If j andk are positive integers, what is the remainder when (8 x 10power k + j)is divided by 9?
(1) k = 13
(2) j = 1
(sorry, i could not able write better the "10 power k")
DS - Integer
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- karthikpandian19
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sum of the digits = divisible by 9 such as 9,18,81 and so.
8*10^k + j means for k=1,2,3 the values are 80,800,8000 and so
so it depends on the value of j,such as j=1,2 giving 81,82.
a for j=1,2 it is both divisible and not divisible. not sufficient.
b for j=1,since k > 0 (positive integer) meaning the 8*10^k + j = 81,801 and so on.
divisible by 9 always.
hence B it is.
8*10^k + j means for k=1,2,3 the values are 80,800,8000 and so
so it depends on the value of j,such as j=1,2 giving 81,82.
a for j=1,2 it is both divisible and not divisible. not sufficient.
b for j=1,since k > 0 (positive integer) meaning the 8*10^k + j = 81,801 and so on.
divisible by 9 always.
hence B it is.
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Any number is divisible by 9 if the sum of digits of the number is divisible by 9.karthikpandian19 wrote:If j and k are positive integers, what is the remainder when (8 x 10power k + j)is divided by 9?
(1) k = 13
(2) j = 1
(sorry, i could not able write better the "10 power k")
Now 8 * 10^k + j, where k is any positive integer.
For any positive integer k, the sum of the digits of the expression 8 * 10^k will always be 8.
If k = 1, 8 * 10^1 = 80; sum of digits = 8
If k = 3, 8 * 10^3 = 8000; sum of digits = 8
(1) k = 13 implies 8 * 10^13 + j or the sum of the digits of 8 * 10^13 will be 8, but we do not know the value of j. Hence NOT sufficient.
(2) j = 1 clearly implies that the sum of the digits of 8 * 10^k + 1 will always be 9; SUFFICIENT.
Th correct answer is B.
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