Find three consecutive even integers such that the sum of twice the smallest number and three times the largest is 42.

The difference between two consecutive integers is 2.
Let n be the smallest of the three even integers.
Then the next larger even integer is n + 2.
The largest even integer is n + 4.
Now we follow what the problem tells us.

Twice the smallest is 2n
3 times the larges is 3(n + 4)

2n + 3(n + 4) = 42

2n + 3n + 12 = 42

5n + 12 = 42

5n = 30

n = 6

n + 2 = 8

n + 4 = 10

The integers are 6, 8, 10

pretty sure its b

step-by-step explanation:

why? because i noticed a pattern. 60 120. those numbers without the zero, make the 6 times tables leaving your answer to be b since 6 times 4 is 24 and adding a zero making it 240.