Match the following with its prime factors.

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

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5 \times 5 \times 5

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3 \times 3 \times 3 \times 3

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Solution

Given, $300, 125 & 81$

Lets factorise:

$300 = 3 \times 2 \times 2 \times 5 \times 5$

$125 = 5 \times 5 \times 5$

$81 = 3 \times 3 \times 3 \times 3$

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\frac{2}{5} + \frac{3}{5^{2}} + \frac{2}{5^{3}} + \frac{3}{5^{4}} + \frac{2}{5^{5}} + \frac{3}{5^{6}} + . . . . . .

(\frac{1}{5} + \frac{2}{5^{2}} + \frac{3}{5^{3}}) + (\frac{1}{5^{4}} + \frac{2}{5^{5}} + \frac{3}{5^{6}}) + . . . . . . . .

7

The number of distinct prime factors in (216) $^{3 / 5}$ (2500) $^{2 / 5}$ (300) $^{1 / 5}$ is

In which of the following expressions, prime factorization has been done:

(i) $24 = 2 \times 3 \times 4$
(ii) $56 = 7 \times 2 \times 2 \times 2$
(iii) $70 = 2 \times 5 \times 7$
(iv) $54 = 2 \times 3 \times 9$

If the prime factorization of a number is 3x3x3x5x5x5, then its cube root is:

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