To determine how many zeros are at the end of the number $45^9 \times 88^5$ without actually computing the number, we need to consider the prime factors involved.

Step 1: Prime Factorization

• Prime factorization of 45: $45 = 3^2 \times 5$ So, $45^9 = (3^2 \times 5)^9 = 3^{18} \times 5^9$

• Prime factorization of 88: $88 = 2^3 \times 11$ So, $88^5 = (2^3 \times 11)^5 = 2^{15} \times 11^5$

Step 2: Combine the Factors

Now, multiply the factorizations: $45^9 \times 88^5 = (3^{18} \times 5^9) \times (2^{15} \times 11^5) = 2^{15} \times 3^{18} \times 5^9 \times 11^5$

Step 3: Determine the Number of Zeros

The number of zeros at the end of a number is determined by the number of factors of 10 in its factorization. Since $10 = 2 \times 5$, the number of zeros at the end of the product is the minimum of the powers of 2 and 5 in its factorization.

• The power of 2 in the factorization is $15$.
• The power of 5 in the factorization is $9$.

Conclusion:

The number of zeros at the end of $45^9 \times 88^5$ is the minimum of these two, which is 9.

Would you like more details or have any further questions?

Relative Questions:

1. How would the number of zeros change if we computed $45^{10} \times 88^5$?
2. What would happen to the number of zeros if the exponents of 45 and 88 were both doubled?
3. How does prime factorization help in finding the number of zeros in any product?
4. If you multiply $45^9 \times 88^5$ by $10^3$, how many zeros would the result have?
5. How would you find the number of trailing zeros in $n!$ for any positive integer $n$?

Tip: When calculating the number of trailing zeros in a product or factorial, always focus on the factors of 2 and 5, as these together make a 10.