Let p and p + 2 be prime numbers and let Δ = | p! (p+1)! (p+2)! | | (p+1)! (p+2)! (p+3)! | | (p+2)!

Q: Wait, why can't we just expand the determinant directly? It's only a 3x3 matrix!

While it is technically a 3x3 matrix, expanding it directly involves multiplying massive factorial terms like $(p+4)! \cdot (p+2)!$. This leads to an algebraic nightmare that is prone to errors. In JEE Advanced, the goal is to find the 'elegant path'—using row operations to simplify the matrix first is the hallmark of a master problem solver.

Q: Why is the power of $p$ in $p!$ just 1? Shouldn't it be higher?

Great question! We use Legendre's Formula, which states that the exponent of a prime $p$ in $n!$ is $\lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor + \dots$. For $n=p$, this is simply $\lfloor \frac{p}{p} \rfloor = 1$. Since $p$ is prime, it doesn't appear in any other factors of $p!$.

Q: Why is $\beta = 1$? I thought $(p+2)$ might appear more often.

Since $p$ and $p+2$ are twin primes (or simply primes), $p+2$ is a prime number greater than $p$. It cannot divide $p!$ or $(p+1)!$ because those products only contain factors smaller than $p+2$. Therefore, the only term in our final expression $\Delta = 2 \cdot (p!)^3 \cdot (p+1)^2 \cdot (p+2)$ that contains the factor $(p+2)$ is the $(p+2)$ term itself, which has an exponent of 1.

Question

Let $p$ and $p + 2$ be prime numbers and let $\Delta = \begin{vmatrix} p! & (p+1)! & (p+2)! \ (p+1)! & (p+2)! & (p+3)! \ (p+2)! & (p+3)! & (p+4)! \end{vmatrix}$. Then the sum of the maximum values of $\alpha$ and $\beta$, such that $p^\alpha$ and $(p + 2)^\beta$ divide $\Delta$, is ________.

Sigma Percentile · Accepted Answer

### The Beauty of Determinants: Unmasking the Factorial Beast

Welcome, fellow traveler on the road to JEE Advanced. Today, we are not just solving a problem; we are peeling back the layers of a mathematical onion.

When you first look at this determinant,

$$ \Delta = \begin{vmatrix} p! & (p+1)! & (p+2)! \ (p+1)! & (p+2)! & (p+3)! \ (p+2)! & (p+3)! & (p+4)! \end{vmatrix} $$

it is natural to feel a surge of intimidation. Factorials grow at an explosive rate, and a $3 	imes 3$ matrix filled with them looks like a wall of complexity. But remember, in the world of competitive mathematics, complexity is often a mask for a hidden, elegant simplicity.

### Phase 1: The Art of Factoring

Our first instinct might be to expand, but that is a trap. Instead, let us look for the 'DNA' of the matrix.

Notice that every term in the first row contains $p!$. We can write $(p+1)!$ as $(p+1) \cdot p!$ and $(p+2)!$ as $(p+2)(p+1) \cdot p!$.

By pulling $p!$ out of the first row, we reduce the 'noise' of the large numbers. We repeat this for the second row, pulling out $(p+1)!$, and the third row, pulling out $(p+2)!$.

Suddenly, the matrix is no longer a collection of massive factorials; it is a clean, manageable grid. We have transformed a mountain into a molehill.

### Phase 2: The Geometry of Operations

Now that we have factored out $p! \cdot (p+1)! \cdot (p+2)!$, we are left with a matrix where the first column is entirely ones:

$$ \begin{vmatrix} 1 & p+1 & (p+1)(p+2) \ 1 & p+2 & (p+2)(p+3) \ 1 & p+3 & (p+3)(p+4) \end{vmatrix} $$

This is a moment of pure joy for a mathematician. A column of ones is an invitation to perform row operations.

By applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_2$, we create zeros. These zeros are our best friends—they make the final expansion trivial.

As we subtract these rows, watch how the algebra collapses. The terms $(p+2)(p+3) - (p+1)(p+2)$ simplify beautifully to $2(p+2)$. The complexity vanishes, leaving us with a simple constant.

