A Long Equation II

Algebra Level 3

2 × 6 × 12 × 20 × × 72 = ? \large \sqrt{2}\times \sqrt{6}\times \sqrt{12}\times \sqrt{20}\times \cdots \times \sqrt{72}=?


The answer is 120960.

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2 solutions

P = 2 × 6 × 12 × 20 × × 72 = 2 × 6 × 12 × 20 × 72 = ( 1 × 2 ) × ( 2 × 3 ) × ( 3 × 4 ) × ( 4 × 5 ) × × ( 8 × 9 ) = 1 × 2 2 × 3 2 × 4 2 × 5 2 × × 8 2 × 9 = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 3 = 120960 \begin{aligned} P & = \sqrt 2 \times \sqrt 6 \times \sqrt{12} \times \sqrt{20} \times \cdots \times \sqrt{72} \\ & = \sqrt {2 \times 6 \times 12 \times 20 \times \cdots 72} \\ & = \sqrt {(1 \times 2) \times (2\times 3) \times (3 \times 4) \times (4 \times 5) \times \cdots \times (8 \times 9)} \\ & = \sqrt {1 \times 2^2 \times 3^2 \times 4^2 \times 5^2 \times \cdots \times 8^2 \times 9} \\ & = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 3 \\ & = \boxed{120960} \end{aligned}

nice solution

Razing Thunder - 9 months, 1 week ago

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Glad that you like it.

Chew-Seong Cheong - 9 months, 1 week ago
Edward Christian
Aug 8, 2019

It’s easy to see the pattern n × ( n + 1 ) \sqrt{n\times (n+1)} , therefore we use it to search the regulations. P = 2 × 6 × 12 × 20 × × 72 = 1 × ( 1 + 1 ) × 2 × ( 2 + 1 ) × 3 × ( 3 + 1 ) × × 8 × ( 8 + 1 ) = 1 × 2 × 2 × 3 × 3 × 4 × × 8 × 9 = 1 × 2 × 3 × 4 × × 8 × 3 = 120960 \begin{aligned} P &=\sqrt{2}\times \sqrt{6}\times \sqrt{12}\times \sqrt{20}\times \cdots \times \sqrt{72} \\ &=\sqrt{1\times (1+1)\times 2\times (2+1)\times 3\times (3+1)\times \cdots \times 8 \times (8+1)} \\ &=\sqrt{1\times 2\times 2\times 3\times 3\times 4\times \cdots \times 8\times 9} \\ &=1\times 2\times 3\times 4\times \cdots \times 8\times 3 \\ &=120960 \end{aligned}

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