How to display all positive even numbers less than n

Question

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Using for statement, write a program that prompts the user for a positive integer n then display all positive even numbers less than n then display their sums and product.

Please tell me how to display even number

C++

#include <iostream>

using namespace std;

int main() {

	int n;
	int sum;
	int product=1;
	int countNumber = 0;
	cout << "Please enter the number of  positive integers  = ";
	cin >> n;
	if (n > 0) {
		for (sum = 0;n > countNumber;)

		{
			sum = sum + n;
			product = product * n;
			n = n - 1;

		}
	}
	cout << "The sum the integers number = " << sum << endl;
	cout << " The product number of integers = " << product << endl;

	system("pause");
	return 0;
}

Posted 1-Nov-15 10:20am

Assam ALzookery

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Comments

Sergey Alexandrovich Kryukov 1-Nov-15 16:29pm

How? By thinking just a bit.
—SA

Assam ALzookery 1-Nov-15 16:49pm

can you tell me where's my mistake

#include <iostream>

using namespace std;

int main() {

int n;
int sum;
int product=1;
int countNumber = 0;
cout << "Please enter the number of positive integers = ";
cin >> n;
if (n % 2 == 0) {
for (sum = 0; n >= 0;)

{

sum = sum + n;
product = product * n;
n = n + 2;
}
}
cout << "The even Numbers are = " << n << endl;
cout << "The sum the integers number = " << sum << endl;
cout << " The product number of integers = " << product << endl;

system("pause");
return 0;
}

Sergey Alexandrovich Kryukov 1-Nov-15 17:11pm

No valid loop. Should be three terms separated by ';': starting initialization, end condition, step statement.
Don't checkup n%2, instead, calculate new ending value valid no matter if initial value is odd or even. Use different loop variable and declare it in loop statement (not only initialize, declare), and never modify it inside loop.

In other words, you need to get familiar with loops from ground zero.

—SA

Matthew Dennis 1-Nov-15 17:45pm

You too.
An empty iteration statement (the third part) is a valid expression. This can be zero or more expressions separated by the comma operator, and don't have to do anything related to the termination condition.

Sergey Alexandrovich Kryukov 1-Nov-15 20:57pm

You are right. I don't think it would make much sense in this particular case, but yes, it's possible to implement things without this third term. Anyway, I would advise to master the most widely used 3-term form of "for" loop before experimenting with more specialized forms.

Thank you for this comment.

—SA

Assam ALzookery 1-Nov-15 19:59pm

is it better like this ??

#include <iostream>

using namespace std;

int main() {

int n;
int sum=0;
int product = 1;
int countNumber = 0;
cout << "Please enter the number of positive integers = ";
cin >> n;

for (int current = 2; current < n; current += 2)
cout << current << endl;

{
if (n % 2 == 0) {
sum = sum + n;
product = product * n;
n = n - 1;

}

cout << " The even Numbers are = " << n << endl;
cout << "The sum of the integers number = " << sum << endl;
cout << " The product number of integers = " << product << endl;

}
system("pause");
return 0;
}

Sergey Alexandrovich Kryukov 1-Nov-15 20:58pm

Before going any further, could I ask you to avoid putting any code in comments? Instead, update the body of your question, add your code in well-formatted fashion.
Did you notice that your first line reads "#include". Please fix it all.
—SA

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Sergey Alexandrovich Kryukov · Accepted Answer · 2015-11-01T10:33:00

Please see my comment to the question.

First, an even number is the number N with N % 2 == 0; second, having an even number N, you can obtain "next" even number, which is N + 2, which will let you to use some loop starting from zero. These considerations, which you should have conducted by yourself in first place, should be enough for you to write this trivial piece of code. Now it's your turn.

—SA

Assam ALzookery · Accepted Answer · 2015-11-01T14:22:00

C++

#include <iostream>

using namespace std;

int main() {

	int n;
	int sum=0;
	int product = 1;
	int countNumber = 0;
	cout << "Please enter the number of  positive integers  = ";
	cin >> n;

	for (countNumber = 2; n>countNumber; countNumber += 2) // everything in the for(;;) 
	{
		cout << countNumber << endl;
		sum = sum + countNumber;
		product = product * countNumber;
	}
	{
		cout << " The even Numbers are = " << n << endl;
		cout << "The sum of the integers number = " << sum << endl;
		cout << " The product number of integers = " << product << endl;

	}
	system("pause");
	return 0;
}</iostream>

How to display all positive even numbers less than n

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