Finishing a prime factorization

Question

If I am given an integer like $$46^{3}17^{4}15^{2}$$ how would I approach finding its prime factorization without a calculator. I think I need congruence but am stuck on how to apply. Thanks

Answer 1

For any number of the form $x^ay^bz^c$, it is pretty simple to find the prime factorization, at least when the numbers have four digits or less. For now, we'll ignore the exponents, and we'll come back to them later.

Because you have small numbers, you might already know their factorizations, but I'll use a method to test for prime factors to generalize this result. We just simply test divisibility by each prime. Starting with $46$, we try dividing it by $2$. An immediate success; we get $23$. Now we have a new number! Is it divisible by $2$? No. $3$? No.

We can stop here now, because any prime factors of a number have to be less than their square root. Because if it was larger, it would have to multiplied by a number smaller than itself to make the number, but we've already tested all the smaller primes. And because the next prime, $5$, has a square of $25$, we know that $25 > 23$. And because we never tested $1$, the only option left is that $23$ is prime.

So the prime factorization of $46$ is $2\cdot{23}$. Cubing this, we know that $(xy)^3 = x^3y^3$, so the prime factorization of $46^3$ is $2^3 \cdot 23^3$.

Let's continue in this manner! Applying this process, we find that $17$ is prime, and that the prime factorization of $15$ is $5\cdot3$. And if we apply the exponents, and multiply them all together, the prime factorization of $46^{3}17^{4}15^{2}$ is:

$$2^3\cdot3^2\cdot5^2\cdot17^4\cdot23^3.$$