A Problem on Geometric Progression

Question

Let $a_1,a_2,...$ be in G.P. where $a_1 = a$ and common ratio $r$ are positive integers.

If $\log_8 a_1 + \log_8 a_2 + ... + \log_8 a_{12} = 2014$, the the number of order pairs $(a,r)$ is

(A) 44 (B) 45 (C) 46 (D) 47

I have tried the following but could not continue further.

$$\log_8 a_1 + \log_8 a_2 + ... + \log_8 a_{12} = 2014$$ $$ \Rightarrow \log_8 (a_1 *a_2 * ...*a_{12} ) = 2014$$ $$ \Rightarrow \log_8 (a^{12}*r^{1+2+...+11}) = 2014$$ $$ \Rightarrow \log_8 (a^{12}*r^{66}) = 2014$$ $$ \Rightarrow 12 \log_8 a + 66 \log_8 r = 2014$$ $$ \Rightarrow 3 (2 \log_8 a + 11 \log_8 r) = 1007$$

If you're using the convention that $a_n = a r^n$, then in the third line you should have in the exponent of $r$ the sum up to and including $12$. This change has a cascading effect on the remainder of the calculation. — Sammy Black, Nov 18 '13 at 22:23
No I am not using such convention because $a_1 = a$ as stated in the problem. — aghost, Nov 18 '13 at 22:24

score 2 · Accepted Answer · answered Nov 18 '13 at 22:33

2

Hint: from $ \log_8 (a^{12}*r^{66}) = 2014$, it follows that $8^{2014} = a^{12}. r^{66}$, so $2^{2014} = a^{4}. r^{22}$. Given the fact that $a$ and $r$ are positive integers, can you take it from here?

answered Nov 18 '13 at 22:33

Nicky Hekster

49,281

Thanks. Here is what I did.
$$2^{1007} = a^2*r^{11}$$ Now let $a = 2^x$ and $r = 2^y$. This implies that $2x + 11y = 1007$. There are 45 positive integers less than 1007 which is even and divisible by 11. Thus, my answer is 45.
– aghost Nov 18 '13 at 22:44
1

Yes, this implies that both $a$ and $r$ must be a power of 2, so ... – Nicky Hekster Nov 18 '13 at 22:46
1

Anirban, well done, you got it! – Nicky Hekster Nov 18 '13 at 22:49
1

Hm, slight detail, I count 46 pairs (x,y): (3,91), (14,89), (25,87), ... (493,1) – Nicky Hekster Nov 18 '13 at 22:58

A Problem on Geometric Progression

1 Answers1