Abundant work

Computer Science Level 2

We call a positive integer n n abundant if the sum of its proper divisors is greater than itself.

For example, 18 is abundant because the sum of its proper divisors ( 1 + 2 + 3 + 6 + 9 ) (1+2+3+6+9) is greater than itself.

What is the smallest odd abundant number?


The answer is 945.

Pi Han Goh by Pi Han Goh , 30, Malaysia

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import java.lang.reflect.Array;
import java.util.Arrays;
public class Main {
    public static int divisors(int a, int b){
        if(a%b==0){
            return b;
        }else {
            return 0;
        }
    }
    public static void main(String[] args) {
        int no;
        int[] sum =new int[100000];
        Arrays.fill(sum, 0);
        for(int i=2;true;i++) {
            for (int j = 1; j < i; j++) {
                sum[i - 1] += divisors(i, j);
            }
            if (sum[i - 1] > i && sum[i - 1] != 0 && i % 2 != 0) {
                System.out.println("The sum of Divisors of " + i+ " is " + sum[i - 1]);
                break;
            }
        }
    }
}

1 Helpful 0 Interesting 0 Brilliant 1 Confused

There is an easier approach that doesn't require testing all cases.

HInt: How is the sum of all divisors calculated?

Calvin Lin Staff - 4 years ago
Valerie Wong
Jan 17, 2021

While the first even abundant number is 12 = 2 2 ⋅  3, with σ (12) = 2 3  −  1 2  −  1  ⋅  (3 + 1) = 7 ⋅  4 = 28 > 24 = 2 ⋅  12 , the first odd abundant number (which happens to be the 232nd abundant number) is: 945 = 3 3 ⋅  5 ⋅  7 = 1 ⋅  3 ⋅  5 ⋅  7 ⋅  9 = 9!! (the double factorial of 9), with σ (945) = 3 4  −  1 3  −  1  ⋅  (5 + 1) ⋅  (7 + 1) = 40 ⋅  6 ⋅  8 = 1920 > 1890 = 2 ⋅  945 . {945, 1575, 2205, 2835, 3465, 4095, 4725, 5355, 5775, 5985, 6435, 6615, 6825, 7245, 7425, 7875, 8085, 8415, 8505, 8925, 9135, 9555, 9765, 10395, 11025, 11655, 12285, 12705, 12915, 13545, 14175, ...}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

But how do you get this value (and the subsequent smallest odd abundant numbers) without asking Google?

Pi Han Goh - 4 months, 3 weeks ago
Hasmik Garyaka
Sep 22, 2017

If n = p q n=p^q sumofalldivs= 1 + p + . . . + p q − 1 1+p+...+p^{q-1} If n=mk and m and k are relatively prime, f(m) f(k)=f(mk) So abundant number needs to be product of several primes. 945=3 3 3 5*7

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Do you know that the smallest possible odd abundant number is 3 * 3 * 3 * 5 * 7?

Pi Han Goh - 3 years, 8 months ago

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