1. Key Concepts and Formulas

• Determinant Properties:
  • Factoring Common Terms: A common factor from any row or column can be factored out of the determinant. If a factor $k$ is common to all elements of a row/column, then $\det(A) = k \cdot \det(A')$.
  • Row/Column Operations: The value of a determinant remains unchanged if we apply the operation $R_i \to R_i + k R_j$ (or $C_i \to C_i + k C_j$). This is particularly useful for creating zeros.
  • Vandermonde Determinant: A determinant of the form $\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{array}\right| = (y-x)(z-x)(z-y)$. Related forms can often be simplified to this.
• Factorials: The definition $n! = n \times (n-1) \times \dots \times 1$. Also, $(n+k)! = (n+k)(n+k-1)\dots(n+1)n!$.
• Legendre's Formula: The exponent of a prime $q$ in the prime factorization of $n!$ is given by $E_q(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{q^k} \right\rfloor$. This formula helps determine the maximum power of a prime that divides a factorial.
• Prime Numbers: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The condition "p and p+2 are prime numbers" implies $p$ must be an odd prime (since if $p=2$, $p+2=4$ is not prime), so $p \ge 3$. This means $p$ and $p+2$ are distinct odd primes.

2. Step-by-Step Solution

Step 1: Simplify the Determinant by Factoring

The given determinant is: $\Delta=\left|\begin{array}{ccc} \mathrm{p} ! & (\mathrm{p}+1) ! & (\mathrm{p}+2) ! \\ (\mathrm{p}+1) ! & (\mathrm{p}+2) ! & (\mathrm{p}+3) ! \\ (\mathrm{p}+2) ! & (\mathrm{p}+3) ! & (\mathrm{p}+4) ! \end{array}\right|$ We observe common factors in each column. We factor out $p!$ from $C_1$, $(p+1)!$ from $C_2$, and $(p+2)!$ from $C_3$.

• From Column 1 ($C_1$): $p! \begin{pmatrix} 1 \\ p+1 \\ (p+1)(p+2) \end{pmatrix}$
• From Column 2 ($C_2$): $(p+1)! \begin{pmatrix} 1 \\ p+2 \\ (p+2)(p+3) \end{pmatrix}$
• From Column 3 ($C_3$): $(p+2)! \begin{pmatrix} 1 \\ p+3 \\ (p+3)(p+4) \end{pmatrix}$

Applying the determinant property of factoring common terms from columns, we get: $\Delta = p! \cdot (p+1)! \cdot (p+2)! \left|\begin{array}{ccc} 1 & 1 & 1 \\ p+1 & p+2 & p+3 \\ (p+1)(p+2) & (p+2)(p+3) & (p+3)(p+4) \end{array}\right|$ Let $K = p! (p+1)! (p+2)!$ and let $D$ be the remaining determinant.

Step 2: Evaluate the Simplified Determinant $D$

The determinant $D$ is: $D = \left|\begin{array}{ccc} 1 & 1 & 1 \\ p+1 & p+2 & p+3 \\ (p+1)(p+2) & (p+2)(p+3) & (p+3)(p+4) \end{array}\right|$ To simplify $D$, we recognize it as a form related to a Vandermonde determinant. Let $x=p+1, y=p+2, z=p+3$. Then $D$ becomes: $D = \left|\begin{array}{ccc} 1 & 1 & 1 \\ x & y & z \\ x(x+1) & y(y+1) & z(z+1) \end{array}\right|$ We can simplify this by performing a row operation $R_3 \to R_3 - R_2$: $D = \left|\begin{array}{ccc} 1 & 1 & 1 \\ x & y & z \\ x^2+x-x & y^2+y-y & z^2+z-z \end{array}\right| = \left|\begin{array}{ccc} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{array}\right|$ This is a standard Vandermonde determinant, which evaluates to $(y-x)(z-x)(z-y)$. Substitute back $x=p+1, y=p+2, z=p+3$:

• $y-x = (p+2)-(p+1) = 1$
• $z-x = (p+3)-(p+1) = 2$
• $z-y = (p+3)-(p+2) = 1$ So, $D = (1)(2)(1) = 2$.

Thus, the determinant $\Delta$ simplifies to: $\Delta = 2 \cdot p! \cdot (p+1)! \cdot (p+2)!$

Step 3: Determine the Exponent of $p$ ($\alpha$)

We are given that $p$ and $p+2$ are prime numbers.

• If $p=2$, then $p+2=4$, which is not prime. Therefore, $p$ must be an odd prime.
• This implies $p \ge 3$.
• Since $p \ge 3$, $p$ and $p+2$ are distinct odd prime numbers.

