Find the product of the prime factors of 50 and 4410 .

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We start by finding the prime factorization of each number.

1. For $$50$$:
   - $$50 = 2 \times 25$$
   - $$25 = 5 \times 5$$
   
   Thus, the prime factors of $$50$$ are $$2$$ and $$5$$.

2. For $$4410$$:
   - Divide by $$2$$: $$4410 = 2 \times 2205$$
   - $$2205$$ is divisible by $$3$$ since $$2+2+0+5=9$$ (divisible by $$3$$):
     - $$2205 = 3 \times 735$$
   - $$735$$ is divisible by $$3$$:
     - $$735 = 3 \times 245$$
   - $$245$$ can be factored as:
     - $$245 = 5 \times 49$$
   - And $$49 = 7 \times 7$$.
   
   Thus, the prime factorization of $$4410$$ is:
   $$
   4410 = 2 \times 3^2 \times 5 \times 7^2
   $$
   The distinct prime factors of $$4410$$ are $$2$$, $$3$$, $$5$$, and $$7$$.

3. Next, we take the union of the prime factors from both numbers. The primes from $$50$$ are $$\{2, 5\}$$ and from $$4410$$ are $$\{2, 3, 5, 7\}$$. Their union is:
   $$
   \{2, 3, 5, 7\}
   $$

4. Finally, we find the product of these prime factors:
   $$
   2 \times 3 \times 5 \times 7 = 210
   $$

Thus, the product of the prime factors is $$210$$.
The product of the prime factors of 50 and 4410 is 210.

Find the product of the prime factors of 50
and 4410 .

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