Let p and p + 2 be prime numbers and let \Delta=|\beginarrayccc p ! & (p+1) ! & (p+2) ! (p+1) ! & (p+2) ! & (p+3) ! (p+2) ! & (p+3) ! & (p+4) ! \endarray| Then the sum of the maximum values of \alpha and \beta , such that p^\alpha and (p+2)^\beta divide \Delta , is \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2022 · 29 Jul · Shift 1 · Q45 with Solution

View written solutionFree

Correct answer: 4

Write the determinant in a factored form

Given

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

Use

(p+k)! = p!\,(p+1)(p+2)\cdots(p+k).

So each entry can be written as $p!$ times a polynomial in $p$ .

Let

a_0=1,\quad a_1=p+1,\quad a_2=(p+1)(p+2),\quad a_3=(p+1)(p+2)(p+3),\quad a_4=(p+1)(p+2)(p+3)(p+4).

Then

\Delta=(p!)^3\begin{vmatrix} a_0 & a_1 & a_2\\ a_1 & a_2 & a_3\\ a_2 & a_3 & a_4 \end{vmatrix}.

So it remains to compute

D=\begin{vmatrix} 1 & p+1 & (p+1)(p+2)\\ p+1 & (p+1)(p+2) & (p+1)(p+2)(p+3)\\ (p+1)(p+2) & (p+1)(p+2)(p+3) & (p+1)(p+2)(p+3)(p+4) \end{vmatrix}.

Simplify $D$ by row operations

Let

x=p+1,\quad y=p+2,\quad z=p+3,\quad w=p+4.

Then

D=\begin{vmatrix} 1 & x & xy\\ x & xy & xyz\\ xy & xyz & xyzw \end{vmatrix}.

Now perform row operations:

$R_2 \to R_2 - xR_1$
$R_3 \to R_3 - xyR_1$

Then

R_2=[0,\ x(y-1),\ xyz-x^2y],

R_3=[0,\ xy(z-1),\ xyzw-x^2y^2].

Since $y-1=x$ and $z-1=y$ ,

R_2=[0,x^2,xy(z-x)]=[0,x^2,2xy],

because $z-x=(p+3)-(p+1)=2$ .

Also,

R_3=[0,xy^2,xy(zw-xy)].

Now

zw-xy=(p+3)(p+4)-(p+1)(p+2)=4p+10=2(2p+5).

R_3=[0,xy^2,2xy(2p+5)].

Hence

D=\begin{vmatrix} 1 & x & xy\\ 0 & x^2 & 2xy\\ 0 & xy^2 & 2xy(2p+5) \end{vmatrix}.

Expanding along the first column,

D=\begin{vmatrix} x^2 & 2xy\\ xy^2 & 2xy(2p+5) \end{vmatrix}.

Therefore

D=2x^3y(2p+5)-2xy\cdot xy^2 =2xy\big(x^2(2p+5)-xy^2\big).

Factor $x$ inside:

D=2x^2y\big(x(2p+5)-y^2\big).

Now substitute $x=p+1$ , $y=p+2$ :

x(2p+5)-y^2=(p+1)(2p+5)-(p+2)^2.

Compute:

(p+1)(2p+5)=2p^2+7p+5,

(p+2)^2=p^2+4p+4.

x(2p+5)-y^2=p^2+3p+1.

Thus

D=2(p+1)^2(p+2)(p^2+3p+1).

Hence

\Delta=2(p!)^3(p+1)^2(p+2)(p^2+3p+1).

Use the condition that $p$ and $p+2$ are both prime

If $p$ and $p+2$ are both prime, they are twin primes.

The only even prime is $2$ . If $p>2$ , then $p$ is odd and $p+2$ is also odd. So in general twin primes are possible for odd $p$ .

We need the highest powers of $p$ and $p+2$ dividing $\Delta$ .

Find the exponent $\alpha$ of $p$ in $\Delta$

From

\Delta=2(p!)^3(p+1)^2(p+2)(p^2+3p+1),

count powers of $p$ .

