Find three consecutive odd integers such that the sum of the smallest and 7 times the largest is 68

First we must assign variables to the integers.

x = smallest
y = middle
z = largest

x + 7z = 68    ← This equation represents the given information
 
We can assume that "z" is less than 11 because 7*11 = 77, which is greater than 68.

Our possible consecutive odd integers could be:

a) 1, 3, 5
b) 3, 5, 7
c) 5,7, 9

Lets to plug in the "x" and "z" values

a) 1 + 7(5) = 1+ 35 = 36       This is incorrect, because 36 ≠ 68

b) 3 + 7(7) = 3 + 49 = 52      This is incorrect as well, because 52 ≠ 68

That leaves our final choice. Let's check to be certain

c) 5 + 7(9) = 5 + 63 = 68

5, 7, 9 is the 3 consecutive integers such that the sum of the smallest and 7 times the largest is 68.

Megan will say more numbers.

Step-by-step explanation:

Given: Megan counts by ones to 10.

So, the numbers counted by Megan=10 times 1

Therefore, Megan will say 10 numbers.

Lee counts by fives to 20.

So, the numbers counted by Lee=4 times 5 [∵20 is a multiple of 5, 5×4=20 ]

Therefore, Lee will say 4 numbers.

And 10>4

Therefore, Megan will say more numbers than Lee.