Here is a Python function that implements the logic described in the problem:

```python
def DetectPerfectNumber(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
if sum == n:
return 1
else:
return sum
```

This function works by iterating over all numbers from 1 to `n-1` (the proper divisors of `n`). If `n` is divisible by the current number `i` (i.e., `n % i == 0`), then `i` is a proper divisor of `n`, and we add it to the `sum`.

After the loop, we check if the `sum` of the proper divisors is equal to `n`. If it is, `n` is a perfect number, and we return `1`. Otherwise, we return the `sum` of the proper divisors.

Question

Here is a Python function that implements the logic described in the problem:

```python
def DetectPerfectNumber(n):
    sum = 0
    for i in range(1, n):
        if n % i == 0:
            sum += i
    if sum == n:
        return 1
    else:
        return sum
```

This function works by iterating over all numbers from 1 to `n-1` (the proper divisors of `n`). If `n` is divisible by the current number `i` (i.e., `n % i == 0`), then `i` is a proper divisor of `n`, and we add it to the `sum`.

After the loop, we check if the `sum` of the proper divisors is equal to `n`. If it is, `n` is a perfect number, and we return `1`. Otherwise, we return the `sum` of the proper divisors.

Knowee AI · Accepted Answer

Here is a Python function that implements the logic described in the problem:

```python
def DetectPerfectNumber(n):
    sum = 0
    for i in range(1, n):
        if n % i == 0:
            sum += i
    if sum == n:
        return 1
    else:
        return sum
```

This function works by iterating over all numbers from 1 to `n-1` (the proper divisors of `n`). If `n` is divisible by the current number `i` (i.e., `n % i == 0`), then `i` is a proper divisor of `n`, and we add it to the `sum`.

After the loop, we check if the `sum` of the proper divisors is equal to `n`. If it is, `n` is a perfect number, and we return `1`. Otherwise, we return the `sum` of the proper divisors.

Question

Solution

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