=begin
 This is a number analogy to a famous card trick.
 Ask the user to enter a three-digit number. Think
 of the number as ABC (where A, B, C are the three 
 digits of the number). Now, find the remainders of 
 the numbers formed by ABC, BCA, and CAB when divided 
 by 11. We will call these remainders X, Y, Z. Add 
 them up as X+Y, Y+Z, Z+X. If any of the sums are odd,
 increase or decrease it by 11 (whichever operation 
 results in a positive number less than 20; note if 
 the sum is 9, just report this and stop the process). 
 Finally, divide each of the sums in half. The 
 resulting digits are A, B, C. Write a program that 
 implements this algorithm.
=end

def first_number(x)
  p = x.to_i % 100
  q = ((x.to_i - p.to_i) / 100)
  return q
end

def second_number(x)
  p = ((x.to_i % 100) * 10) / 100
  return p
end

def third_number(x)
  p = x.to_i % 10
  return p
end

def find_remainder(hundreds, tens, ones)
  hundreds = hundreds.to_i * 100
  tens = tens.to_i * 10
  new_number = (hundreds + tens + ones) % 11
  return new_number
end

def is_odd(x)
  if (x % 2) == 0 
    return false
  else return true
  end
end

def less_than_20(x)
  if x < 9  
    x = (x + 11) / 2
  elsif x > 9
    x = (x-11) / 2
  else
    return x
  end
  return x
end

def equation(x, y, z)
  u = x + y
  p = y + z
  s = z + x
  if is_odd(u) == true 
    u = less_than_20(u)
  else
    u = u / 2
  end
  
  if(is_odd(p) == true)
    p = less_than_20(p)
  else
    p = p / 2
  end
  
  if(is_odd(s) == true)
    s = less_than_20(s)
  else
    s = s / 2
  end
  
  puts  "A: " << u.to_s
  puts  "B: " << p.to_s
  puts  "C: " << s.to_s
end

puts "Enter a 3 digit number: "
abc = gets.chomp
if (abc.to_i > 99 && abc.to_i < 1000) 
  a = first_number(abc)
  b = second_number(abc)
  c = third_number(abc)
  x = find_remainder(a,b,c)
  y = find_remainder(b,c,a)
  z = find_remainder(c,a,b)
  equation(x,y,z)
else
  puts "The integer must be 3 digits!"
end