1. **State the problem:** Find how many zeros are at the end of the number $5^4 \times 3^5 \times 4^5$. 2. **Recall the rule for trailing zeros:** The number of trailing zeros in a number is determined by the number of factors of 10 it contains. Since $10 = 2 \times 5$, count the pairs of 2s and 5s in the prime factorization. 3. **Prime factorize each term:** - $5^4$ is already prime factors of 5. - $3^5$ is prime factors of 3, no 2 or 5 here. - $4^5 = (2^2)^5 = 2^{10}$. 4. **Combine all prime factors:** $$5^4 \times 3^5 \times 2^{10}$$ 5. **Count pairs of 2 and 5:** - Number of 5s: 4 - Number of 2s: 10 6. **Number of trailing zeros:** The minimum of the count of 2s and 5s is $\min(4,10) = 4$. **Final answer:** The number $5^4 \times 3^5 \times 4^5$ ends with **4 zeros**.