C++ Solutions
Note the similarities in the loops.
1a. int i, sum;
sum = 0;
for (i = 1; i < 11; i++) // or use i <= 10
{
sum += i;
cout << "\nCurrent sum is " << sum);
}
1b. Change for statement to for (i = 1; i < 51; i++)
1c. Change for statement to for (i = 1; i < 101; i++)
2. int i, sum;
sum = 0;
for (i = 2; i < 10; i++)
{
sum += i;
cout << "\nCurrent sum is " << sum;
}
3a. int i;
for (i = 3; i <= 20; i+=3)
{
cout << "\n" << i;
}
Here you start with the first multiple and increment by the number that you are finding multiples of. You must know the first multiple.
OR
for (i = 1; i <= 20; i++)
{
if (i % 3 == 0)
cout << "\n" << i;
}
Here you increment by 1, but only print it if it is a multiple.
3b. int i;
for (i = 100; i <= 200; i++)
{
if (i % 3 == 0)
cout << " " << i);
}
By using the 2nd version, you do not need to know the first multiple.
3c. int i, begin, end;
cout << "\nEnter the beginning of the search range: "; cin >> begin; cout << "\nEnter the end of the search range: "; cin >> end;
for (i = begin; i <= end; i++)
{
if (i % 3 == 0)
cout << " " << i;
}
4. #include ...
#include ...
void main()
{
int i, begin, end, base;
cout << "\nEnter the beginning of the search range: "; cin >> begin; cout << "\nEnter the end of the search range: "; cin >> end; cout << "\nEnter the integer whose multiples you want: "; cin >> base;
for (i = begin; i <= end; i++)
{
if (i % base == 0)
cout << " " << i;
}
}