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 nn, we need to find the number of factors of 10 in nn. Each factor of 10 is composed of a factor of 2 and a factor of 5.

Given:

n=23×34×54×7n = 2^3 \times 3^4 \times 5^4 \times 7

The number of factors of 2 in nn is 3 (from 232^3), and the number of factors of 5 in nn is 4 (from 545^4). 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:

min(3,4)=3\min(3, 4) = 3

Thus, the number of consecutive zeroes at the end of nn is 3.

Would you like further clarification, or do you have any other questions?

Here are 5 related questions:

  1. What would be the number of trailing zeroes if n=25×33×52×11n = 2^5 \times 3^3 \times 5^2 \times 11?
  2. How do the exponents of 2 and 5 affect the number of trailing zeroes in a product?
  3. If you had the number n=26×53×72n = 2^6 \times 5^3 \times 7^2, how many trailing zeroes would there be?
  4. What if the problem included additional prime factors other than 2 and 5, how would that affect the number of trailing zeros?
  5. How do you calculate the number of trailing zeroes in a factorial, such as 100!100!?

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