5 digit numbers that can be formed by using digits 0,1,2,3,4,5,6 and are divisible by 8

Question

How many 5 digit numbers can be formed by using digits 0,1,2,3,4,5&6 and are divisible by 8?

Answer 1

Note that $100\equiv4$ modulo $8$. We therefore can distinguish the following two cases:

If the middle digit is one of $0$, $2$, $4$, $6$ the last two digits can be one of $$00, 16, 24,32,40,56, 64\ ,$$ and if the middle digit is one of $1$, $3$, $5$ the last two digits can be one of $$04, 12, 20, 36, 44, 52, 60\ .$$ This gives $4\cdot7+3\cdot7=49$ possibilities for the last three digits. For the first two digits there are $6\cdot 7=42$ possibilities (if a leading zero is forbidden), so that we obtain $42\cdot 49=2058$ such numbers in total.

Answer 2

To be divisible by 8, the number's e last three digits form a number divisible by 8 In this case, they are these;

000 016 024 032 040 056 064

104 112 120 136 144 152 160

200 216 224 232 240 256 264

304 312 320 336 344 352 360

400 416 424 432 440 456 464

504 512 520 536 544 552 560

600 616 624 632 640 656 664

The total is 49

the first two digits can be anything, so the combinations are 6*6=36 so, the whole combinations possible are 49*36=1764

1764 possible ways

(assuming that, we use a digit more than once like 10008)

Edit; I forgot 0 at the part where I get the first two digits.

the first two digits can be anything(the first digit cannot be 0), so the combinations are 6*7=42 so, the whole combinations possible are 49*42=2058

2058 ways