What is the smallest of four consecutive odd integers whose sum is 968?
A. 235
B. 239
C. 243
D. 245
Let the smallest of the four consecutive odd integers be m. Then, m + (m + 2) + (m + 4) + (m + 6) = 968 or 4m + 12 = 968.
Thus, m = 956 / 4 = 239.
The four consecutive odd integers are 239, 241, 243 and 245.