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Question

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 12k=1log8ak=2010 is

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Solution

Given: 12k=1log8ak=2010
log8a1+log8a2+log8a3++log8a12=2010log8[a1a2a3a12]=2010a.ar.ar2.ar3ar11=82010a12r66=26030a2r11=21005

Let a=2α, r=2β, where α,β are non negative integers
2α+11β=1005
If α=0, then
β=[100511]=91
β91
Also, 11β=10052α
Which is odd, so β is also odd, so
β=1,3,5,7,...,91
Therefore, the number of terms
1+(n1)×2=91n=46

Hence, there are 46 pairs of (a,r) possible.

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