Zeroes of a Series

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 2^{1^{0}} × 3^{2^{1}} × 4^{3^{2}} × 5^{4^{3}} × \cdots × 10^{9^{8}}


The answer is 43046785.

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2 solutions

Chew-Seong Cheong
Sep 14, 2019

Let P = 2 1 0 × 3 2 1 × 4 3 2 × 5 4 3 × × 1 0 9 8 P = 2^{1^0} \times 3^{2^1} \times 4^{3^2} \times 5^{4^3} \times \cdots \times 10^{9^8} . We know that 1 0 9 8 10^{9^8} contributes 9 8 9^8 trailing zeroes to P P . The other source is 2 N × 5 4 3 2^N \times 5^{4^3} , where N = 1 0 + 2 × 3 2 + 5 4 + 3 × 7 6 N = 1^0 + 2\times 3^2+5^4+3\times 7^6 is the total power of 2 2 in P P . Since N > 4 3 N > 4^3 , the additional trailing zeroes is from 2 4 3 × 5 4 3 2^{4^3} \times 5^{4^3} which is 4 3 4^3 . Therefore, the total number of trailing zeroes is 9 8 + 4 3 = 43046721 + 64 = 43046785 9^8 + 4^3 = 43046721 + 64 = \boxed {43046785} .

Kaizen Cyrus
Sep 14, 2019

The number of trailing zeroes can be identified on how many tens a number has as a factor or how many pairs of 2 2 and 5 5 .

2 a × 5 b 2^{a}×5^{b}

In the expression above, its number of trailing zeroes is a a if a < b a<b or b b if a > b a>b . The expression can be multiplied to any positive integer (that doesn't have 2 2 and 5 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 2 and then the terms that are divisible by 5 5 .

2 1 0 × 4 3 2 × 5 4 3 × 6 5 4 × 8 7 6 × 1 0 9 8 \small \textcolor{#D61F06}{2^{1^{0}}} × \textcolor{#EC7300}{4^{3^{2}}} × \textcolor{#CEBB00}{5^{4^{3}}} × \textcolor{#20A900}{6^{5^{4}}} × \textcolor{#3D99F6}{8^{7^{6}}} × \textcolor{#302B94}{10^{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 \small \textcolor{#D61F06}{2^{1^{0}}} × \textcolor{#EC7300}{2^{3^{2}}} × \textcolor{#EC7300}{2^{3^{2}}} × \textcolor{#CEBB00}{5^{4^{3}}} × \textcolor{#20A900}{2^{5^{4}}} × \textcolor{#3D99F6}{2^{7^{6}}} × \textcolor{#3D99F6}{2^{7^{6}}} × \textcolor{#3D99F6}{2^{7^{6}}} × \textcolor{#302B94}{2^{9^{8}}} × \textcolor{#302B94}{5^{9^{8}}}

By using the Product of Powers Property, we seperate the exponents of 2 2 and 5 5 and add them accordingly.

1 0 + 3 2 + 3 2 + 5 4 + 7 6 + 7 6 + 7 6 + 9 8 = 43400312 \small \textcolor{#D61F06}{1^{0}} + \textcolor{#EC7300}{3^{2}} + \textcolor{#EC7300}{3^{2}} + \textcolor{#20A900}{5^{4}} + \textcolor{#3D99F6}{7^{6}} + \textcolor{#3D99F6}{7^{6}} + \textcolor{#3D99F6}{7^{6}} + \textcolor{#302B94}{9^{8}} = 43400312

4 3 + 9 8 = 43046785 \small \textcolor{#CEBB00}{4^{3}} + \textcolor{#302B94}{9^{8}} = 43046785

Since the sum of the exponents of 5 5 is smaller than 2 2 's, it is the number of trailing zeroes for our problem.

2 43400312 × 5 43046785 2^{43400312} × 5^{43046785}

43400312 > 43046785 43400312 > \boxed{43046785}

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