Question

write a program to print first 50 multiple of 7. Q basic​

Answer 1

Answer:

Given a positive integer n, find count of all multiples of 7 less than or equal to n.

Examples :

Input : n = 10

Output : Count = 4

The multiples are 3, 6, 7 and 9

Input : n = 25

Output : Count = 10

The multiples are 3, 6, 7, 9, 12, 14, 15, 18, 21 and 24

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A Simple Solution is to iterate over all numbers from 1 to n and increment count whenever a number is a multiple of 3 or 7 or both.

// A Simple C++ program to find count of all

// numbers that multiples

#include<iostream>

using namespace std;

// Returns count of all numbers smaller than

// or equal to n and multples of 3 or 7 or both

int countMultiples(int n)

{

int res = 0;

for (int i=1; i<=n; i++)

if (i%3==0 || i%7 == 0)

res++;

return res;

}

// Driver code

int main()

{

cout << "Count = " << countMultiples(25);

}

Output :

Count = 10

Time Complexity : O(n)

An efficient solution can solve the above problem in O(1) time. The idea is to count multiples of 3 and add multiples of 7, then subtract multiples of 21 because these are counted twice.

count = n/3 + n/7 - n/21

// A better C++ program to find count of all

// numbers that multiples

#include<iostream>

using namespace std;

// Returns count of all numbers smaller than

// or equal to n and multples of 3 or 7 or both

int countMultiples(int n)

{

return n/3 + n/7 -n/21;

}

// Driver code

int main()

{

cout << "Count = " << countMultiples(25);

}

Explanation:

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Answer 2

Answer:

first 50 multiple of 7. Q

Explanation:

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