If x and n are positive integers (or integers?), and when (n+1)(n-1) is divided by 24, the quotient is x and the remainder is r. r=?
1) 2 is not the factor of n
If 2 is not a factor of n, implies n is odd
If n = 3 then (n+1)(n-1) = 8, when divided by 24 leaves a remainder 8
If n = 7 then (n+1)(n-1) = 48, when divided by 24 leaves a remainder 0
Two different answers, Insufficient!
2) 3 is not the factor of n
If n = 2 then (n+1)(n-1) = 2, when divided by 24 leaves a remainder 2
If n = 7 then (n+1)(n-1) = 48, when divided by 24 leaves a remainder 0
from 1 and 2,
If n = 1 then (n+1)(n-1) = 0, when divided by 24 leaves a remainder 0
If n = 5 then (n+1)(n-1) = 24, when divided by 24 leaves a remainder 0
Hence
C
To explain it further,
From 1 - If a number is not divisible by 2 then it should be in the form 2k+1 or 2k-1. So,
If n = 2k+1,
(n+1)*(n-1) = (2k+1-1)*(2k+1+1) = 4k*(k+1) and
If n = 2k-1,
(n+1)*(n-1) = (2k-1-1)*(2k-1+1) = 4k*(k-1).
So (n+1)*(n-1) may or may not be divisible by 24.i.e. (n+1)*(n-1) may or may not leave a remainder when divided by 24
From 2 - If a number is not divisible by 3 then it should be in the form 3k+1 or 3k-1. So,
If n = 3k+1,
(n+1)*(n-1) = (3k+1-1)*(3k+1+1) = 3k*(3k+2) and
If n = 3k-1,
(n+1)*(n-1) = (3k-1-1)*(3k-1+1) = 3k*(3k-2).
So (n+1)*(n-1) may or may not be divisible by 24.i.e. (n+1)*(n-1) may or may not leave a remainder when divided by 24
From 1 and 2 - If a number is not divisible by 6 then it should be in the form 6k+1 or 6k-1. So,
If n = 6k+1,
(n+1)*(n-1) = (6k+1-1)*(6k+1+1) = 6k*(6k+2)= 12k*(3k+1)
If n = 6k-1,
(n+1)*(n-1) = (6k-1-1)*(6k-1+1) = 6k*(6k-2)= 12k*(3k-1)
If k = odd, 3k+1 and 3k-1 are even.
If k = even, 3k+1 and 3k-1 are odd.
So k*(3k+1) and k*(3k-1) are always EVEN
(n+1)*(n-1) = 12*even integer= 24 * Integer
So (n+1)*(n-1)is always divisible by 24 and leaves a remainder 0.
