If k^4 is divisible by 81 then which of the following could be the remainder when k is divided by 9. I think the answer choices were 1,2,4,6,8
Though I am also not able to understand the problem But i thought there is no harm in giving a try,
K^4 is divisible by 81 ,
It means K must have a least one 3 in its factor, Because only then 3 ^ 4 will be divisible by 81.
Now k ^ 2 must have been divisible by 9 { Because at least two 3's are there in it }
Now K must have at least one 3 in it.
Now if we write first 10 multiples of 3 , it will be
0
3
6
9
12
15
18
21
24
27
30
Now 0 / 9 ===Remainder 0
3 / 9 ===remainder will be 3
6 / 9 ===remainder will be 6
9 / 9 ===remainder will be 0
12/9 ===remainder will be 3
15 / 9 ===remainder will be 6
18 / 9 === remainder will be 0
21/9 ===remainder will be 0
like this it will be continue...
remainders will be in the form of 0 , 3 , 6 , 0 , 3 , 6 ........................
we don't have 0 3 in the answer choice so my take would be 6 ...................
