Problem 101 - Algebra for JEE ADVANCED

Question

The total number of divisors of 480 , that are of the form \(4 n+2, n \geq\) 0 , is equal to a. 2 b. 3 c. 4 d. none of these

Step-by-Step Solution

Answer

The total number of divisors of 480 of the form \(4n + 2\) is 4.

Step 1: Prime Factorization

To find the number of divisors of 480, start with its prime factorization. The prime factorization of 480 is obtained as follows:\[ 480 = 2^5 \times 3^1 \times 5^1 \]

Step 2: Identify Divisors of the Form \(4n + 2\)

A divisor of the form \(4n + 2\) can be rewritten as \(2(2n + 1)\), meaning the divisor is even but not divisible by 4. Such a number only contains the factor 2 raised to the power of 1. Therefore, the candidate divisors are of the form \(2 \times k\), where \(k\) must be an odd divisor of the factor \(3^1 \times 5^1\).

Step 3: Calculate Odd Divisors of Remaining Factors

To find \(k\), list out the odd divisors of \(3^1 \times 5^1\). The number \(3^1 \times 5^1 = 15\) has divisors 1, 3, 5, and 15. Thus, the candidate divisors of 480 of the form \(4n + 2\) are \(2 \times 1 = 2\), \(2 \times 3 = 6\), \(2 \times 5 = 10\), and \(2 \times 15 = 30\).

Step 4: Count Valid Divisors

Check which of the calculated divisors are actually divisors of 480. \(2, 6, 10,\) and \(30\) are all divisors of 480 as they divide evenly into 480. Therefore, there are 4 valid divisors of 480 of the form \(4n + 2\).

Key Concepts

Prime Factorization

To begin with, prime factorization helps in breaking down a number into its basic building blocks: prime numbers. For example, the prime factorization of 480 is \(480 = 2^5 \times 3^1 \times 5^1\). This means 480 can be expressed as a product of the prime number 2 raised to the 5th power, 3 raised to the 1st power, and 5 raised to the 1st power.

Understanding prime factorization is crucial because it lets you determine all possible divisors of the number. Each divisor corresponds to a unique combination of the prime factors, raised to allowable powers. Knowing the prime factors and their powers, one can easily calculate potential divisors, which is quite handy in solving problems like this.

Divisors of the Form 4n+2

Divisors of the form \(4n + 2\) have a special structure. They can be rewritten as \(2(2n + 1)\), indicating that these numbers are even but not multiples of 4. This is because they have only one factor of 2.

For our example with the number 480, knowing that \(480 = 2^5 \times 3^1 \times 5^1\) helps. In order to have a divisor of the form \(4n + 2\), we need it to fit \(2 \times k\), where \(k\) must be odd. This is derived from isolating where the single factor of 2 lies and making sure no more than one power of 2 can remain.

Odd Divisors

Odd divisors play a key role when identifying numbers of the form \(4n + 2\). To find odd divisors of 480, we focus on the non-\(2^5\) portion of the factorization. This is \(3^1 \times 5^1\), simplifying to 15.

The odd numbers, or odd divisors, of 15 are 1, 3, 5, and 15. These are important because they are used as the \(k\) in \(2 \times k\) from the explanation of divisors of the form \(4n + 2\). This means each odd divisor multiplied by 2 will potentially be a divisor of 480 that fits the form \(4n + 2\).

Even Divisors

Even divisors are all divisors that include at least one factor of 2, thereby making them divisible by 2. For a number like 480, any divisor which results from the factorization involving at least one power of 2 will be even.

However, when we specifically talk about divisors of the form \(4n + 2\), we're interested in even divisors that aren't multiples of 4, making them distinct from general even divisors. In this context, we're only considering numbers with a single power of 2 within their factorization. In our problem, these divisors include 2, 6, 10, and 30, verified through division of the original number, confirming their even status and special form.