If $2^{p} {.3}^{q} {.5}^{r} = 60$ , which of the following is true? Given; p, q, r are natural numbers.

p + q + r = 3

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p = q = r

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p = 2 , q = r = 1

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p + q + r = 0

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Solution

The correct option is C
p = 2 , q = r = 1

$60 = 2^{2} .3 .5$
Fundamental theorem of arithmetic states that every composite number can be expressed as product of prime numbers and this factorisation is unique.

$\Rightarrow$ $60 = 2^{2} .3 .5 = (2^{p} {.3}^{q} {.5}^{r}$ )
Hence comparing , we get
$∴$ p = 2, q = 1 and r =1

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If $2^{p} {.3}^{q} {.5}^{r} = 60$ , which of the following is true? Given; p, q, r are natural numbers.

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