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Question

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 5k=1 AkG122kH12k 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 7
A1H11=A2H10=A3H9=A4H8=A5H7=9G2G10=G4G8=(G6)2=95k=1 AkG122kH12k=(A1H11)(A2H10)(A3H9)(A4H8)(A5H7)(G2G10)(G4G8)(G6)=95×92×9N=315

So, the number of positive divisors of N is 16

315=397=3(101)7=10k3=10(k1)+7
So, the remainder is 7

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