Problem:
The sequence a0​,a1​,a2​,⋯ is a strictly increasing arithmetic sequence of positive integers such that
2a7​= 227⋅a7​
What is the minimum possible value of a2​ ?
Answer Choices:
A. 8
B. 12
C. 16
D. 17
E. 22
Solution:
There are integers a0​ and d such that an​=a0​+dn. Thus 2a0​+7d=227⋅(a0​+7d). It follows that a0​+7d is a power of 2 ; that is, a0​+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 a0​+7d=2k=32, the pair (a0​,d) must be one of (4,4),(11,3),(18,2), or (25,1). The corresponding values of a2​ are, respectively, 12,17,22, and 27 . The requested minimum value is (B)12​ .
The problems on this page are the property of the MAA's American Mathematics Competitions