CODEWITHSUNDEEP
python
psquare
psquare

Reputation: 25

My Code for Perfect Number

So far, I've learned about for loops, if statements, and counting so it would be great to get feedback using only these topics. I was trying to solve the Perfect Number problem where the divisors excluding the number add up to that number. So for 6, The divisors 1,2,3 add up to 6. I've tried looking at other posts, but I wanted to know why my code specifically isn't working. I've tried solving the problem, and I feel that I'm almost there. I just have one issue. This is what I have so far:

#

num = 30

for i in range(1,num+1):
    count = 0
    for k in range(1,num+1):
        if i%k == 0 and i!=k:
            count += k
            if count == i:
                print(k,'Perfect',i)

#

The output on shell gives me this:

3 Perfect 6
8 Perfect 24
14 Perfect 28

#

I know that 24 is not a perfect number. The highest divisor for 24 excluding 24 should be 12, but I'm getting 8. This is why it's showing me that 24 is perfect. Can anyone clarify why I'm not able to get 12?

Upvotes: 2

Views: 326

Answers (3)

Nozar Safari
Nozar Safari

Reputation: 505

12 is not perfect 1 + 2 + 3+ 4 + 6 != 12 second for also have problem in count (change it to i+1) and critical problem was las check was in for

num = 30

for i in range(1,num+1):
    count = 0
    last = 0
    for k in range(1,i):
        if i%k == 0:
            count += k
            last = k
    if count == i:
        print(last,'Perfect',i)

result :

3 Perfect 6

14 Perfect 28

Upvotes: 0

Graham
Graham

Reputation: 3873

Your problem is that you're checking whether count == i in the wrong place. You need to check after the entire count summation is finished, not after each addition to the count sum.

num = 30

for i in range(1,num+1):
    count = 0
    for k in range(1,num+1):
        if i%k == 0 and i != k:
            count += k
    if count == i:
         print(k,'Perfect',i)

Also, if you want slightly more efficient code, you could do the subtraction afterwards instead of checking each value. Also, as lenik suggested, you only need to count to i (since i can't have any divisors larger than itself.) Even more efficiently, one could just use a Sieve of Eratosthenes-like approach, and count to sqrt(i)+1, finding the corresponding factors along the way:

from math import sqrt    

num = 30

for i in range(1,num+1):
    count = 0
    for k in range(1,sqrt(i) + 1):
        if i%k == 0:
            count += k
            factor = count/k
            # The conditional is necessary in case i is a perfect square.
            if k != factor:
                count += factor
    if count - i == i:
         print(k,'Perfect',i)

Upvotes: 1

lenik
lenik

Reputation: 23556

You have to count all divisors, before you make a conclusion, your code goes up to 8 for 24, then sums up all divisors so far and declares 24 a perfect number, not waiting for other divisors to appear.

Your code should looks like this:

for i in range(1,num+1):
    count = 0
    for k in range(1,i):  # no point searching for numbers greater than i
        if i%k == 0 :
            count += k
    if count == i:
        print('Perfect',i)

and produce:

('Perfect', 6)
('Perfect', 28)

Upvotes: 2

Related Questions