Find the H.C.F of three monomials $56 m^{2} n, 12 m^{2} n, 10 m^{3} n^{2}$ .

2 m^{2} n

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2 m n

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2

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2 m^{3} n^{2}

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Solution

The correct option is A $2 m^{2} n$
$56 m n, 12 m^{2} n, 10 m^{3} n^{2}$
$56 m n = 2 \times 4 \times 7 \times m^{2} n$
$12 m^{2} n = 2 \times 2 \times 3 \times m^{2} n$
$10 m^{3} n^{2} = 2 \times 5 \times m^{3} n^{2}$
So, the common factor is $2 m^{2} n$ .
Therefore, H.C.F is $2 m^{2} n$ .

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