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 integer difference between adjacent numbers ai and ai+1. Then the difference between the 1st and 15th numbers is a15−a1=14d. Given the range values for a1 and a15, the quantity 14d must lie in the following interval:
241−10≤14d≤250−1,231≤14d≤249.
The only multiple of 14 in this interval is 238=14⋅17. Thus the difference d equals 17 and the numbers a2 and a15 can be expressed in terms of a1 as followa:
a2=a1+d=a1+17,a15=a1+14d=a1+238.
Next observe that the greatest possible value of a1 is bounded by the maximum value of a2.
a2≤20⇒a1+17≤20⇒a1≤3.
Similarly the least possible value of a1 is bounded by the minimum value of a15.
a15≥241⇒a1+238≥241⇒a1≥3.
The only value of a1 that satisfies both inequalities is a1=3. Therefore
a2=a1+17=20,a15=a1+238=241.
and the value of a14 is
a14=a15−d=241−17=224.
Thus the sum of the digits of a14 is 2+2+4=8.
Answer: A.
The problems on this page are the property of the MAA's American Mathematics Competitions