Let a1,a2,a3,… be a G.P. with a1=a and common ratio r, where a and r positive integers, then the number of ordered pairs (a,r) such that 12∑r=1log8ar=2010 is
(correct answer + 1, wrong answer - 0.25)
A
40
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B
42
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C
44
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D
46
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Solution
The correct option is D46 Given: 12∑r=1log8ar=2010 ⇒log8a1+log8a2+log8a3+⋯+log8a12=2010 ⇒log8[a1a2a3⋯a12]=2010 ⇒a.ar.ar2.ar3⋯ar11=82010 ⇒a12r66=26030 ⇒(a2r11)6=(21005)6 ⇒a2r11=21005
Let a=2α,r=2β, where α,β are non negative integers ⇒2α+11β=1005
If α=0, then β=[100511]=91 ⇒β≤91
Also, 11β=1005−2α
which is odd, so β is also odd. ∴β=1,3,5,7,...,91, which are 46.