Let's try to find some primes that divide this number.
The number is not divisible by 2, because it is odd.
The number is divisible by 3 though, because the sum of its digits is:
![6+5+1+6+9=27=3\cdot 9](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ygc7o8yqnmxcci8kzdmp965ue6oam6newp.png)
So, we can divide the number by 3 and keep going with the factorization:
![65169/ 3 = 21723](https://img.qammunity.org/2019/formulas/mathematics/middle-school/qttgidsltikp22n3zwyve3mlxjs0uk3xh4.png)
This number is again divisible by 3, because
![2+1+7+2+3 = 15 = 3\cdot 5](https://img.qammunity.org/2019/formulas/mathematics/middle-school/u3v6htm52kwfo1k09t54gbn7w5sebnvvum.png)
We have
![21723/ 3 = 7241](https://img.qammunity.org/2019/formulas/mathematics/middle-school/lks4bpbxtcubixb16hnhfrgqfdtzdzt20t.png)
This number is no longer divisible by 3. Let's go on looking for primes that divide it: 5 doesn't because the number doesn't end in 0 nor 5. This number is not divisible by 7 or 11 either (just try). It is divisible by 13 though: we have
![7241/ 13 = 557](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ohnbej0m7m7zns9ltcmam2a91npa2neuv1.png)
And 557 is prime, so we're done. This means that the prime factorization of 65169 is
![3^2\cdot 13 \cdot 557](https://img.qammunity.org/2019/formulas/mathematics/middle-school/3gy16c4b3jpy7ice9c3u0yq4soq7geyyq9.png)