What are the prime factors of 102?

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Answer 1 · 2016-10-20T07:40:51

Prime factors of $102$ are $2xx3xx17$

It is observed that the last digit in $102$ is even number, so it is divisible by $2$ .

Further, sum of the digits of $102$ is $3$ and hence, $102$ is also divisible by $3$ .

Therefore, $102$

= $2xx51$

= $2xx3xx17$

Now as $17$ is a prime number,

Prime factors of $102$ are $2xx3xx17$