How to resolve the algorithm Amicable pairs step by step in the Nim programming language

Published on 12 May 2024 09:40 PM
#Nim

How to resolve the algorithm Amicable pairs step by step in the Nim programming language

Table of Contents

Problem Statement

Two integers

N

{\displaystyle N}

and

M

{\displaystyle M}

are said to be amicable pairs if

N ≠ M

{\displaystyle N\neq M}

and the sum of the proper divisors of

N

{\displaystyle N}

(

s u m

(

p r o p D i v s

( N ) )

{\displaystyle \mathrm {sum} (\mathrm {propDivs} (N))}

)

= M

{\displaystyle =M}

as well as

s u m

(

p r o p D i v s

( M ) )

N

{\displaystyle \mathrm {sum} (\mathrm {propDivs} (M))=N}

.

1184 and 1210 are an amicable pair, with proper divisors:

Calculate and show here the Amicable pairs below 20,000; (there are eight).

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Amicable pairs step by step in the Nim programming language

Source code in the nim programming language

from math import sqrt

const N = 524_000_000.int32

proc sumProperDivisors(someNum: int32, chk4less: bool): int32 =
    result = 1
    let maxPD = sqrt(someNum.float).int32
    let offset = someNum mod 2
    for divNum in countup(2 + offset, maxPD, 1 + offset):
        if someNum mod divNum == 0:
            result += divNum + someNum div divNum
            if chk4less and result >= someNum:
                return 0

for n in countdown(N, 2):
    let m = sumProperDivisors(n, true)
    if m != 0 and n == sumProperDivisors(m, false):
        echo $n, " ", $m


from math import sqrt

const N = 524_000_000.int32
var x = newSeq[int32](N+1)

for i in 2..sqrt(N.float).int32:
  var p = i*i
  x[p] += i
  var j = i + i
  while (p += i; p <= N):
    j.inc
    x[p] += j

for m in 4..N:
  let n = x[m] + 1
  if n < m and n != 0 and m == x[n] + 1:
      echo n, " ", m

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