The HCF of $420$ and $130$ is

10

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15

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5

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3

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Solution

The correct option is A $10$
HCF is the product of the common factors of the given numbers.

First, we need to find the prime factors of the given numbers:

$420 = 2 \times 2 \times 3 \times 5 \times 7 = 2^{2} \times 3 \times 7 \times 5$

$130 = 2 \times 5 \times 13$

So, $H . C . F . = 2 \times 5 = 10$

Hence, HCF if the given number is $10.$

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Similar questions

Use Euclid's division algorithm to find HCF of $420$ and $130$ , hence find LCM.

Q. If HCF of

(420, x) = 60

and LCM of

(420, x) = 1260

, then

x

Question 3

HCF of co-prime numbers 4 and 15 was found as follows by factorization: $4 = 2 \times 2$ and $15 = 3 \times 5$ . Since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

Q. Use Euclid’s algorithm to find the H.C.F. of 420 and 130.

Q. Find the H.C.F of 420 and 130 by iterative division method.

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The HCF of 420 and 130 is

The correct option is A 10HCF is the product of the common factors of the given numbers. First, we need to find the prime factors of the given numbers: 420=2×2×3×5×7=22×3×7×5 130=2×5×13 So, H.C.F.=2×5=10 Hence, HCF if the given number is 10.

The HCF of $420$ and $130$ is