Problem: Let P equal the product of any three consecutive positive odd integers. The largest integer dividing all such P is:
Answer Choices:
A. 15
B. 6
C. 5
D. 3
E. 1
Solution:
Method I. Set P=(2k−1)(2k+1)(2k+3). If 2k−1=3m then 3∣P. If 2k−1 =3m+1, then 2k+1=3m+3 and so 3∣P. If 2k−1=3m−1, then 2k+3=3 m+3 and so 3∣P.
Method II. Set P=(2k−1)(2k+1)(2k+3)=8k3+12k2−2k−3. ∴P=(9k3+12k2−3k−3)−k3+k=3Q−(k−1)(k)(k+1). Since the product of three consecutive integers is divisible by 3,P=3Q−3R. Hence, 3∣P.
Method III. Consider the three consecutive integers k,k+1,k+2; one of these is divisible by 3 (see Method II). If k+1 is divisible by 3 , then so is k+1+3=k+4. Therefore, one of k,k+2,k+4 is divisible by 3 . Therefore, if P=k(k+2)(k+4) with k odd, P is divisible by 3 .
Note. For all three methods, since the greatest common factor of 1.3.5 and 7⋅9⋅11 is 3 , no number >3 is an exact divisor for all P.