Question

what are the fractors of 140​

Answer 1

Factors of 140 = 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70 and 140

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Answer 2

Factors of 140

First we have to do prime factorization of 140.

$\Large \begin{tabular}{c|l}2&140\\ \cline{2-}2&70\\ \cline{2-}5&35\\ \cline{2-}7&7\\ \cline{2-}&1\\ \cline{2-}\end{tabular}$

Thus,  140 = 2² × 5¹ × 7¹

Hence, there are a total of (2 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12 factors  140 has.

And those 12 factors are given below:

$1.\ 2^0 \times 5^0 \times 7^0 = 1 \times 1 \times 1 = \bold{1} \\ \\ 2.\ 2^0 \times 5^0 \times 7^1 = 1 \times 1 \times 7 = \bold{7} \\ \\ 3.\ 2^0 \times 5^1 \times 7^0 = 1 \times 5 \times 1 = \bold{5} \\ \\ 4.\ 2^0 \times 5^1 \times 7^1 = 1 \times 5 \times 7 = \bold{35}$

$5.\ 2^1 \times 5^0 \times 7^0 = 2 \times 1 \times 1 = \bold{2} \\ \\ 6.\ 2^1 \times 5^0 \times 7^1 = 2 \times 1 \times 7 = \bold{14} \\ \\ 7.\ 2^1 \times 5^1 \times 7^0 = 2 \times 5 \times 1 = \bold{10} \\ \\ 8.\ 2^1 \times 5^1 \times 7^1 = 2 \times 5 \times 7 = \bold{70}$

$9.\ 2^2 \times 5^0 \times 7^0 = 4 \times 1 \times 1 = \bold{4} \\ \\ 10.\ 2^2 \times 5^0 \times 7^1 = 4 \times 1 \times 7 = \bold{28} \\ \\ 11.\ 2^2 \times 5^1 \times 7^0 = 4 \times 5 \times 1 = \bold{20} \\ \\ 12.\ 2^2 \times 5^1 \times 7^1 = 1 \times 5 \times 7 = \bold{140}$