What is the sum of all positive whole-number divisors of 720?

Solutions

Solutions

Step 1

Find all the prime numbers in 720. 720 should be the prodcut of all the numbers you will have found.

\[720 = 72 \times 10 = 8 \times 9 \times 2 \times 5 = 2^4 \times 3^2 \times 5.\]

Step 2

We calculate a number for each prime factor.

For 2, we calculate:

\[1 + 2^1 + 2^2 + 2^3 + 2^4 = 31.\]

You may have noticed that this is the sum of first 5 terms in a geometric sequence. The sequence starts from 1. Multiplying a term by 2 (the prime factor wer are considering now) gives us the next term. The last term we include is \(2^4\), which is what we found in Step 1.

For 3, we calculate:

\[1 + 3^1 + 3^2 = 13.\]

Similarly, this is the sum of first 3 terms of a geometric squence. The last term \(3^2\) is what we found in Step 1.

For 5, we calculate:

\[1 + 5^1 = 6.\]

Step 3

Now we multiply together the three numbers we have calculated in Step 2:

\[31 \times 13 \times 6 = 2418.\]

The sum of all positive whole-number divisors of 720 is 2418.

Discussions

As we have learned in problems of counting the number of divisors, any integer \(n\) that is grater than 1 (staring from 2) can be reprsented with a prodcut of prime factors, like

\[n = p_1 ^ {e_1} \times p_2 ^ {e_2} \times p_3 ^ {e_3} \ldots\]

where \(p_1, p_2, p_3, \ldots\) are distinct prime numbers and \(e_1, e_2, e_3, \ldots\) are their exponents. And there is only one (unique) way to do that.

The sum of all divisors of \(n\) can be calculated as:

\[(1 + p_1 + p_1^2 + \ldots + p_1^{e_1}) \cdot (1 + p_2 + p_2^2 + \ldots + p_2^{e_2}) \cdot (1 + p_3 + p_3^2 + \ldots + p_3^{e_3}) \ldots\]

Although the proof is not simple, here is a more detailed explanation.

Also, the formula for calculating the sum of first \(n\) terms of a geometric sequence can be handy if a prime factor has a large exponent.

\[(1 + p + p^2 + \ldots + p^{n}) = \frac{p^{n+1} - 1}{p - 1}.\]

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