The number of trailing zeroes in a factorial is determined by the number of times 10 is a factor in the numbers being multiplied. Since 10 is a product of 2 and 5, we need to count the number of times 2 and 5 appear as factors.

However, since 2 is a more common factor than 5, we only need to count the number of times 5 appears as a factor.

In the case of multiplying consecutive positive integers starting from 120, we need to find when we will have 4 zeroes at the end, which means we need to find when we will have multiplied by 5 four times.

120 is divisible by 5, so that's our first 5. The next number that is divisible by 5 is 125, then 130, then 135.

So, the person needs to multiply 16 numbers (from 120 to 135 inclusive) to get a result with 4 zeroes at the end.

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The number of trailing zeroes in a factorial is determined by the number of times 10 is a factor in the numbers being multiplied. Since 10 is a product of 2 and 5, we need to count the number of times 2 and 5 appear as factors.

However, since 2 is a more common factor than 5, we only need to count the number of times 5 appears as a factor.

In the case of multiplying consecutive positive integers starting from 120, we need to find when we will have 4 zeroes at the end, which means we need to find when we will have multiplied by 5 four times.

120 is divisible by 5, so that's our first 5. The next number that is divisible by 5 is 125, then 130, then 135.

So, the person needs to multiply 16 numbers (from 120 to 135 inclusive) to get a result with 4 zeroes at the end.

Knowee AI · Accepted Answer

The number of trailing zeroes in a factorial is determined by the number of times 10 is a factor in the numbers being multiplied. Since 10 is a product of 2 and 5, we need to count the number of times 2 and 5 appear as factors.

However, since 2 is a more common factor than 5, we only need to count the number of times 5 appears as a factor.

In the case of multiplying consecutive positive integers starting from 120, we need to find when we will have 4 zeroes at the end, which means we need to find when we will have multiplied by 5 four times.

120 is divisible by 5, so that's our first 5. The next number that is divisible by 5 is 125, then 130, then 135.

So, the person needs to multiply 16 numbers (from 120 to 135 inclusive) to get a result with 4 zeroes at the end.

A person starts multiplying consecutive positive integers from 120. How many numbers should he multiply before he will have a result of 4 zeroes?*1 point610115

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