How to resolve the algorithm Arithmetic numbers step by step in the XPL0 programming language

Published on 12 May 2024 09:40 PM
#Xpl0

How to resolve the algorithm Arithmetic numbers step by step in the XPL0 programming language

Table of Contents

Problem Statement

A positive integer n is an arithmetic number if the average of its positive divisors is also an integer. Clearly all odd primes p must be arithmetic numbers because their only divisors are 1 and p whose sum is even and hence their average must be an integer. However, the prime number 2 is not an arithmetic number because the average of its divisors is 1.5. 30 is an arithmetic number because its 7 divisors are: [1, 2, 3, 5, 6, 10, 15, 30], their sum is 72 and average 9 which is an integer. Calculate and show here:

  1. The first 100 arithmetic numbers.
  2. The xth arithmetic number where x = 1,000 and x = 10,000.
  3. How many of the first x arithmetic numbers are composite. Note that, technically, the arithmetic number 1 is neither prime nor composite. Carry out the same exercise in 2. and 3. above for x = 100,000 and x = 1,000,000.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Arithmetic numbers step by step in the XPL0 programming language

Source code in the xpl0 programming language

int N, ArithCnt, CompCnt, Div, DivCnt, Sum, Quot;
[Format(4, 0);
N:= 1;  ArithCnt:= 0;  CompCnt:= 0;
repeat  Div:= 1;  DivCnt:= 0;  Sum:= 0;
        loop    [Quot:= N/Div;
                if Quot < Div then quit;
                if Quot = Div and rem(0) = 0 then \N is a square
                    [Sum:= Sum+Quot;  DivCnt:= DivCnt+1;  quit];
                if rem(0) = 0 then
                    [Sum:= Sum + Div + Quot;
                    DivCnt:= DivCnt+2;
                    ];
                Div:= Div+1;
                ];
        if rem(Sum/DivCnt) = 0 then \N is arithmetic
            [ArithCnt:= ArithCnt+1;
            if ArithCnt <= 100 then
                [RlOut(0, float(N));
                if rem(ArithCnt/20) = 0 then CrLf(0);
                ];
            if DivCnt > 2 then CompCnt:= CompCnt+1;
            case ArithCnt of 1000, 10_000, 100_000, 1_000_000:
                [CrLf(0);
                IntOut(0, N);  ChOut(0, 9\tab\);
                IntOut(0, CompCnt);
                ]
            other;
            ];
        N:= N+1;
until   ArithCnt >= 1_000_000;
]

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