Problem:
The sequence a0β,a1β,a2β,β― is a strictly increasing arithmetic sequence of positive integers such that
2a7β= 227β
a7β
What is the minimum possible value of a2β ?
Answer Choices:
A. 8
B. 12
C. 16
D. 17
E. 22
Solution:
There are integers a0β and d such that anβ=a0β+dn. Thus 2a0β+7d=227β
(a0β+7d). It follows that a0β+7d is a power of 2 ; that is, a0β+7d=2k for some nonnegative integer k. Then 22k=227+k, so 2k=27+k. Note that k=5 is the only solution (because the value of the exponential function y=2x is greater than the value of the linear function y=27+x for x>5 ). Because a0β+7d=2k=32, the pair (a0β,d) must be one of (4,4),(11,3),(18,2), or (25,1). The corresponding values of a2β are, respectively, 12,17,22, and 27 . The requested minimum value is (B)12β .
The problems on this page are the property of the MAA's American Mathematics Competitions