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Find M for which A, B, C form an A.P. in given order if any one is multiplied by M

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  • Last Updated : 21 Feb, 2022
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Given 3 positive integers A, B, and C. Choose a positive integer M and multiply any one of A, B, or C with M. The task is to decide whether A, B and C would comply with Arithmetic Progression Mean (A.P.) Mean after performing the above-given operation once. The order of A, B, C cannot be changed.

Examples:

Input: A = 30, B = 5, C = 10
Output: YES
Explanation: B = 5 can be multiplied by M = 4. 
Then 2 * B = A + C (2 * 20 = 30 + 10).

Input: A = 2, B = 1, C = 1
Output: NO
Explanation: A + C = 3 which is odd. B = 1 which is odd. 
Both these values cannot be made even (because 2*B is always even so need for making both of them even) and equal simultaneously.

 

Approach: 

  • Three integers are said to be in A.P. if

2 * B = A + C             -(1)

This equation can also be written as 

2 * B – C = A              -(2)

and 

2 * B – A = C              -(3)

Below are the steps that use this relationship to determine whether the given three numbers can form an A.P. or not-

  • Declare a variable new_b and initiate it equal to 2 * B.
  • Return true if A, B, C are already in AP. Any of them can be multiplied by M=1 and the answer will always be “YES“.
  • Return true if B can be multiplied by any number to make it equal to A+C i.e. (A + C) % new_b = 0. The remainder will always be 0 irrespective of the value of M.
  • Now the only options remaining are to multiply either A or C to make their sum equal to new_b.
    • If (A + C) > new_b, return false. In this case, multiplying A or C with M will only increase its value. The answer will always be “NO“.
    • Else if (new_b – C) % A = 0, return true. In this case, A can be multiplied by some integer M. In this case, (new_b – C) can be seen as a multiple of A. Hence, on dividing (new_b – C) by A, the remainder will always be 0 irrespective of the value of M. The answer will be “YES” (a per equation (2)).
    • Else if  (new_b – A) % B = 0, return true. In this case, C can be multiplied by some integer M. In this case, (new_b – A) can be seen as a multiple of C. Hence, on dividing (new_b – A) by C, the remainder will always be 0 irrespective of the value of M. The answer will be “YES” (as per equation (3)).
    • Else return false.

Illustration:

Input: A = 30, B = 5, C = 10 
Output: YES
Explanation:
new_b = 2 * B
           = 2 * 5
           = 10
(A + C) % new_b 
(30 + 10) % 10 
40 % 10 = 0
This proves that the numbers 30 5 10 are in A.P. after multiplying with some number M.

Below is the implementation for the above approach-

C++




// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if A, B, C
// can be in A.P. after multiplying
// by any integer M
bool isAP(int A, int B, int C)
{
    int new_b = B * 2;
 
    // Check if A, B, C are already
    // in A.P.
    if (new_b == (A + C))
        return true;
 
    // Check if multiplying B will
    // give A.P.
    if ((A + C) % new_b == 0)
        return true;
 
    if ((A + C) > new_b)
        return false;
 
    // Check if multiplying A will
    // give A.P.
    if ((new_b - C) % A == 0)
        return true;
 
    // Check if multiplying C will
    // give A.P.
    if ((new_b - A) % C == 0)
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    int A, B, C;
    A = 30;
    B = 5;
    C = 10;
 
    // Function call
    bool ans = isAP(A, B, C);
 
    // Displaying the answer
    if (ans)
        cout << "YES";
    else
        cout << "NO";
    return 0;
}


Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to check if A, B, C
  // can be in A.P. after multiplying
  // by any integer M
  static Boolean isAP(int A, int B, int C)
  {
    int new_b = B * 2;
 
    // Check if A, B, C are already
    // in A.P.
    if (new_b == (A + C))
      return true;
 
    // Check if multiplying B will
    // give A.P.
    if ((A + C) % new_b == 0)
      return true;
 
    if ((A + C) > new_b)
      return false;
 
