Three numbers which are relatively prime to each other has a product of $900,$ then their sum is

10

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28

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38

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45

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Solution

The correct option is C $38$
Prime factorization of $900$ is $2^{2} \times 3^{2} \times 5^{2}$
$= 4 \times 9 \times 25$
They are relatively co-prime to each other.
Sum $= 4 + 9 + 25 = 38$

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Three numbers which are relatively prime to each other has a product of 900, then their sum is

The correct option is C 38Prime factorization of 900 is 22×32×52 =4×9×25 They are relatively co-prime to each other. Sum =4+9+25=38

Three numbers which are relatively prime to each other has a product of $900,$ then their sum is

The correct option is C $38$
Prime factorization of $900$ is $2^{2} \times 3^{2} \times 5^{2}$
$= 4 \times 9 \times 25$
They are relatively co-prime to each other.
Sum $= 4 + 9 + 25 = 38$