- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

# Find three integers less than or equal to N such that their LCM is maximum - C++

In this tutorial, we are going to write a program that is based on the concepts of LCM. As the title says, we have to find three numbers that are less than or equal to the given number whose LCM is maximum.

Let's see an example.

Before diving into the problem let's see what LCM is and write program for it.

**LCM** is the least common multiple of a number. It is also known as Least common divisor. For two positive number a and b, the LCM is the smallest integer that is evenly divisible by both a and b.

If the given integers don't have common factor, then the **LCM** is the product of the given numbers.

## Example

Let's write the program to find LCM of any two positive numbers given.

#include <iostream> using namespace std; int main() { int a = 4, b = 5; int maximum = max(a, b); while (true) { if (maximum % a == 0 && maximum % b == 0) { cout << "LCM: " << maximum << endl; break; } maximum++; } }

## Output

If you run the above program, you will get the following output.

20

We have seen what LCM is and program to find the LCM of two positive numbers.

Let's see the steps to solve the problem.

If the number is odd then three number with max LCM are

**n, n - 1,**and**n - 3**.If the number is even and the GCM of

**n**and**n - 3**is**1**then the three numbers with max LCM are**n, n - 1**, and**n - 3**.Else the three numbers with max LCM are

**n - 1, n - 2**, and**n - 3**.

## Example

Let's see the code.

#include <bits/stdc++.h> using namespace std; void threeNumbersWithMaxLCM(int n) { if (n % 2 != 0) { cout << n << " " << (n - 1) << " " << (n - 2); } else if (__gcd(n, (n - 3)) == 1) { cout << n << " " << (n - 1) << " " << (n - 3); } else { cout << (n - 1) << " " << (n - 2) << " " << (n - 3); } cout << endl; } int main() { int n = 18; threeNumbersWithMaxLCM(n); return 0; }

## Output

If you execute the above program, then you will get the following result.

17 16 15

## Conclusion

If you have any queries in the tutorial, mention them in the comment section.

- Related Questions & Answers
- Find three integers less than or equal to N such that their LCM is maximum in C++
- Find maximum number of elements such that their absolute difference is less than or equal to 1 in C++
- Find unique pairs such that each element is less than or equal to N in C++
- C++ program to find unique pairs such that each element is less than or equal to N
- Find maximum product of digits among numbers less than or equal to N in C++
- Find all factorial numbers less than or equal to n in C++
- Find Multiples of 2 or 3 or 5 less than or equal to N in C++
- Find maximum sum array of length less than or equal to m in C++
- Print all prime numbers less than or equal to N in C++
- Maximum sum of distinct numbers such that LCM of these numbers is N in C++
- Maximum product from array such that frequency sum of all repeating elements in product is less than or equal to 2 * k in C++
- Print all Semi-Prime Numbers less than or equal to N in C++
- Maximum sum subarray having sum less than or equal to given sums in C++
- Program to Find Out Integers Less than n Containing Multiple Similar Digits in C++
- Maximum subarray size, such that all subarrays of that size have sum less than k in C++