If you multiply 100!*200!*300!, how many 0s are there?

Think of all the factors that will produce a zero

1. clearly all multiples of 10 will add a zero
that is, numbers like 30, 60, 210, etc

2. also every factor ending in 2 multiplied by a factor ending in 5 will produce a zero

so zeros are produced by the following
* 2 * * 5 * * * * 10 * 12 * * 15 * * * * 20 * ....

so for the first 10! we should have 2 zeros, test with your calculator to get 3628800
by the time we get to 20! we should have 4 zeros
at 30! we should have 6 zeros
etc
so after 100! we should have 20 zeros

by the same reasoning:
200! should have 40 zeros
300! should have 60 zeros

so we have (a x 10^20)(b x 10^40)(c x 10^60)
= abc x 10^120

check my thinking.

Thank you!

To find out how many zeros there are when you multiply 100!, 200!, and 300!, we need to determine the number of factors of 10 in the multiplication. Since 10 is composed of 2 and 5, we need to count the number of pairs of 2 and 5.

First, consider the number of 2s in the prime factorization of each factorial. For this, we can use the concept of the exponent of a prime number in the prime factorization of a factorial. The exponent of 2 in n! is given by n divided by 2 + n divided by 2^2 + n divided by 2^3 + ... (continuing until the power of 2 in the prime factorization becomes zero).

Similarly, the exponent of 5 in n! is given by n divided by 5 + n divided by 5^2 + n divided by 5^3 + ... (continuing until the power of 5 becomes zero).

From the above formulas, we can calculate the number of 2s and 5s in 100!, 200!, and 300!:

- Number of 2s: (100 divided by 2) + (100 divided by 2^2) + (100 divided by 2^3) + ... = 50 + 25 + 12 + 6 + 3 + 1 = 97
- Number of 5s: (100 divided by 5) + (100 divided by 5^2) + (100 divided by 5^3) + ... = 20 + 4 + 0 = 24

- Number of 2s: (200 divided by 2) + (200 divided by 2^2) + (200 divided by 2^3) + ... = 100 + 50 + 25 + 12 + 6 + 3 + 1 = 197
- Number of 5s: (200 divided by 5) + (200 divided by 5^2) + (200 divided by 5^3) + ... = 40 + 8 + 1 = 49

- Number of 2s: (300 divided by 2) + (300 divided by 2^2) + (300 divided by 2^3) + ... = 150 + 75 + 37 + 18 + 9 + 4 + 2 + 1 = 296
- Number of 5s: (300 divided by 5) + (300 divided by 5^2) + (300 divided by 5^3) + ... = 60 + 12 + 2 = 74

Since each pair of 2 and 5 creates a trailing zero, the number of trailing zeros in the multiplication is determined by the smaller of the two counts. Therefore, in this case, there are 24 trailing zeros in the multiplication of 100!, 200!, and 300!.