Discussion
How many zeroes are there at the end of the following product?
5 × 5 × 10 × 15 × 15 × 20 × 35 × 40 × 45 × 55 × 75 × 80 × 85 × 95 × 100 × 100
Answer: B
The product ends in 14 zeroes.
Each trailing zero needs one factor 10 = 2 × 5, so count the 2s and 5s and take the smaller count.
Factors of 5: 5, 5, 10, 15, 15, 20, 35, 40, 45, 55, 80, 85 and 95 give one each (13), and 75, 100 and 100 give two each (6), so there are 19 fives. Factors of 2: 10 gives 1, 20 gives 2, 40 gives 3, 80 gives 4, and each 100 gives 2, so there are \(1 + 2 + 3 + 4 + 2 + 2 = 14\) twos.
The 2s run out first, so only 14 tens can be formed. 18 or 19 comes from counting the 5s, which is the habit from factorial questions where 2s are plentiful; here they are not.
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