To find the number of trailing zeros in the value of 120!, we need to find the number of times 10 is a factor in the number. Since 10 is the product of 2 and 5, we need to find the number of times 2 and 5 are factors in 120!.

However, since 2 is a factor more often than 5 in any given number (every even number is divisible by 2, but only every fifth number is divisible by 5), we only need to count the number of times 5 is a factor in 120!.

1. Divide 120 by 5, which equals 24. This means there are 24 numbers less than or equal to 120 that have 5 as a factor.

2. But some numbers may have more than one 5 as a factor, like 25, 50, 75, 100, etc. So, we divide 120 by 25 (5^2), which equals 4.8. We only consider the whole number part, so there are 4 numbers that have at least two 5s as a factor.

3. We continue this process with 125 (5^3), which equals 0.96. We only consider the whole number part, so there are no numbers that have at least three 5s as a factor.

4. We add up all these counts: 24 (from step 1) + 4 (from step 2) = 28.

So, there are 28 trailing zeros in the value of 120!.

Question

To find the number of trailing zeros in the value of 120!, we need to find the number of times 10 is a factor in the number. Since 10 is the product of 2 and 5, we need to find the number of times 2 and 5 are factors in 120!.

However, since 2 is a factor more often than 5 in any given number (every even number is divisible by 2, but only every fifth number is divisible by 5), we only need to count the number of times 5 is a factor in 120!.

1. Divide 120 by 5, which equals 24. This means there are 24 numbers less than or equal to 120 that have 5 as a factor.

2. But some numbers may have more than one 5 as a factor, like 25, 50, 75, 100, etc. So, we divide 120 by 25 (5^2), which equals 4.8. We only consider the whole number part, so there are 4 numbers that have at least two 5s as a factor.

3. We continue this process with 125 (5^3), which equals 0.96. We only consider the whole number part, so there are no numbers that have at least three 5s as a factor.

4. We add up all these counts: 24 (from step 1) + 4 (from step 2) = 28.

So, there are 28 trailing zeros in the value of 120!.

Knowee AI · Accepted Answer

To find the number of trailing zeros in the value of 120!, we need to find the number of times 10 is a factor in the number. Since 10 is the product of 2 and 5, we need to find the number of times 2 and 5 are factors in 120!.

However, since 2 is a factor more often than 5 in any given number (every even number is divisible by 2, but only every fifth number is divisible by 5), we only need to count the number of times 5 is a factor in 120!.

1. Divide 120 by 5, which equals 24. This means there are 24 numbers less than or equal to 120 that have 5 as a factor.

2. But some numbers may have more than one 5 as a factor, like 25, 50, 75, 100, etc. So, we divide 120 by 25 (5^2), which equals 4.8. We only consider the whole number part, so there are 4 numbers that have at least two 5s as a factor.

3. We continue this process with 125 (5^3), which equals 0.96. We only consider the whole number part, so there are no numbers that have at least three 5s as a factor.

4. We add up all these counts: 24 (from step 1) + 4 (from step 2) = 28.

So, there are 28 trailing zeros in the value of 120!.

Find the number of trailing zeros in the value of 120! ?

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