### Phase 3: The Prime Number Mystery

After expanding, we arrive at $\Delta = 2 \cdot p! \cdot (p+1)! \cdot (p+2)!$. Now, we must find the maximum powers $\alpha$ and $\beta$ such that $p^\alpha$ and $(p+2)^\beta$ divide $\Delta$.

This is where we shift from algebra to number theory. We use Legendre's Formula to find the power of $p$ in $p!$, which is $1$.

Since we have $(p!)^3$ in our expression (after accounting for the factors), the total power of $p$ is $3 	imes 1 = 3$. For $(p+2)$, we recognize that since it is a prime greater than $p$, it cannot divide $p!$ or $(p+1)$.

It only appears once in the final expression. Thus, $\alpha = 3$ and $\beta = 1$.

### The Victory

Adding these together, we get $3 + 1 = 4$.

We started with a terrifying determinant and ended with a simple integer. This is the essence of JEE Advanced mathematics: it is not about brute force; it is about finding the structure, simplifying the chaos, and trusting the logic.

You have the tools. You have the insight. Now, go forth and conquer. The final answer is **4**.

Enter Numerical Value:

Visualized Solution

Analyzing the Determinant $Δ$

Factoring the First Row

Factoring the Second Row

Factoring the Third Row

Applying Row Operations

Executing $R_{2} - R_{1}$

Executing $R_{3} - R_{2}$

Expanding the Determinant

Simplifying the Final Expression

Finding the Power of $p$ ( $α$ )

Finding the Power of $p + 2$ ( $β$ )

Final Sum and Conclusion

The Sigma Insight: Properties of Determinants

The Beauty of Determinants

Phase 1

Phase 2

Phase 3

The Victory

Similar Questions

For a fixed positive integer $n$ , if $D = n! (n + 1)! (n + 2)! (n + 1)! (n + 2)! (n + 3)! (n + 2)! (n + 3)! (n + 4)!$ , then show that $[\frac{D}{( n ! ) ^{3}} - 4]$ is divisible by $n$ .

Let $A = 2631232112$ and $P = 157201025$ . The sum of the prime factors of $∣ P^{- 1} A P - 2 I ∣$ is equal to

Let $a > 0, d > 0$ . Find the value of the determinant $\frac{1}{a} \frac{1}{( a + d )} \frac{1}{( a + 2 d )} \frac{1}{a ( a + d )} \frac{1}{( a + d ) ( a + 2 d )} \frac{1}{( a + 2 d ) ( a + 3 d )} \frac{1}{( a + d ) ( a + 2 d )} \frac{1}{( a + 2 d ) ( a + 3 d )} \frac{1}{( a + 3 d ) ( a + 4 d )}$ .

Let $P$ be a matrix of order $3 \times 3$ such that all the entries in $P$ are from the set ${- 1, 0, 1}$ . Then, the maximum possible value of the determinant of $P$ is ________.

If $a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots$ are in G.P., then the determinant $Δ = lo g a_{n} lo g a_{n + 3} lo g a_{n + 6} lo g a_{n + 1} lo g a_{n + 4} lo g a_{n + 7} lo g a_{n + 2} lo g a_{n + 5} lo g a_{n + 8}$ is equal to

Let $a, b, c, d$ be in arithmetic progression with common difference $λ$ . If $x + a - c x - 1 x - b + d x + b x + c x + d x + a x + b x + c = 2$ , then value of $λ^{2}$ is equal to

Which of the following values of $α$ satisfy the equation $(1 + α)^{2} (2 + α)^{2} (3 + α)^{2} (1 + 2 α)^{2} (2 + 2 α)^{2} (3 + 2 α)^{2} (1 + 3 α)^{2} (2 + 3 α)^{2} (3 + 3 α)^{2} = - 648 α$ ?

For $α, β \in R$ and a natural number $n$ , let $A_{r} = r 2 r 3 r - 2 123 \frac{n ^{2}}{2} + α n^{2} - β \frac{n ( 3 n - 1 )}{2}$ . Then $2 A_{10} - A_{8}$ is equal to :

If $1, lo g_{10} (4^{x} - 2)$ and $lo g_{10} (4^{x} + \frac{18}{5})$ are in arithmetic progression for a real number $x$ , then the value of the determinant $210 (x - 1) 01 x^{2} 1 x$ is equal to :

The determinant $a b a α + b b c b α + c a α + b b α + c 0$ is equal to zero, if