Problem:
Fifteen integers a1β,a2β,a3β,β¦,a15β are arranged in order on a number line. The integers are equally spaced and have the property that
1β€a1ββ€10,13β€a2ββ€20,and241β€a15ββ€250.
What is the sum of the digits of a14β?
Answer Choices:
A. 8
B. 9
C. 10
D. 11
E. 12
Solution:
Let d be the common difference. If we let a1β=10 and a15β=241, we see that dβ₯14241β10β=6.5. Since all the numbers are integers, d must be at least 17. Also, if a1β=1 and a15β=250, we get that dβ€14250β1ββ17.8. Once again, since all the numbers are integers, d is at most 17. This tells us that d=17. Note that 17β
14=238.
This means that a1β must be at least 3 for a15β to be within the desired range. If a1β>3, however, a2β becomes greater than 20, which is not allowed. Now we know that a1β=3 and d=17. This tells us that a14β=3+13β
17=224.
Therefore, the sum of the digits is 2+2+4=8.
Thus, A is the correct answer.
Answer: Aβ.
The problems on this page are the property of the MAA's American Mathematics Competitions