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# Let A1,A2,A3,...,A11 be 11 arithmetic means and H1,H2,H3,...,H11 be 11 harmonic means and G1,G2,G3,...,G11 be 11 geometric means between 1 and 9, and the value of 5∏k=1 Ak⋅G12−2k⋅H12−k is N. Then which of the following is\are correct?

A
Number of positive divisors of N are 16
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B
Number of positive divisors of N are 15
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C
If N is divided by 10, then the remainder is 7
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D
If N is divided by 10, then the remainder is 3
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Solution

## The correct options are A Number of positive divisors of N are 16 C If N is divided by 10, then the remainder is 7A1⋅H11=A2⋅H10=A3⋅H9=A4⋅H8=A5⋅H7=9G2⋅G10=G4⋅G8=(G6)2=95∏k=1 Ak⋅G12−2k⋅H12−k=(A1⋅H11)⋅(A2⋅H10)⋅(A3⋅H9)⋅(A4⋅H8)⋅(A5⋅H7)⋅(G2⋅G10)⋅(G4⋅G8)⋅(G6)=95×92×9⇒N=315 So, the number of positive divisors of N is 16 315=3⋅97=3(10−1)7=10k−3=10(k−1)+7 So, the remainder is 7  Suggest Corrections  1      Similar questions  Related Videos   Relation between AM, GM and HM
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