Math Problem Statement
If n = 2 ^ 3 * 3 ^ 4 * 5 ^ 4 * 7 , then the number of consecutive zeroes in n, where n is a natural number, is
Solution
To determine the number of consecutive zeroes at the end of the number , we need to find the number of factors of 10 in . Each factor of 10 is composed of a factor of 2 and a factor of 5.
Given:
The number of factors of 2 in is 3 (from ), and the number of factors of 5 in is 4 (from ). Since each pair of a factor of 2 and a factor of 5 forms a factor of 10, the number of trailing zeros is determined by the minimum of the exponents of 2 and 5.
Here, the number of factors of 2 is 3, and the number of factors of 5 is 4. Therefore, the number of factors of 10 is:
Thus, the number of consecutive zeroes at the end of is 3.
Would you like further clarification, or do you have any other questions?
Here are 5 related questions:
- What would be the number of trailing zeroes if ?
- How do the exponents of 2 and 5 affect the number of trailing zeroes in a product?
- If you had the number , how many trailing zeroes would there be?
- What if the problem included additional prime factors other than 2 and 5, how would that affect the number of trailing zeros?
- How do you calculate the number of trailing zeroes in a factorial, such as ?
Tip: Remember that for trailing zeroes, you only need to count the pairs of factors of 2 and 5. The limiting factor is the lesser exponent between them.
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Math Problem Analysis
Mathematical Concepts
Number Theory
Prime Factorization
Exponents
Formulas
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Theorems
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Suitable Grade Level
Grades 9-12