EKOK - Programlama (Seçenekler Yanlış)

Hazırlayanın da yanlış çözdüğü için şıkları hatalı verdiği bir ÖZDEBİR sorusu:

https://i.ibb.co/LdV1fQVK/EKOK.png
https://www.facebook.com/photo/?fbid=10162806493872579&set=g.137856289571386&locale=tr_TR

b = 24 Sayısının Pozitif Tam Sayı Bölenleri
b = {1, 2, 3, 4, 6, 8, 12, 24}

Kod:

 b        a = b + 2       a / (24 / b) Sayısı Pozitif Bir Tam Sayı mı?
---       ---------       --------------------------------------------
 1            3                 3 / (24 / 1) = 1 / 8 ∉ N^(+), Hayır

 2            4                 4 / (24 / 2) = 1 / 3 ∉ N^(+), Hayır

 3            5                 5 / (24 / 3) = 5 / 8 ∉ N^(+), Hayır

 4            6                 6 / (24 / 4) = 1 ∈ N^(+), Evet

 6            8                 8 / (24 / 6) = 2 ∈ N^(+), Evet

 8           10                10 / (24 / 8) = 10 / 3 ∉ N^(+), Hayır

12           14                14 / (24 / 12) = 7 ∈ N^(+), Evet

24           26                26 / (24 / 24) = 26 ∈ N^(+), Evet

Σa = 6 + 8 + 14 + 26 = 54.

Bilgisayar Programlamayla İlgilenen Öğrenciler İçin Fortran 2003 Uyarlaması:

https://i.ibb.co/DDhvd32t/EKOK-Fortran.png

Program:

Kod:

! https://app.codeconvert.ai/code-generator?outputLang=Fortran sitesindeki
! yapay zekâ tarafından yazılmış Fortran 2003 Programının biraz düzeltip
! değiştirdiğim hali:
program find_a
  implicit none
  integer :: a, b, c, toplam = 0
  integer :: possible_b

  print *, "Finding possible values for integer a..."
  print *, "Based on: LCM(a, 24 / b) = b + 2 and LCM(a, 24 / b) = a"
  print *, "Where a and b are different and positive integers."
  print *

  print *, "Possible values for a:"
  
  ! From the given equations:
  ! LCM(a, 24 / b) = b + 2
  ! LCM(a, 24 / b) = a
  ! This implies a = b + 2.
  ! Also, LCM(a, 24 / b) = a implies that a must be a multiple of (24 / b).
  ! Substituting a = b + 2, we need (b + 2) to be a multiple of (24 / b).
  ! This means LCM(b + 2, 24 / b) = b + 2.

  ! b must be a positive integer divisor of 24 for 24/b to be an integer.
  ! The positive divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.
  ! We iterate through these possible values for b.
  do possible_b = 1, 24
    if (mod(24, possible_b) == 0) then
      b = possible_b
      a = b + 2

      ! Check conditions:
      ! 1. a and b are positive integers: b is positive from the loop, a = b + 2 is positive.
      ! 2. a and b are different: a = b + 2 implies a /= b for positive b.
      ! 3. 24 / b is an integer: Ensured by checking mod(24, possible_b) == 0.
      c = 24 / b

      ! Check the derived condition: LCM(a, c) = a
      if (lcm(a, c) == a) then
        print *, a; toplam=toplam+a
      end if
    end if
  end do

print*,""
write(6,10)"The sum of values for a = ", toplam
10 format (a,i2)

contains

  ! GCD function (Euclidean algorithm)
  integer function gcd(x, y) result(res)
    integer, intent(in) :: x, y
    !integer :: res
    integer :: temp_x, temp_y, remainder

    temp_x = abs(x)
    temp_y = abs(y)

    do while (temp_y /= 0)
      remainder = mod(temp_x, temp_y)
      temp_x = temp_y
      temp_y = remainder
    end do
    res = temp_x
  end function gcd

  ! LCM function
  integer function lcm(x, y) result(res)
    integer, intent(in) :: x, y
    ! integer :: res
    integer :: common_divisor

    ! Handle edge cases where x or y is zero (LCM is 0)
    if (x == 0 .or. y == 0) then
      res = 0
    else
      common_divisor = gcd(x, y)
      ! LCM(x, y) = (x * y) / GCD(x, y)
      ! Calculate as (x / GCD(x, y)) * y to avoid potential overflow
      ! Note: For this problem, x and y (a and c) will be positive.
      res = (x / common_divisor) * y
    end if
  end function lcm

end program find_a