N is a natural number such that when N3 is divided
by 9, it leaves remainder a. It can be concluded that
(a) a is a perfect square.
(b) a is a perfect cube.
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Answers
Given : N is a natural number such that when N³ is divided by 9, it leaves remainder a.
To find : Choose correct option
(a) a is a perfect square.
(b) a is a perfect cube.
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Solution:
Any number N can be represented as
3k , 3k+ 1 , 3k+2
(3k)³
= 27k³
= 9(3k³)
Divisible by 9
hence remainder = 0 (0 = 0² , 0³)
(3k + 1)³
= 27k³ + 1 + 27k² + 9k
= 9k(3k² + 3k + 1) + 1
Hence remainder = 1 when divided by 9 (1= 1² , 1³)
(3k + 2)³
= 27k³ + 8 + 54k² + 36k
= 9k(3k² + 6k + 4) + 8
Hence remainder = 8 when divided by 9 ( 8 = 2³)
0 is a perfect square & perfect cube
1 is a perfect square & perfect cube
8 is perfect cube only
Hence a is a perfect cube.
remainder a is a perfect cube
option b is correct answer
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