9-5=4 7-4=3 3*8=24
write the largest number you can make using each of the digits 7,1,0,2, and 9 just once
There can be only 1 combination since, in a combination, the order of the digits does not matter. So, since you must have each digit once and only once, the only combination is 3456.
7,610 or (6)710
The closest you can get by: Using Each Number Once With Powers: 10 th the second power - 4 th the third power=46 Using each Number Once Without Powers:(4*10)-3+2 Using an indefinite amount of each number:4-*10+(3*2)-(2+2+2)
6^(7-5) = 36
(9-5)*(7-1)
96 x 1/4 = 24
(-6)*(-5+(-7)/(-7)) = 24
It is simply: 1*2*3*4 = 24
It is: (9-4-2) times 8 = 24
78 - (6 x 9) = 78 - 54 = 24
Try: (5-1) times (8-2) = 24
It is: (5+13)/6 times 8 = 24
It can be done by: (9-5) times 6 times 1 = 24
It is absolutely certain that a number containing each of the numbers 1 - 9 exactly once will be divisble by 9. Any number whose digits add up equal to 9 will also be equal to 9 - and any number with all the numbers 1 - 9 in them will have a digit sum equal to 45.
(9 - 1) x (7 - 4) = 8 x 3 = 24