21 and three left over
A) 24 * (9-8)
(5+7)*(5-3)=12*2=24
(4+5-6)x8=24
24 of them.
24
5! / 4 - 2 * 3
4*3*2 = 24 of them.
(7 + 5 - 9) x 8 = 24
24 different numbers.
3= 21/7 21/7 + 3/7 =24/7(7)= 24
If you want 4-digit numbers, there are 24 of them.
Assuming it's required that all the numbers must be used once and only once: (2 + 8 - 9) x 4! = 4! = 24 Factorials are always handy! I imagine other solutions will follow.