How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the MAD programming language

Published on 12 May 2024 09:40 PM
#Mad

How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the MAD programming language

Table of Contents

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:

Step by Step solution about How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the MAD programming language

Source code in the mad programming language

            NORMAL MODE IS INTEGER
            DIMENSION P(20000)
            MAX = 20000
            THROUGH INIT, FOR I=1, 1, I.G.MAX
INIT        P(I) = 0
            THROUGH CALC, FOR I=1, 1, I.G.MAX/2
            THROUGH CALC, FOR J=I+I, I, J.G.MAX
CALC        P(J) = P(J)+I
            DEF = 0
            PER = 0
            AB = 0
            THROUGH CLSFY, FOR N=1, 1, N.G.MAX
            WHENEVER P(N).L.N, DEF = DEF+1
            WHENEVER P(N).E.N, PER = PER+1
CLSFY       WHENEVER P(N).G.N, AB = AB+1
            PRINT FORMAT FDEF,DEF
            PRINT FORMAT FPER,PER
            PRINT FORMAT FAB,AB
            VECTOR VALUES FDEF = $I5,S1,9HDEFICIENT*$
            VECTOR VALUES FPER = $I5,S1,7HPERFECT*$
            VECTOR VALUES FAB =  $I5,S1,8HABUNDANT*$
            END OF PROGRAM

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