How many zeroes are there at the end of the following product? 1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60
- A10Correct
- B12
- C14
- D15
Explanation
To find the number of zeroes at the end of a product, we need to count the number of pairs of (2 x 5) in its prime factorization. The number of zeroes will be limited by the factor (2 or 5) that appears fewer times.
Let's list the prime factors of 2 and 5 for each number in the product:
1: No 2s, No 5s 5: One 5 10 = 2 x 5: One 2, One 5 15 = 3 x 5: One 5 20 = 2 x 2 x 5: Two 2s, One 5 25 = 5 x 5: Two 5s 30 = 2 x 3 x 5: One 2, One 5 35 = 5 x 7: One 5 40 = 2 x 2 x 2 x 5: Three 2s, One 5 45 = 3 x 3 x 5: One 5 50 = 2 x 5 x 5: One 2, Two 5s 55 = 5 x 11: One 5 60 = 2 x 2 x 3 x 5: Two 2s, One 5
Now, let's sum up the total number of factors of 2 and 5:
Total number of factors of 2: From 10 (1) + From 20 (2) + From 30 (1) + From 40 (3) + From 50 (1) + From 60 (2) = 1 + 2 + 1 + 3 + 1 + 2 = 10
Total number of factors of 5: From 5 (1) + From 10 (1) + From 15 (1) + From 20 (1) + From 25 (2) + From 30 (1) + From 35 (1) + From 40 (1) + From 45 (1) + From 50 (2) + From 55 (1) + From 60 (1) = 1 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 1 + 2 + 1 + 1 = 15
The number of zeroes is determined by the smaller count of the factors of 2 or 5. Minimum (Number of 2s, Number of 5s) = Minimum (10, 15) = 10.
Therefore, there are 10 zeroes at the end of the product.
The final answer is A) 10.

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