Solution:

To get the number of zeroes at the end of the product of the first 40 natural numbers, we need to find the number of factors of 10 in the product. Since 10 is the product of 2 and 5, we need to count the number of pairs of 2 and 5 factors.


Step 1: Prime factorization of the first 40 natural numbers
- We can write the first 40 natural numbers as follows:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40.

- Prime factorization of each of the numbers is shown below:

- 1 = 1^1
- 2 = 2^1
- 3 = 3^1
- 4 = 2^2
- 5 = 5^1
- 6 = 2^1 x 3^1
- 7 = 7^1
- 8 = 2^3
- 9 = 3^2
- 10 = 2^1 x 5^1
- 11 = 11^1
- 12 = 2^2 x 3^1
- 13 = 13^1
- 14 = 2^1 x 7^1
- 15 = 3^1 x 5^1
- 16 = 2^4
- 17 = 17^1
- 18 = 2^1 x 3^2
- 19 = 19^1
- 20 = 2^2 x 5^1
- 21 = 3^1 x 7^1
- 22 = 2^1 x 11^1
- 23 = 23^1
- 24 = 2^3 x 3^1
- 25 = 5^2
- 26 = 2^1 x 13^1
- 27 = 3^3
- 28 = 2^2 x 7^1
- 29 = 29^1
- 30 = 2^1 x 3^1 x 5^1
- 31 = 31^1
- 32 = 2^5
- 33 = 3^1 x 11^1
- 34 = 2^1 x 17^1
- 35 = 5^1 x 7^1
- 36 = 2^2 x 3^2
- 37 = 37^1
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