You are given a positive integer K. Your task is to find out whether K is a perfect number or not.
Note :Perfect numbers are those numbers that are equal to the sum of all its proper divisors.
Proper divisors are all the divisors of a number excluding the number itself.
For example :
For K = 6, the proper divisors of 6 are 1, 2, 3 and its sum is 6 so 6 is a perfect number. For K = 8, the proper divisors of 8 are 1, 2,4 and its sum is 7 so 8 is not a perfect number.
The first line of input contains a single integer T, representing the number of test cases.
The first and the only line of each test case contains a single positive integer K, as described in the problem statement.
Output Format :
For each test case, return true if K is a perfect number or else return false.
Note :
You do not need to print anything. It has already been taken care of. Just implement the given function.
Constraint :
1 <= T <= 100
2 <= K <= 10^8
Time Limit: 1 sec
Follow Up :
Can you do this in O(sqrt(N)) time and constant space?
3
2
6
8
False
True
False
In the first test case, the only proper divisor of 2 is 1 so 2 is not a perfect number.
The test cases 2 and 3 are already explained above.
3
10
28
12
False
True
False
Can you think of iterating through all numbers?
O(N), where N is the given number.
As we iterate through all numbers from 1 to n - 1.
O(1)
Since we are using constant extra memory.