question stem: xy+1 = 3k + 0stubbornp wrote:14. If x and y are integers, is xy + 1 divisible by 3?
(1) When x is divided by 3, the remainder is 1.
(2) When y is divided by 9, the remainder is 8.
Statement I
x = 3k + 1
x= 4,7,13. We dont know anything about y.
Statement II
y = 9t + 8.
y= 17, 26. We dont know anything about x.
Combining I & II
x = 3k + 1
y = 9t + 8
xy +1 = (3k + 1)(9t + 8) + 1
(3k + 1)(9t + 8) + 1
= 27kt + 24k + 9t + 8 + 1
=27kt + 24k + 9t + 9
27kt is divisible by 3
24k is divisible by 3
9t is divisible by 3
9 is divisible by 3
Therefore, xy + 1 will be divisible by 3.
Sufficient.
Hence C.
OA?