We need to find the maximum value of $\alpha$ such that $p^{\alpha}$ divides $\Delta$. This is equivalent to finding the exponent of prime $p$ in the prime factorization of $\Delta$, denoted as $E_p(\Delta)$. $\Delta = 2 \cdot p! \cdot (p+1)! \cdot (p+2)!$ Since $p$ is an odd prime, $p \ne 2$, so $p$ does not divide 2. Thus, $E_p(2) = 0$. The total exponent of $p$ in $\Delta$ is the sum of exponents from its factors: $\alpha = E_p(\Delta) = E_p(2) + E_p(p!) + E_p((p+1)!) + E_p((p+2)!)$ Using Legendre's formula $E_q(n!) = \sum_{k=1}^{\infty} \lfloor n/q^k \rfloor$:

• $E_p(p!) = \lfloor p/p \rfloor + \lfloor p/p^2 \rfloor + \dots = 1 + 0 + \dots = 1$. (Since $p \ge 3$, $p^2 > p$, so higher powers of $p$ don't contribute).
• $E_p((p+1)!) = \lfloor (p+1)/p \rfloor + \lfloor (p+1)/p^2 \rfloor + \dots = 1 + 0 + \dots = 1$. (Since $p \ge 3$, $p+1 < 2p$, so $\lfloor (p+1)/p \rfloor = 1$. Also $p^2 > p+1$, so higher powers of $p$ don't contribute).
• $E_p((p+2)!) = \lfloor (p+2)/p \rfloor + \lfloor (p+2)/p^2 \rfloor + \dots = 1 + 0 + \dots = 1$. (Since $p \ge 3$, $p+2 < 2p$, so $\lfloor (p+2)/p \rfloor = 1$. Also $p^2 > p+2$, so higher powers of $p$ don't contribute).

Summing these exponents: $\alpha = 0 + 1 + 1 + 1 = 3$

Step 4: Determine the Exponent of $(p+2)$ ($\beta$)

Similarly, we need to find the maximum value of $\beta$ such that $(p+2)^{\beta}$ divides $\Delta$. This is $E_{p+2}(\Delta)$. Since $p+2$ is an odd prime, $p+2 \ne 2$, so $p+2$ does not divide 2. Thus, $E_{p+2}(2) = 0$. $\beta = E_{p+2}(\Delta) = E_{p+2}(2) + E_{p+2}(p!) + E_{p+2}((p+1)!) + E_{p+2}((p+2)!)$ Using Legendre's formula:

• $E_{p+2}(p!) = 0$. (Since $p+2 > p$, $p+2$ cannot be a factor of $p!$).
• $E_{p+2}((p+1)!) = 0$. (Since $p+2 > p+1$, $p+2$ cannot be a factor of $(p+1)!$).
• $E_{p+2}((p+2)!) = \lfloor (p+2)/(p+2) \rfloor + \lfloor (p+2)/(p+2)^2 \rfloor + \dots = 1 + 0 + \dots = 1$. (Since $p+2 \ge 5$, $(p+2)^2 > p+2$, so higher powers don't contribute).

Summing these exponents: $\beta = 0 + 0 + 0 + 1 = 1$

Step 5: Calculate the Sum $\alpha + \beta$

We found $\alpha = 3$ and $\beta = 1$. The sum of the maximum values of $\alpha$ and $\beta$ is: $\alpha + \beta = 3 + 1 = 4$

3. Common Mistakes & Tips

• Forgetting to factor out all common terms: Ensure all possible common factors are extracted from rows/columns to simplify the determinant as much as possible.
• Errors in Vandermonde determinant calculation: Be careful with the order of terms in $(y-x)(z-x)(z-y)$.
• Incorrect application of Legendre's Formula: Remember that $E_q(n!) = \sum \lfloor n/q^k \rfloor$. For $q > n$, $E_q(n!) = 0$. For $q \le n$, ensure all relevant powers of $q$ are considered, though for small $n$ relative to $q$, often only $\lfloor n/q \rfloor$ is non-zero.
• Misinterpreting "p and p+2 are prime numbers": This condition is crucial for determining the behavior of $p$ and $p+2$ as prime factors (e.g., $p \ne 2$, $p$ and $p+2$ are distinct primes).

4. Summary

The problem required us to first simplify a determinant involving factorials and then find the maximum powers of primes $p$ and $p+2$ that divide the simplified determinant. We factored out common factorial terms from the determinant, reducing it to a product of factorials and a simpler $3 \times 3$ determinant. This simpler determinant was identified as a form of a Vandermonde determinant, which evaluated to 2. Thus, $\Delta = 2 \cdot p! (p+1)! (p+2)!$. Using Legendre's formula and the property that $p$ and $p+2$ are distinct odd primes, we calculated the exponent of $p$ in $\Delta$ as $\alpha=3$ and the exponent of $p+2$ in $\Delta$ as $\beta=1$. The sum of these maximum values is $\alpha + \beta = 3 + 1 = 4$.

5. Final Answer

The final answer is $\boxed{4}$.