    // Check if multiplying A will
    // give A.P.
    if ((new_b - C) % A == 0)
      return true;
 
    // Check if multiplying C will
    // give A.P.
    if ((new_b - A) % C == 0)
      return true;
 
    return false;
  }
 
  // Driver code
  public static void main (String[] args) {
    int A, B, C;
    A = 30;
    B = 5;
    C = 10;
 
    // Function call
    Boolean ans = isAP(A, B, C);
 
    // Displaying the answer
    if (ans)
      System.out.print("YES");
    else
      System.out.print("NO");
  }
}
 
// This code is contributed by hrithikgarg03188


Python3




# Python code for the above approach
 
# Function to check if A, B, C
# can be in A.P. after multiplying
# by any integer M
def isAP(A, B, C):
    new_b = B * 2
 
    # Check if A, B, C are already
    # in A.P.
    if (new_b == (A + C)):
        return True
 
    # Check if multiplying B will
    # give A.P.
    if ((A + C) % new_b == 0):
        return True
 
    if ((A + C) > new_b):
        return False
 
    # Check if multiplying A will
    # give A.P.
    if ((new_b - C) % A == 0):
        return True
 
    # Check if multiplying C will
    # give A.P.
    if ((new_b - A) % C == 0):
        return True
 
    return False
 
# Driver code
A = 30
B = 5
C = 10
 
# Function call
ans = isAP(A, B, C)
 
# Displaying the answer
if (ans):
    print("YES")
else:
    print("NO")
 
# This code is contributed by Saurabh Jaiswal


C#




// C# program to implement
// the above approach
using System;
class GFG
{
 
  // Function to check if A, B, C
  // can be in A.P. after multiplying
  // by any integer M
  static bool isAP(int A, int B, int C)
  {
    int new_b = B * 2;
 
    // Check if A, B, C are already
    // in A.P.
    if (new_b == (A + C))
      return true;
 
    // Check if multiplying B will
    // give A.P.
    if ((A + C) % new_b == 0)
      return true;
 
    if ((A + C) > new_b)
      return false;
 
    // Check if multiplying A will
    // give A.P.
    if ((new_b - C) % A == 0)
      return true;
 
    // Check if multiplying C will
    // give A.P.
    if ((new_b - A) % C == 0)
      return true;
 
    return false;
  }
 
  // Driver code
  public static int Main()
  {
    int A, B, C;
    A = 30;
    B = 5;
    C = 10;
 
    // Function call
    bool ans = isAP(A, B, C);
 
    // Displaying the answer
    if (ans)
      Console.Write("YES");
    else
      Console.Write("NO");
    return 0;
  }
}
 
// This code is contributed by Taranpreet


Javascript




<script>
        // JavaScript code for the above approach
 
        // Function to check if A, B, C
        // can be in A.P. after multiplying
        // by any integer M
        function isAP(A, B, C)
        {
            let new_b = B * 2;
 
            // Check if A, B, C are already
            // in A.P.
            if (new_b == (A + C))
                return true;
 
            // Check if multiplying B will
            // give A.P.
            if ((A + C) % new_b == 0)
                return true;
 
            if ((A + C) > new_b)
                return false;
 
            // Check if multiplying A will
            // give A.P.
            if ((new_b - C) % A == 0)
                return true;
 
            // Check if multiplying C will
            // give A.P.
            if ((new_b - A) % C == 0)
                return true;
 
            return false;
        }
 
        // Driver code
        let A, B, C;
        A = 30;
        B = 5;
        C = 10;
 
        // Function call
        let ans = isAP(A, B, C);
 
        // Displaying the answer
        if (ans)
            document.write("YES");
        else
            document.write("NO");
 
       // This code is contributed by Potta Lokesh
    </script>


 
 

Output

YES

 

Time Complexity: O(1) 
Auxiliary Space: O(1)

 


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