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Question : 16 of 86

Marks: +1, -0

\left(216\right)^{\frac{3}{5}} \times \left(2500\right)^{\frac{2}{5}} \times \left(300\right)^{\frac{1}{5}}

Solution:

Concept:

Express each term in prime factor form, then combine exponents to find the total count of prime factors (including multiplicity).

Explanation:

216 = 2^3 \times 3^3

, so

(216)^{\frac{3}{5}} = 2^{\frac{9}{5}} \times 3^{\frac{9}{5}}

2500 = 2^2 \times 5^4

, so

(2500)^{\frac{2}{5}} = 2^{\frac{4}{5}} \times 5^{\frac{8}{5}}

300 = 2^2 \times 3 \times 5^2

, so

(300)^{\frac{1}{5}} = 2^{\frac{2}{5}} \times 3^{\frac{1}{5}} \times 5^{\frac{2}{5}}

Multiplying, for prime

2

\frac{9}{5} + \frac{4}{5} + \frac{2}{5} = \frac{15}{5} = 3

For prime

3

\frac{9}{5} + \frac{1}{5} = \frac{10}{5} = 2

For prime

5

\frac{8}{5} + \frac{2}{5} = \frac{10}{5} = 2

Thus the product equals

2^3 \times 3^2 \times 5^2

Total prime factors (with multiplicity) =

3 + 2 + 2 = 7

Answer:

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