yezz wrote:

how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40^5*45^5*50^5?

A) 30

B) 60

C)40

D)45

E)53

Let’s first rewrite the given expression:

N = (5^5)(10^5)(15^5)(20^5)(25^5)(30^5)(35^5)(40^5)(45^5)(50^5)

To determine the number of trailing zeros in a number, we need to determine the number of 5- and-2 pairs within the prime factorization of that number. Each 5-and-2 pair yields 10, and each 10 contributes one trailing zero to the product.

5^5 = 5^5

10^5 = 2^5 x 5^5

15^5 = 3^5 x 5^5

20^5 = (4 x 5)^5 = (2^2 x 5^1)^5 = 2^10 x 5^5

25^5 = (5^2)^5 = 5^10

30^5 = (5 x 3 x 2)^5 = 5^5 x 3^5 x 2^5

35^5 = 5^5 x 7^5

40^5 = (8 x 5)^5 = (2^3 x 5^1)^5 = 2^15 x 5^5

45^5 = (9 x 5)^5 = (3^2 x 5^1)^5 = 3^10 x 5^5

50^5 = (25 x 2)^5 = (5^2 x 2^1)^5 = 5^10 x 2^5

Since we see there are fewer 2s than 5s in the factorization above, we simply count the number of 2s and thus we can determine the number of 5-and-2 pairs.

We have:

2^5, 2^10, 2^5, 2^15, 2^5

Since we have 40 2s, we have 40 5-and-2 pairs and thus there are 40 trailing zeros in the product.

Answer: C

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