Write the numbers of zeroes in the end of a number whose prime factorization is `2^2 times 5^3 times 3^2 times 17`

Answer

To find the number of zeros at the end of a number whose prime factorization is \(2^2 \times 5^3 \times 3^2 \times 17\), we can follow these steps:

### Step 1: Understand the formation of zeros
A zero at the end of a number is formed by the factors of 10. Each 10 is made up of one factor of 2 and one factor of 5.

### Step 2: Identify the number of 2s and 5s in the prime factorization
From the given prime factorization:
- The number of 2s is \(2^2\), which means there are 2 factors of 2.
- The number of 5s is \(5^3\), which means there are 3 factors of 5.

### Step 3: Determine the limiting factor
To form a 10, we need one 2 and one 5. Therefore, the number of 10s (and thus the number of zeros at the end of the number) is determined by the smaller count of 2s or 5s.

### Step 4: Calculate the number of zeros
We have:
- 2 factors of 2
- 3 factors of 5

The limiting factor is the number of 2s, which is 2. Thus, the number of zeros at the end of the number is 2.

### Final Answer
The number of zeros at the end of the number is **2**.

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