An odd integer can be expressed as x = 2n+1 where n is an integer. (Therefore making 2n an even number and 2n+1 an odd number).
Three consecutive odd integers can then be expressed as (2n+1), (2n+3), and (2n+5).
The sum of the first two integers is (2n+1) + (2n+3) = 4n+4 = 4(n+1). It is supposed to be 24 less than 4 times the third integer, or
4(2n+5)-24.
Setting both equal and solving for n:
4(n+1) = 4(2n+5)-24
24 = 4·6, so you can factor out 4 in the right-hand expression.
4(n+1) = 4(2n+5-6) or 4(n+1) = 4(2n-1)
Cancel out factor 4 and solve for n.
n+1 = 2n-1 this yields n=2
Now plug in into your three consecutive odd integer expressions: (2n+1) = 5, (2n+3) = 7, (2n+5) = 9. The sum is therefore 5+7+9 = 21.
Double check your answer:
Sum of first two integers: 5+7 = 12
24 less than 4 times third integer: 4·9 - 24 = 36-24 = 12.
