Problem 21
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
Let d(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.
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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