Find  the  number  of  last  zeros  in  the  product  of  the expression (23× 34× 56× 35× 57× 79× 810+ 45× 56× 67× 78-102× 153× 204)?

To find the number of trailing zeros in a product, we need to count how many times 10 divides into the number. Each 10 is a product of 2 and 5, so we need to determine how many factors of 5 and 2 appear in the factorization of the entire expression.

Let’s break down the expression:

\[
(23 \times 34 \times 56 \times 35 \times 57 \times 79 \times 810) + (45 \times 56 \times 67 \times 78) - (102 \times 153 \times 204)
\]

### Step 1: Prime Factorization
To find trailing zeros, we focus on the prime factorizations of each term and count the factors of 2 and 5.

#### Term 1: \( 23 \times 34 \times 56 \times 35 \times 57 \times 79 \times 810 \)
Prime factorizations:
- \( 23 \) is prime.
- \( 34 = 2 \times 17 \)
- \( 56 = 2^3 \times 7 \)
- \( 35 = 5 \times 7 \)
- \( 57 = 3 \times 19 \)
- \( 79 \) is prime.
- \( 810 = 2 \times 3^4 \times 5 \)

Factors of 2: \( 2 \times 2^3 \times 2 = 2^5 \)

Factors of 5: \( 5 \times 5 = 5^2 \)

#### Term 2: \( 45 \times 56 \times 67 \times 78 \)
Prime factorizations:
- \( 45 = 3^2 \times 5 \)
- \( 56 = 2^3 \times 7 \)
- \( 67 \) is prime.
- \( 78 = 2 \times 3 \times 13 \)

Factors of 2: \( 2^3 \times 2 = 2^4 \)

Factors of 5: \( 5 \)

#### Term 3: \( 102 \times 153 \times 204 \)
Prime factorizations:
- \( 102 = 2 \times 3 \times 17 \)
- \( 153 = 3^2 \times 17 \)
- \( 204 = 2^2 \times 3 \times 17 \)

Factors of 2: \( 2 \times 2^2 = 2^3 \)

No factors of 5.

### Step 2: Combine the Factors
Now, we combine the factors of 2 and 5 from each term.

- **Total factors of 2**:
  - From Term 1: \( 2^5 \)
  - From Term 2: \( 2^4 \)
  - From Term 3: \( 2^3 \)

Total factors of 2: \( 2^{5+4+3} = 2^{12} \)

- **Total factors of 5**:
  - From Term 1: \( 5^2 \)
  - From Term 2: \( 5 \)
  - From Term 3: \( 0 \)

Total factors of 5: \( 5^{2+1} = 5^3 \)

### Step 3: Determine the Number of Trailing Zeros
The number of trailing zeros is determined by the number of pairs of factors of 2 and 5. Since the number of factors of 5 is smaller, the number of trailing zeros is equal to the number of factors of 5.

Thus, the product has **3 trailing zeros**.

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Would you like further details or have any questions about this solution?

Here are 5 related questions for practice:
1. How do you determine the number of trailing zeros in a factorial?
2. What is the prime factorization of a large number like 810?
3. How do trailing zeros relate to divisibility by powers of 10?
4. What role does the highest power of 5 play in determining trailing zeros in large numbers?
5. How can we extend this technique to find trailing zeros in decimal fractions?

**Tip:** When calculating trailing zeros in large products, always focus on counting factors of 5 and 2 in the prime factorization of the numbers involved.

Find the number of last zeros in the product of the expression (23× 34× 56× 35× 57× 79× 810 + 45× 56× 67× 78 - 102× 153× 204)?

Learn how to find the number of trailing zeros in the product of a complex expression using prime factorization. This problem breaks down factors of 2 and 5, key concepts in number theory, and is ideal for students in Grades 9-12.

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