Problem:
Let a,b,c,d, and e be positive integers with a+b+c+d+e=2010, and let M be the largest of the sums a+b,b+c,c+d and d+e. What is the smallest possible value of M ?
Answer Choices:
A. 670
B. 671
C. 802
D. 803
E. 804
Solution:
Note that 3M>(a+b)+c+(d+e)=2010, so M>670. Because M is an integer M≥671. The value of 671​ is achieved if (a,b,c,d,e)= (669,1,670,1,669).
The problems on this page are the property of the MAA's American Mathematics Competitions