If for any positive integer x , d[x] denotes its smallest odd divisor and D[x] denotes its largest odd divisor, is x even?
1. D[x] - d[x] = 0
2. D[3x] = 3
1. D[x] - d[x] = 0
2. D[3x] = 3
Smallest odd divisor of any positive integer is 1. Thus, d[x] = 1 for any integer x > 0replayyyy wrote:If for any positive integer x , d[x] denotes its smallest odd divisor and D[x] denotes its largest odd divisor, is x even?
1. D[x] - d[x] = 0
2. D[3x] = 3
Statement 1: D[x] - d[x] = 0 => D[x] = d[x] = 1
Now, there may be 2 cases:
- (1) x = 1 => x odd
(2) x = 1*(Any even integer) => x even.
Statement 2: D[3x] = 3 => Largest odd divisor of 3x is 3
Now, there may be 2 cases:
- (1) 3x = 3*1 => x = 1 => x odd
(2) 3x = 3*(Any even integer) => x even.
The correct answer is E.
Note: This problem is solved assuming that 1 is a divisor of any number. To avoid ambiguity 1, -1, n and -n are termed as Trivial Divisors of n, whereas other divisors are termed as Non-trivial Divisors of n.
