Problem:
The sequence a1,a2,… is geometric with a1=a and common ratio r, where a and r are positive integers. Given that log8a1+log8a2+⋯+log8a12=2006, find the number of possible ordered pairs (a,r).
Solution:
Note that
loga1+loga2+⋯+loga12=log(a1a2⋯a12)=log(a⋅ar⋯ar11)=log(a12r66)
where the base of the logarithms is 8. Therefore a12r66=82006=23⋅2006, so a2r11=21003. Because a and r are positive integers, each must be a factor of 21003. Thus a=2x and r=2y for nonnegative integers x and y. Hence 2x+11y=1003, and each ordered pair (a,r) corresponds to exactly one ordered pair (x,y) that satisfies this equation. Because 2x is even and 1003 is odd, y must be odd, so y has the form 2k−1, where k is a positive integer. Then 1003=2x+11y=2x+22k−11, so x=507−11k. Therefore 507−11k≥0, and so 1≤k≤⌊507/11⌋=46. Thus there are 46 possible ordered pairs (a,r).
The problems on this page are the property of the MAA's American Mathematics Competitions