Problem:
Fifteen integers a1​,a2​,a3​,…,a15​ are arranged in order on a number line. The integers are equally spaced and have the property that
1≤a1​≤10,13≤a2​≤20,and241≤a15​≤250.
What is the sum of the digits of a14​?
Answer Choices:
A. 8
B. 9
C. 10
D. 11
E. 12
Solution:
Let d be the common difference. If we let a1​=10 and a15​=241, we see that d≥14241−10​=6.5. Since all the numbers are integers, d must be at least 17. Also, if a1​=1 and a15​=250, we get that d≤14250−1​≈17.8. Once again, since all the numbers are integers, d is at most 17. This tells us that d=17. Note that 17⋅14=238.
This means that a1​ must be at least 3 for a15​ to be within the desired range. If a1​>3, however, a2​ becomes greater than 20, which is not allowed. Now we know that a1​=3 and d=17. This tells us that a14​=3+13⋅17=224.
Therefore, the sum of the digits is 2+2+4=8.
Thus, A is the correct answer.
Answer: A​.
The problems on this page are the property of the MAA's American Mathematics Competitions