10 combinations- 4&7, 4&0, 4&8, 4&2, 7&0, 7&8, 7&2, 0&8, 0&2, and 8&2
0 and 8
If digits can't be repeated, then there are (7 x 7 x 6 x 5 x 4 x 3 x 2) = 35,280If digits can be repeated, then there are (7 x 8 x 8 x 8 x 8 x 8 x 8) = 1,835,008
That is 8001. For a number is divisible by 9, the digits must sum up to 9 (or a multiple of 9). So for 8000, the digits sum to 8 [8 + 0 + 0 + 0 = 8]. So you're only 1 away from 9 [8 + 0 + 0 + 1 = 9]
8 Digits!
-5
Without repeating digits (not digets!) and without leading 0s, 600 of them.
10 combinations- 4&7, 4&0, 4&8, 4&2, 7&0, 7&8, 7&2, 0&8, 0&2, and 8&2
Ten different digits can be used to make 10C4 = 10*9*8*7/(4*3*2*1) = 210 four-digit numbers. Either numbers starting with 0 are permitted or the 10 digits do not contain a 0.
8,764,320
10 digits, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0
11 digits. (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
012258
Ten digits. They are: 0 1 2 3 4 5 6 7 8 9
The answer is 10P3 = 10!/(10-3)! = 10*9*8 = 720
In binary system there are two digits: 0 and 1, or false and true01010101 (8 binary digits or 8 bits) are 1 byte.
0 and 8