You can arrange them to make a cube.12 edges, 6 faces.
There are no such numbers. The only two numbers that sum to 12 and whose product is -84 are irrational, and so cannot be considered as factors of 84.
12 and 1
Well, honey, if you want a set of numbers that fit those criteria, how about 7, 7, 7, 12, 12, 12, 12, 12, 12, 12, 18, 18? You've got your mode of 7, median of 12, maximum of 18, and a range of 13. Looks like I just served you a math cocktail with a side of sass. Cheers!
The bottoms number like 3/4 4 would be the denominator heres a problem 6/12 + 2/12 the bottom numbers are the same
Assuming that the only permitted operation is addition, then there are 4 combinations.
5 + 7 + (6 x √4) = 12 + (6 x 2) = 12 + 12 = 24
12+5+5+1
One example. 12*8+3+7-6 = 100 The key is finding sets of 10s
7+5=12 2-9=3 3x3=9
(5+7)*(5-3)=12*2=24
No. By adding 4+4+4 the answer comes to 12. To subtract the answer comes to 4. By multiplying it is 64.
(5-2)*4 = 12
2x2x3 = 12
You cannot! Answer dated: 13 January 2012
Impossible, as there are not enough numbers to cover all squares.
You can make 4*3*2/2 = 12 numbers.