find the sum of all the positive integers which cannot be written as the sum of two abundant numbers

euler23.rb

class Integer
    def isabundant?
        sum_of_proper_divisors(self) > self
    end
end

def sum_of_proper_divisors(num) #borrowed from github user mlsimpson
    
    orig = num
    
    if num == 1
        return 0
    end
    
    sum = 1
    p = 2
    while p*p <= num && num > 1
        if (num%p).zero?
            j = p*p
            num = (num / p)
            while (num%p).zero?
                j = j*p
                num = (num / p)
            end
            sum = sum*(j-1)
            sum = sum / (p-1)
            else
        end
        if p == 2
            p = 3
            else
            p = p+2
        end
    end
    
    if num > 1
        sum = sum * (num+1)
    end
    
    sum - orig
end

def timing_method(block=nil, *args) #borrowed from github user mlsimpson
    start = Time.now
    if block_given?
        yield
        else
        self.send(block, args)
    end
    finish = Time.now
    print "Elapsed time:  #{(finish - start)*1000} milliseconds\n"
end

def euler23
    maxnum = 20161 #Wolfram Mathworld's discussion on Abundant Numbers, "Every number greater than 20161 can be expressed as a sum of two abundant numbers. " So our upper bound is 20161 instead of 28123

    allnums = (1..maxnum).to_a

    sumarray = []
    
    abarray = allnums.select{|i| i if i.isabundant? }
    
    for i in 0..abarray.length-1
        for j in i..abarray.length-1
            sum = abarray[i] + abarray[j]
            sumarray << sum unless sum > maxnum
        end
    end
    
    sumarray = sumarray.uniq
    (allnums - sumarray).inject(:+)
end

timing_method do
    puts euler23
end

# Problem Statement:
#perfect number: sum of proper divisors equal to number
#
#sum of proper divisors of 28 is 1+2+4+7+14=28, 28 is a perfect number
#
#n is deficient is s.o.p.d. is less than n, abundant if more
#
#12 smallest abundant number, 1+2+3+4+6=16
#
#smallest number that can be written as the sum of two abundant numbers is 24 (12+12)
#
#all integers > 28123 can be written as the sum of two abundant numbers
#
#the greatest number than cannot be expressed as the sum of two abundant numbers is less than this limit
#
#find the sum of all the positive integers which cannot be written as the sum of two abundant numbers
#
#
#
#Algorithm Discussion:
# This is a very simple and slow solution that borrow a fast isabundant? function from a different solution
#
# First we generate all abundant numbers between 1 and maxnum, then we create a sumarray of all the possible sums that can be created from it, then we take the unique of that and remove that array from the all numbers array and take the sum