This may not be exactly what you want, but it does come close.
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for (x = 2; x < 30; x += 5)
{
for (i = 2; i < x; i += 2)
{
conversion = i*3.28;
cout << right << setw(5) << i << setw(15) << fixed << setprecision(2) << conversion << endl;
++count;
if(count == 5 || count == 10 || count == 15)
cout << endl;
}
cout << endl;
count = 0;
}
Consider the first time through the loops when x = 2 the inner for loop only prints 3 lines then prints a blank line before going through the inner for loop a second time.
If this does not work let me know. I have another idea I have not tried yet.