Problem 89 - Fundamentals of Mathematics

Question

Write all divisors of \(2^{3} \cdot 5^{2} \cdot 7\).

Step-by-Step Solution

Answer

The divisors are: 1, 2, 4, 5, 7, 8, 10, 14, 20, 25, 28, 35, 40, 50, 56, 70, 100, 140, 175, 200, 280, 350, 700, 1400.

Step 1: Understand the Problem

We need to find all the divisors of the expression \(2^{3} \cdot 5^{2} \cdot 7\). A divisor is a number that divides another number completely without leaving a remainder.

Step 2: Identify the Prime Factorization

The expression \(2^{3} \cdot 5^{2} \cdot 7\) represents the number expressed as a product of prime factors. Here, the number is \(2\) raised to the power of \(3\), \(5\) to the power of \(2\), and \(7^{1}\).

Step 3: List Possible Combinations

To find all divisors, we must consider all combinations of the powers of prime factors. This means considering each prime raised to its power from \(0\) up to its maximum in the factorization. For example:- For \(2\), we consider powers \(0\), \(1\), \(2\), \(3\).- For \(5\), we consider powers \(0\), \(1\), \(2\).- For \(7\), we consider powers \(0\), \(1\).

Step 4: Calculate Divisors

We calculate the product of these powers to form divisors. The general form of a divisor based on the combination \(2^{a} \cdot 5^{b} \cdot 7^{c}\) where \(0 \leq a \leq 3\), \(0 \leq b \leq 2\), and \(0 \leq c \leq 1\).The divisors are:- \(2^{0} \cdot 5^{0} \cdot 7^{0} = 1\)- \(2^{0} \cdot 5^{0} \cdot 7^{1} = 7\)- \(2^{0} \cdot 5^{1} \cdot 7^{0} = 5\)- \(2^{0} \cdot 5^{1} \cdot 7^{1} = 35\)- \(2^{0} \cdot 5^{2} \cdot 7^{0} = 25\)- \(2^{0} \cdot 5^{2} \cdot 7^{1} = 175\)- \(2^{1} \cdot 5^{0} \cdot 7^{0} = 2\)- \(2^{1} \cdot 5^{0} \cdot 7^{1} = 14\)- \(2^{1} \cdot 5^{1} \cdot 7^{0} = 10\)- \(2^{1} \cdot 5^{1} \cdot 7^{1} = 70\)- \(2^{1} \cdot 5^{2} \cdot 7^{0} = 50\)- \(2^{1} \cdot 5^{2} \cdot 7^{1} = 350\)- \(2^{2} \cdot 5^{0} \cdot 7^{0} = 4\)- \(2^{2} \cdot 5^{0} \cdot 7^{1} = 28\)- \(2^{2} \cdot 5^{1} \cdot 7^{0} = 20\)- \(2^{2} \cdot 5^{1} \cdot 7^{1} = 140\)- \(2^{2} \cdot 5^{2} \cdot 7^{0} = 100\)- \(2^{2} \cdot 5^{2} \cdot 7^{1} = 700\)- \(2^{3} \cdot 5^{0} \cdot 7^{0} = 8\)- \(2^{3} \cdot 5^{0} \cdot 7^{1} = 56\)- \(2^{3} \cdot 5^{1} \cdot 7^{0} = 40\)- \(2^{3} \cdot 5^{1} \cdot 7^{1} = 280\)- \(2^{3} \cdot 5^{2} \cdot 7^{0} = 200\)- \(2^{3} \cdot 5^{2} \cdot 7^{1} = 1400\).

Step 5: Finalize the Solution

List the complete set of divisors in increasing order: 1, 2, 4, 5, 7, 8, 10, 14, 20, 25, 28, 35, 40, 50, 56, 70, 100, 140, 175, 200, 280, 350, 700, 1400.

Key Concepts

Prime Factorization

Prime factorization is a method used to express a number as a product of prime numbers. Primes are numbers greater than 1 that have no divisors other than 1 and themselves, such as 2, 3, 5, and 7. Every integer greater than 1 can be uniquely written in terms of its prime factors. For instance, suppose you want to write 1400 in terms of its prime factors. Using factorization, this number breaks down into

• First, divide 1400 by 2, the smallest prime: 1400 ÷ 2 = 700
• Divide 700 by 2 again: 700 ÷ 2 = 350
• Continue this until 350 ÷ 2 = 175, giving you 2 raised to the power of 3.
• Next, use 5: 175 ÷ 5 = 35 and, 35 ÷ 5 = 7 to give 5 squared. Lastly, divide by 7, resulting in 7 to power of 1.

Thus, the prime factorization is expressed as \(2^{3} \times 5^{2} \times 7 \). This process is essential in number theory as it helps simplify complex computations and solve divisor-related problems.

Powers of Prime Factors

Understanding the powers of prime factors is key to determining divisors. In a prime factorization like \(2^{3} \cdot 5^{2} \cdot 7 \), each prime number in the expression is raised to a certain power. These numbers aren’t just random; they represent how many times each prime divides the original number. Let's break it down:- The 3 in \(2^{3}\) means 2 is a factor three times.- The 2 in \(5^{2}\) shows that 5 is a factor twice.- The 1 in \(7\) indicates 7 is included once.These powers enable us to calculate the total number of divisors quickly. We calculate by multiplying each power incremented by one. For example:- For \(2^{3}\), incrementing gives us 4 possibilities: 2 to the power of 0 through 3.- For \(5^{2}\), it's 3 possibilities (since 2 + 1 = 3).- For \(7^{1}\), there are 2 possibilities.The total product of these values, 4 x 3 x 2 = 24, indicates the number of divisors.

Combinatorial Counting

Combinatorial counting is a strategy used to determine all possible combinations that arise from a set of choices. In the case of prime factorization, it involves calculating different combinations of prime powers to find all divisors.
The combinatorial method utilizes a format such as \(2^{a} \cdot 5^{b} \cdot 7^{c} \)where - \(0 \leq a \leq 3\)- \(0 \leq b \leq 2\)- \(0 \leq c \leq 1\).
By selecting variables \(a, b,\) and \(c\) within their permitted ranges, you generate all different divisors. This way:- Every possible combination of a, b, and c creates a unique divisor.- The approach systematically covers every combination, ensuring all divisors get listed.Overall, this method is both systematic and efficient, helping simplify determining divisors from large numbers.

Number Theory

Number theory is a branch of mathematics that deals with the properties and relationships among integers. It focuses on understanding integers through different aspects, such as divisors, prime numbers, and factorization.
When you encounter a problem involving finding divisors, you are diving into basic concepts of number theory. This branch helps us:

• Understand the role and distribution of primes.
• Explore the properties of numbers, such as evenness, oddness, and perfect numbers.
• Develop concepts like divisibility rules and congruences.

In solving exercises like finding the divisors of \(2^{3} \cdot 5^{2} \cdot 7\), we apply these number-theoretical concepts to identify patterns and simplify solutions.
By mastering number theory fundamentals, complex numerical problems become easier to approach and solve.