- Related Questions & Answers
- Maximum positive integer divisible by C and is in the range [A, B] in C++
- Find the minimum positive integer such that it is divisible by A and sum of its digits is equal to B in Python
- Minimum positive integer value possible of X for given A and B in X = P*A + Q*B in C++
- Find minimum positive integer x such that a(x^2) + b(x) + c >= k in C++
- Check if a number is divisible by 23 or not in C++
- Check if a number is divisible by 41 or not in C++
- Find permutation of n which is divisible by 3 but not divisible by 6 in C++
- Count pairs (i,j) such that (i+j) is divisible by both A and B in C++
- Minimum positive integer required to split the array equally in C++
- Program to find Nth term divisible by a or b in C++
- Check if a large number is divisible by 11 or not in C++
- Check if a large number is divisible by 25 or not in C++
- Check if a large number is divisible by 3 or not in C++
- Check if a large number is divisible by 5 or not in C++
- Check if a large number is divisible by 75 or not in C++

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

Suppose we have three integers A, B, and C. We have to find one minimum integer X, such that X mod C = 0, and X is not in the range [A, B]. If the values of A, B and C are 5, 10 and 4 respectively, then the value of X will be 4. Let us see the steps to get the solution −

**Steps **−

- If C is not in the range [A, B], then return C as a result
- Otherwise get the first multiple of C, which is greater than B, then return that value

#include <iostream> using namespace std; int findMinMumber(int a, int b, int c) { if (c < a || c > b) return c; int res = ((b / c) * c) + c; return res; } int main() { int a = 2, b = 4, c = 2; cout << "Minimum number X: " << findMinMumber(a, b, c); }

Minimum number X: 6

Advertisements