How many trailing zeroes does the result of the expression below have?
2 1 0 × 3 2 1 × 4 3 2 × 5 4 3 × ⋯ × 1 0 9 8
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The number of trailing zeroes can be identified on how many tens a number has as a factor or how many pairs of 2 and 5 .
2 a × 5 b
In the expression above, its number of trailing zeroes is a if a < b or b if a > b . The expression can be multiplied to any positive integer (that doesn't have 2 and 5 as their factors) and would still yield the same number of zeroes at the end of the product.
Using the question's expression, we get all terms that are divisible by 2 and then the terms that are divisible by 5 .
2 1 0 × 4 3 2 × 5 4 3 × 6 5 4 × 8 7 6 × 1 0 9 8
Then factorize the terms to get the necessary factors.
2 1 0 × 2 3 2 × 2 3 2 × 5 4 3 × 2 5 4 × 2 7 6 × 2 7 6 × 2 7 6 × 2 9 8 × 5 9 8
By using the Product of Powers Property, we seperate the exponents of 2 and 5 and add them accordingly.
1 0 + 3 2 + 3 2 + 5 4 + 7 6 + 7 6 + 7 6 + 9 8 = 4 3 4 0 0 3 1 2
4 3 + 9 8 = 4 3 0 4 6 7 8 5
Since the sum of the exponents of 5 is smaller than 2 's, it is the number of trailing zeroes for our problem.
2 4 3 4 0 0 3 1 2 × 5 4 3 0 4 6 7 8 5
4 3 4 0 0 3 1 2 > 4 3 0 4 6 7 8 5
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Let P = 2 1 0 × 3 2 1 × 4 3 2 × 5 4 3 × ⋯ × 1 0 9 8 . We know that 1 0 9 8 contributes 9 8 trailing zeroes to P . The other source is 2 N × 5 4 3 , where N = 1 0 + 2 × 3 2 + 5 4 + 3 × 7 6 is the total power of 2 in P . Since N > 4 3 , the additional trailing zeroes is from 2 4 3 × 5 4 3 which is 4 3 . Therefore, the total number of trailing zeroes is 9 8 + 4 3 = 4 3 0 4 6 7 2 1 + 6 4 = 4 3 0 4 6 7 8 5 .