Can you solve this tower?

Algebra Level 3

Find the value of 1 2015 1 2015 2015 1 2015 × 2015 × 2015 × 2015 × 2015 \dfrac{\hspace{18 mm} \dfrac{\hspace{16 mm} \dfrac{\hspace{14 mm} 1 \hspace{14 mm}}{\dfrac{\hspace{5 mm} 2015 \hspace{5 mm}}{\dfrac{1}{2015}}}\hspace{16 mm}}{2015} \hspace{18 mm}}{\dfrac{1}{2015 \times 2015 \times 2015 \times 2015 \times 2015}}

1 1 201 5 2 2015^{-2} 201 5 2 2015^2 201 5 1 2015^{-1} 2015 2015

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1 solution

Rick B
Jan 20, 2015

1 2015 1 2015 2015 1 2015 × 2015 × 2015 × 2015 × 2015 \dfrac{\hspace{18 mm} \dfrac{\hspace{16 mm} \dfrac{\hspace{14 mm} 1 \hspace{14 mm}}{\dfrac{\hspace{5 mm} 2015 \hspace{5 mm}}{\dfrac{1}{2015}}}\hspace{16 mm}}{2015} \hspace{18 mm}}{\dfrac{1}{2015 \times 2015 \times 2015 \times 2015 \times 2015}}

= 1 201 5 2 2015 1 2015 × 2015 × 2015 × 2015 × 2015 = \dfrac{\hspace{18 mm} \dfrac{\hspace{16 mm} \dfrac{\hspace{14 mm} 1 \hspace{14 mm}}{2015^2}\hspace{16 mm}}{2015} \hspace{18 mm}}{\dfrac{1}{2015 \times 2015 \times 2015 \times 2015 \times 2015}}

= 1 201 5 3 1 201 5 5 = \dfrac{\hspace{18 mm} \dfrac{1}{2015^3} \hspace{18 mm}}{\dfrac{1}{2015^5}}

= 1 201 5 3 × 201 5 5 1 = 201 5 5 201 5 3 = 201 5 2 = \dfrac{1}{2015^3} \times \dfrac{2015^5}{1} = \dfrac{2015^5}{2015^3} = \boxed{2015^2}

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