Three of them ... the binary numbers 3, 5, and 6. 011 101 110
8. Algebraic way of solving: 1 * 2 * 2 * 2 * 1 Long way of solving: 10000 10010 10110 11110 10100 11100 11000 11010
Five (5) have one or the other but not both. Six (6) have both. Total of eleven (11).
Three of the bits are already determined. That leaves 5 .25 = 32
5
4? 5?
5
the total 128 ^5 the strings without @ at all 127^5 to get the strings that has at least @ once 128^5 - 127^5
The longest factor string for 45 is 3 x 3 x 5
Yes, many have 5 strings
4 or sometimes 5.
Actually, lengths of a bass does not depend on the number of strings. Basses can be of different length but this has no connection with the number of strings.