The first box is 3 more than the quotient of 6 and a number, the second is 6 less than the product of 3 and a number, the third box is the product of 6 and the sum of a number and 3.
Supposing that the integers are x, y and z, and that their average is 8, then
x+y+z
-------- = 8, or x + y + z = 24.
3
Let's experiment. Making up a table, we get:
x y z sum (sum MUST be 24)
5 9 10 24
4 9 11 24
3 10 11 24
2 11 11 24
1 11 12 24
1 10 13 24
1 1 22 24
Can't choose integers smaller than 1, so it appears that 22 is the largest possible integer that could be one of x, y and z.
Answer:
168
Step-by-step explanation:
7*-4=-28
-28*-6=168
