If $49392 = a^4b^2c^3$, find the values of $a, b$ and $c$, where $a, b$ and $c$ are different positive primes.

Given:

$49392 = a^4b^2c^3$

To do:

We have to find the values of $a, b$ and $c$, where $a, b$ and $c$ are different positive primes.

Solution:

Prime factorisation of 49392 is,

$49392=2^4\times3^2\times7^3$

Therefore,

$2^4\times3^2\times7^3=a^4b^2c^3$

Comparing both the sides, we get,

$a=2, b=3$ and $c=7$.

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Updated on: 10-Oct-2022

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