Without experience that would help you recognize it right away, I think
you would just need to try numbers until you find it.
Pick a number, multiply it by itself with pencil and paper, and see if it works.
(2)² = 4 too small
(3)² = 9 too small, and not growing fast enough; this could take all day
(10)² = 100 still too small
(15)² = 225 oops, too big; go back down
(12)² = 144 too small
(14)² = 196 too big
(13)² = 169 just right; hope the bears don't come back too soon
The number of different groups can be found by finding 9C3 (Using combinations)
We will find combinations from n = 9 to r = 3
Therefore, 9C3 = 9!/6!*3! = (9*8*7*6!)/(6!*3*2)
= 3*4*7
= 84 ways.
10/1 * 1/3 = 30/3
1 1/3 = 4/3
4/3 can only go into 30/3, 7 times, with a remainder of 2/3.
7 2/3
the total cost of order c is 30
3x^3=0
the only possible answer is x=0
factor
3(x^3)=0
x^3=0
x=0
C
