Problem Statement

These define three classifications of positive integers based on their   proper divisors. Let   P(n)   be the sum of the proper divisors of   n   where the proper divisors are all positive divisors of   n   other than   n   itself.

6   has proper divisors of   1,   2,   and   3. 1 + 2 + 3 = 6,   so   6   is classed as a perfect number.

Calculate how many of the integers   1   to   20,000   (inclusive) are in each of the three classes. Show the results here.

Let's start with the solution:

10 M=20000
20 I=1
30 :I)=0
40 I=I+1
50 #=M>I*30
60 I=1
70 J=I*2
80 :J)=:J)+I
90 J=J+I
100 #=M>J*80
110 I=I+1
120 #=M/2>I*70
130 D=0
140 P=0
150 A=0
160 I=1
170 #=:I)
180 #=:I)=I*210
190 A=A+1
200 #=240
210 P=P+1
220 #=240
230 D=D+1
240 I=I+1
250 #=M>I*170
260 ?=D
270 ?=" deficient"
280 ?=P
290 ?=" perfect"
300 ?=A
310 ?=" abundant"