Animated Solution for Mathematics - Permutations and Combinations: Number of divisor of the form $4n+2$ ($n\ge 0$) of the integer 240 is

Unlocking the Secrets of $240$

Imagine you are standing before a massive, complex structure—the integer $240$. To the untrained eye, it is just a number. But to a JEE aspirant, it is a puzzle waiting to be solved, a collection of prime building blocks waiting to be rearranged.

Today, we are not just finding divisors; we are performing a surgical operation on the number $240$ to extract only those that fit a very specific, elegant pattern: $4n+2$. This is the beauty of Number Theory. It is not about brute force; it is about understanding the internal architecture of integers.

The Prime Factorization (The Blueprint)

Before we can manipulate the divisors, we must understand the 'DNA' of $240$. We need to break it down into its fundamental prime components. Think of this as identifying the raw materials before building a house.

We know that $240=24\times 10$. Breaking these down further, $24=8\times 3=2^3\times 3$, and $10=2\times 5$. When we combine these, we get the prime factorization:

This is our master blueprint. Any divisor $D$ of $240$ must be constructed using these specific building blocks. Mathematically, any divisor $D$ can be expressed as:

where the exponents are constrained by our blueprint: $0\le a\le 4$, $0\le b\le 1$, and $0\le c\le 1$. This is the playground where we will find our answers.

Decoding the $4n+2$ Constraint

Now, let us look at the condition imposed by the problem: the divisor must be of the form $4n+2$. This looks like a simple algebraic expression, but it is actually a powerful filter. Let us factor out a $2$ from this expression:

Look closely at the term $(2n+1)$. By definition, any integer multiplied by $2$ and then added to $1$ is an odd number. This is the 'Aha!' moment.

The condition $4n+2$ tells us that our divisor must be an even number, but it must contain exactly one factor of $2$. If it contained more than one factor of $2$ (like $2^2=4$), it would be a multiple of $4$, which would violate our condition. If it contained zero factors of $2$, it would be odd, which also violates the condition.

Therefore, for our divisor $D$, the exponent of $2$ must be exactly $1$. That is, $a=1$. We have just locked one of our variables!

The Combinatorial Dance

With $a$ fixed at $1$, our divisor structure simplifies significantly. We are now looking for divisors of the form:

The power of $2$ is settled. Now, we only need to determine the possible values for $b$ and $c$. From our prime factorization, we know that $b$ can be either $0$ or $1$, and $c$ can be either $0$ or $1$.

This is where the magic of combinatorics comes in. We have two independent choices to make: 1. Choose the power of $3$ ($b$): We have $2$ options ($0$ or $1$). 2. Choose the power of $5$ ($c$): We have $2$ options ($0$ or $1$).

The total number of combinations for the odd part of our divisor is simply the product of these choices:

These four combinations represent the four distinct odd parts that, when multiplied by our fixed $2^1$, will satisfy the $4n+2$ condition.

The Final Reveal

Let us list them out to be absolutely certain. We are multiplying $2^1$ by each of the four combinations of $3^b\times 5^c$:

1. For $b=0,c=0$: $D=2^1\times 3^0\times 5^0=2\times 1\times 1=2$ 2. For $b=1,c=0$: $D=2^1\times 3^1\times 5^0=2\times 3\times 1=6$ 3. For $b=0,c=1$: $D=2^1\times 3^0\times 5^1=2\times 1\times 5=10$ 4. For $b=1,c=1$: $D=2^1\times 3^1\times 5^1=2\times 3\times 5=30$

Our set of valid divisors is $\{2,6,10,30\}$. There are exactly $4$ such divisors.

We have navigated the constraints, applied the logic of prime factorization, and arrived at the solution with precision. Remember, in JEE Advanced, the math is rarely about memorizing formulas; it is about understanding the constraints and using them to narrow down the possibilities. You have just mastered the art of divisor analysis.