how many factors of 1200 power 10 have exactly 1 trailing zeros

To find the number of factors of 1200 raised to the power of 10 that have exactly 1 trailing zero, we need to consider the prime factorization of 1200.

Step 1: Prime factorization of 1200 To find the prime factorization of 1200, we can start by dividing it by the smallest prime number, which is 2. 1200 ÷ 2 = 600

So, 2 is a prime factor of 1200. Now, we divide 600 by 2 again. 600 ÷ 2 = 300

2 is a prime factor of 600 as well. Continuing this process, we divide 300 by 2 again. 300 ÷ 2 = 150

Again, 2 is a prime factor of 300. Dividing 150 by 2 gives us: 150 ÷ 2 = 75

Now, 2 is no longer a factor of 75. We move on to the next prime number, which is 3. 75 ÷ 3 = 25

3 is a prime factor of 75. Dividing 25 by 3 gives us: 25 ÷ 3 = 8 remainder 1

Since 3 is not a factor of 1, we move on to the next prime number, which is 5. 25 ÷ 5 = 5

5 is a prime factor of 25. Dividing 5 by 5 gives us: 5 ÷ 5 = 1

Now, we have reached 1, which means we have found all the prime factors of 1200. Therefore, the prime factorization of 1200 is 2^4 × 3^1 × 5^2.

Step 2: Finding the factors with exactly 1 trailing zero To have exactly 1 trailing zero, a factor must have at least one factor of 5 and no factors of 2 other than the ones already accounted for in the prime factorization of 1200.

Since we have 2^4 × 3^1 × 5^2 as the prime factorization of 1200, we can choose one factor of 5 and any combination of factors of 2 and 3.

For the factor of 5, we have 2 options: 5^1 or 5^2.

For the factors of 2, we have 5 options: 2^0, 2^1, 2^2, 2^3, or 2^4.

For the factors of 3, we have 2 options: 3^0 or 3^1.

Multiplying these options together, we get a total of 2 × 5 × 2 = 20 factors of 1200 raised to the power of 10 that have exactly 1 trailing zero.