**Problem 21**

Let d(

If d(

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

*n*) be defined as the sum of proper divisors of*n*(numbers less than*n*which divide evenly into*n*).If d(

*a*) =*b*and d(*b*) =*a*, where*a**b*, then*a*and*b*are an amicable pair and each of*a*and*b*are called amicable numbers.For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

**Solution (in Ruby)**

For this problem, it is vital to write an efficient method to find the sum of divisors of a number. The divisor finding logic is same as the one stated in Problem 12. Rest is an as-is implementation of the problem statement.

Hover here to see the solution

Cheers!!

Bragaadeesh

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