Problem 92 - Fundamentals of Mathematics

Question

Does 13 divide \(8^{3} \cdot 10^{2} \cdot 11^{4} \cdot 13^{2} \cdot 15 ?\)

Step-by-Step Solution

Answer

Yes, the expression is divisible by 13.

Step 1: Rewrite the Expression with Prime Factors

First, rewrite each number in terms of its prime factors so it's easier to determine divisibility:- \(8 = 2^3\)- \(10 = 2 \times 5\)- \(11 = 11\) (already a prime number)- \(13 = 13\) (already a prime number)- \(15 = 3 \times 5\)Then the expression can be expanded as: \[(2^3)^3 \cdot (2 \times 5)^2 \cdot 11^4 \cdot 13^2 \cdot (3 \times 5)\] which simplifies to:\[2^9 \cdot 5^2 \cdot 11^4 \cdot 13^2 \cdot 3 \cdot 5\] Further simplifying the powers:\[2^9 \cdot 5^3 \cdot 11^4 \cdot 13^2 \cdot 3\]

Step 2: Check for Divisibility by 13

To determine if the entire product is divisible by 13, observe the prime factorization of the entire expression. The expression \(2^9 \cdot 5^3 \cdot 11^4 \cdot 13^2 \cdot 3\) includes \(13^2\).Since this contains at least one factor of 13, the whole product is divisible by 13.

Key Concepts

Prime factorization

Prime factorization is a process of expressing a number as a product of its prime numbers. A prime number is a number greater than 1, which cannot be formed by multiplying two smaller natural numbers. To better understand this concept, let’s break down the given problem.
In the exercise, the numbers given (\(8, 10, 11, 13, \text{and } 15\)) are expressed using their prime factors. For instance:

• \(8\) is expressed as \(2^3\), because \(8 = 2 \times 2 \times 2\).
• \(10\) is rewritten as \(2 \times 5\).
• \(11\) and \(13\) are themselves prime.
• \(15\) can be written as \(3 \times 5\).

This breakdown helps you see the fundamental building blocks of each number. Recognizing these building blocks is important for determining divisibility, as prime factors reveal the hidden structure within the numbers.

Exponentiation

Exponentiation is a short way to denote repeated multiplication of the same base number. Understanding exponentiation can simplify complex calculations, especially when dealing with multiple factors.
In our example, the prime factorization of \(8, 10, 11, 13, \text{and } 15\) are raised to certain powers. When these numbers are part of a multiplication process, writing them with exponents makes the arithmetic easier. Let's look closer:

• \(2^9\) means multiplying the base \(2\) nine times, which comes from simplifying \((2^3)^3\) since \(2^3 \times 2^3 \times 2^3 = 2^9\).
• \(5^3\) results from multiplying \(5\) by itself three times, representing contributions from both \((2 \times 5)^2\) and \((3 \times 5)\).
• \(11^4\) stays unchanged as a power since \(11\) is already a prime and was simply repeated in the original multiplication.
• \(13^2\) signifies that the base \(13\) is multiplied by itself twice.

Grasping these calculations facilitates understanding larger expressions and determining factors like divisibility within these products.

Mathematics Problem Solving

Mathematics problem-solving revolves around assessing and tackling problems systematically using factual data and logical reasoning. In exercises involving divisibility rules, like the one we considered, the goal is to determine if a larger complex expression can be divided by a smaller number without a remainder.
In our case, the challenge was to check if 13 divides the product \(8^3 \cdot 10^2 \cdot 11^4 \cdot 13^2 \cdot 15\). This process involved two primary steps:

• **Breaking Down**: Start by simplifying the numbers into prime factors using prime factorization methods.
• **Observing Divisibility**: Next, look at the presence of the particular factor in the resultant prime factorization. The problem reveals that \(13^2\) is part of the expanded expression, clearly demonstrating the existing divisibility since at least one factor of \(13\) is present.

Through step-by-step analysis and employing principles such as prime factorization and exponentiation, problems that initially seem daunting become more manageable. This skill of problem-solving not only aids in specific scenarios but also enhances analytical thinking in mathematics more broadly.