Question

Let $A_1,A_2,A_3,...,A_{11}$ be $11$ arithmetic means and $H_1,H_2,H_3,...,H_{11}$ be $11$ harmonic means and $G_1,G_2,G_3,...,G_{11}$ be $11$ geometric means between $1$ and $9$, and the value of $\prod _{k=1}^5{A}_k\cdot G_{12-2k}\cdot H_{12-k}$ is $N$. Then which of the following is\are correct?

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

Answer 1

Answer: (A), (C)

The correct options are
A Number of positive divisors of $N$ are $16$
C If $N$ is divided by $10$, then the remainder is $7$

$A_1\cdot H_{11}=A_2\cdot H_{10}=A_3\cdot H_9=A_4\cdot H_8=A_5\cdot H_7=9G_2\cdot G_{10}=G_4\cdot G_8=(G_6{)}^2=9\prod _{k=1}^5{A}_k\cdot G_{12-2k}\cdot H_{12-k}=(A_1\cdot H_{11})\cdot (A_2\cdot H_{10})\cdot (A_3\cdot H_9)\cdot (A_4\cdot H_8)\cdot (A_5\cdot H_7)\cdot (G_2\cdot G_{10})\cdot (G_4\cdot G_8)\cdot (G_6)=9^5\times 9^2\times 9\Rightarrow N=3^{15}$

So, the number of positive divisors of $N$ is $16$

$3^{15}=3\cdot 9^7=3(10-1{)}^7=10k-3=10(k-1)+7$
So, the remainder is $7$