Math Problem Statement

12 × 23 × 34 × ... × 2829. Find the number of zeroes in the expression.

Solution

To find the number of trailing zeroes in the product 12×23×34××282912 \times 23 \times 34 \times \dots \times 2829, we need to focus on how many factors of 10 are present in the product. Each factor of 10 is the result of multiplying a 2 and a 5.

To calculate the number of trailing zeroes in this expression, we need to count how many multiples of 5 are in the given product range.

Step-by-step approach:

  1. Numbers divisible by 5: We need to find how many multiples of 5 are there in the sequence 12,13,14,...,282912, 13, 14, ..., 2829. For this, we divide the largest number (2829) by 5: 28295=565.8565 multiples of 5.\frac{2829}{5} = 565.8 \Rightarrow 565 \text{ multiples of 5}. So there are 565 multiples of 5.

  2. Numbers divisible by 525^2 (25): Some numbers contribute more than one factor of 5, such as multiples of 52=255^2 = 25. We divide by 25: 282925=113.16113 multiples of 25.\frac{2829}{25} = 113.16 \Rightarrow 113 \text{ multiples of 25}.

  3. Numbers divisible by 535^3 (125): We continue by counting the multiples of 53=1255^3 = 125: 2829125=22.63222 multiples of 125.\frac{2829}{125} = 22.632 \Rightarrow 22 \text{ multiples of 125}.

  4. Numbers divisible by 545^4 (625): Multiples of 54=6255^4 = 625 contribute additional factors of 5: 2829625=4.5264 multiples of 625.\frac{2829}{625} = 4.526 \Rightarrow 4 \text{ multiples of 625}.

  5. Numbers divisible by 555^5 (3125): Since 3125>28293125 > 2829, there are no multiples of 31253125.

Total factors of 5:

The total number of factors of 5 is the sum of the above: 565+113+22+4=704.565 + 113 + 22 + 4 = 704.

Thus, there are 704 factors of 5 in the product 12×23××282912 \times 23 \times \dots \times 2829.

Number of trailing zeroes:

Since each trailing zero is produced by a pair of factors of 2 and 5, and there are always more factors of 2 than factors of 5, the number of trailing zeroes is the same as the number of factors of 5. Therefore, the number of trailing zeroes is 704.

However, there seems to be a mismatch between the given options and the actual number calculated. Based on the question and the possible options (which may suggest a different method or focus), I may have misunderstood the problem's intended scope. Let me reanalyze the focus or confirm the method.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Factorization
Trailing Zeroes

Formulas

Count of trailing zeroes = Sum of floor(n/5^i) for i=1,2,...

Theorems

Divisibility Theorem

Suitable Grade Level

Grades 10-12