Divisors + Of + 100
Number Theory
Level
2
What is the sum of all of the positive divisors of 100 (inclusive of 1 and itself)?
The answer is 217.
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Solution 1: From the blog post Divisors of an Integer , we know that each of the divisors of 1 0 0 have the form 2 a 5 b , where a , b are non-negative integers satisfying 0 ≤ a ≤ 2 and 0 ≤ b ≤ 2 . Hence, each term will be added exactly once in the multiplication ( 2 0 + 2 1 + 2 2 ) ( 5 0 + 5 1 + 5 2 ) , so it has a total sum of 217.
Solution 2: We can list out all of the divisors of 100 as 1, 2, 4, 5, 10, 20, 25, 50, 100, and find that they have a sum of 217.