The HCF of $3^{4}, 27^{2}, 81^{2} and 90^{3}$ is

3^{3}

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3^{6}

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3^{4}

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3^{8}

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Solution

The correct option is C $3^{4}$
$3^{4} = 3^{4}$

$27^{2} = [{(3)}^{3}]^{2} = 3^{6}$

$81^{2} = [{(3)}^{4}]^{2} = 3^{8}$

$90^{3} = {(3^{2} \times 2 \times 5)}^{3} = (3^{2})^{3} \times 2^{3} \times 5^{3} = 3^{6} \times 2^{3} \times 5^{3}$

So the HCF of $3^{4}, 27^{2}, 81^{2} and 90^{3} is 3^{4} .$

Hence, the correct answer is option (c).

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Similar questions

Find the HCF of $1 \frac{1}{2}$ and $3 \frac{4}{5}$

The HCF of $2^{2} \times 3^{2} \times 5^{2}, 2^{4} \times 3^{4} \times 5^{3} \times 11, 2 \times 3^{2} \times 5^{2}, 2^{4} \times 3^{4} \times 5^{3} \times 11$ is:

90^{3}

is equal to

Question 1 (f)

Find the HCF of the following numbers:
34, 102

The value of $(90)^{3} - (80)^{3} - 3 \times 90 \times 80 \times 10 =$ ___

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The HCF of 34, 272, 812 and 903 is

The correct option is C 3434=34 272=[(3)3]2=36 812=[(3)4]2=38 903=(32×2×5)3=(32)3×23×53=36×23×53 So the HCF of 34, 272, 812 and 903 is 34. Hence, the correct answer is option (c).

The HCF of $3^{4}, 27^{2}, 81^{2} and 90^{3}$ is