Without hairy calculation

Algebra Level 3

What is the maximum positive integer n n which satisfies ∑ i = 1 n 2 i − 6 > ∏ i = 1 n 2 i − 6 \displaystyle \sum_{i=1}^n 2^{i-6} > \displaystyle \prod_{i=1}^n 2^{i-6} ?

Hint: Think before you get down to calculation.


The answer is 12.

Alice Smith by Alice Smith , 20, China

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1 solution

∑ k = 1 n 2 k − 6 > ∏ k = 1 n 2 k − 6 ∑ k = − 5 n − 6 2 k > ∏ k = − 5 n − 6 2 k 1 32 ∑ k = 0 n − 6 2 k > ∏ k = − 5 5 2 k ∏ k = 6 n − 6 2 k Sum of a geometric progression 2 n − 5 − 1 32 > ∏ k = 6 n − 6 2 k Note that ∏ k = − 5 5 2 k = 1 2 n − 5 − 1 > 32 ⋅ 2 ∑ k = 6 n − 6 k Sum of an arithmetic progression = 2 5 ⋅ 2 n ( n − 11 ) 2 ⟹ 2 n − 5 > 2 ( n − 10 ) ( n − 1 ) 2 + 1 ⟹ n ≤ 12 \begin{aligned} \sum_{k=1}^n 2^{k-6} & > \prod_{k=1}^n 2^{k-6} \\ \sum_{k=-5}^{n-6} 2^k & > \prod_{k=-5}^{n-6} 2^k \\ \frac 1{32} \color{#3D99F6} \sum_{k=0}^{n-6} 2^k & > {\color{#D61F06}\prod_{k=-5}^5 2^k} \prod_{k=6}^{n-6} 2^k & \small \color{#3D99F6} \text{Sum of a geometric progression} \\ \frac {\color{#3D99F6} 2^{n-5}-1}{32} & > \prod_{k=6}^{n-6} 2^k & \small \color{#D61F06} \text{Note that }\prod_{k=-5}^5 2^k = 1 \\ 2^{n-5}-1& > 32 \cdot 2^{\color{#3D99F6} \sum_{k=6}^{n-6} k} & \small \color{#3D99F6} \text{Sum of an arithmetic progression} \\ & = 2^5 \cdot 2^{\color{#3D99F6}\frac {n(n-11)}2} \\ \implies 2^{n-5} & > 2^{\frac {(n-10)(n-1)}2} + 1 \\ \implies n & \le \boxed{12} \end{aligned}

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Mod: Why did the Level change from Pending to 3? Is that usually the practice?

E Koh - 1 year, 9 months ago

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As far as I know. The difficulty rating of the problem is done automatically basing on the number of persons trying the problem and how many get it right.

Chew-Seong Cheong - 1 year, 9 months ago

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