In $(p!)^3$ , since $p!$ contains exactly one factor of $p$ , we get contribution $3$ .
$(p+1)^2$ contributes no factor of $p$ .
$(p+2)$ contributes no factor of $p$ .
$p^2+3p+1 \equiv 1 \pmod p$ , so no factor of $p$ .
The factor $2$ contributes no factor of $p$ for odd prime $p$ .

So for odd twin-prime $p$ ,

\alpha=3.

(If $p=2$ , then $p+2=4$ is not prime, so this case is impossible.)

Find the exponent $\beta$ of $p+2$ in $\Delta$

Now count powers of $p+2$ .

In $p!$ , there is no factor $p+2$ since $p+2>p$ . So $(p!)^3$ contributes $0$ .
The explicit factor $(p+2)$ contributes $1$ .
$(p+1)^2$ contributes $0$ .
Check whether $p+2$ divides $p^2+3p+1$ .

Modulo $p+2$ , we have $p\equiv -2$ , so

p^2+3p+1 \equiv (-2)^2+3(-2)+1=4-6+1=-1 \not\equiv 0 \pmod{p+2}.

So no extra factor of $p+2$ comes from this term.

Also $2$ is not divisible by $p+2$ since $p+2\ge 5$ for twin primes.

Hence

\beta=1.

Required sum

Therefore,

\alpha+\beta=3+1=4.

So the required integer is

\boxed{4}.

Comparison with stored answer

Stored correct answer = $4$ .

Our derived answer is also $4$ , so they agree.

More from Matrices and Determinants

View written solutionFree

Correct answer: 4

Write the determinant in a factored form

Given

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

Use

(p+k)! = p!\,(p+1)(p+2)\cdots(p+k).

So each entry can be written as $p!$ times a polynomial in $p$ .

Let

a_0=1,\quad a_1=p+1,\quad a_2=(p+1)(p+2),\quad a_3=(p+1)(p+2)(p+3),\quad a_4=(p+1)(p+2)(p+3)(p+4).

Then

\Delta=(p!)^3\begin{vmatrix} a_0 & a_1 & a_2\\ a_1 & a_2 & a_3\\ a_2 & a_3 & a_4 \end{vmatrix}.

So it remains to compute

D=\begin{vmatrix} 1 & p+1 & (p+1)(p+2)\\ p+1 & (p+1)(p+2) & (p+1)(p+2)(p+3)\\ (p+1)(p+2) & (p+1)(p+2)(p+3) & (p+1)(p+2)(p+3)(p+4) \end{vmatrix}.

Simplify $D$ by row operations

Let

x=p+1,\quad y=p+2,\quad z=p+3,\quad w=p+4.

Then

D=\begin{vmatrix} 1 & x & xy\\ x & xy & xyz\\ xy & xyz & xyzw \end{vmatrix}.

Now perform row operations:

$R_2 \to R_2 - xR_1$
$R_3 \to R_3 - xyR_1$

Then

R_2=[0,\ x(y-1),\ xyz-x^2y],

R_3=[0,\ xy(z-1),\ xyzw-x^2y^2].

Since $y-1=x$ and $z-1=y$ ,

R_2=[0,x^2,xy(z-x)]=[0,x^2,2xy],

because $z-x=(p+3)-(p+1)=2$ .

Also,

R_3=[0,xy^2,xy(zw-xy)].

Now

zw-xy=(p+3)(p+4)-(p+1)(p+2)=4p+10=2(2p+5).

R_3=[0,xy^2,2xy(2p+5)].

Hence

D=\begin{vmatrix} 1 & x & xy\\ 0 & x^2 & 2xy\\ 0 & xy^2 & 2xy(2p+5) \end{vmatrix}.

Expanding along the first column,

D=\begin{vmatrix} x^2 & 2xy\\ xy^2 & 2xy(2p+5) \end{vmatrix}.

Therefore

D=2x^3y(2p+5)-2xy\cdot xy^2 =2xy\big(x^2(2p+5)-xy^2\big).

Factor $x$ inside:

D=2x^2y\big(x(2p+5)-y^2\big).

Now substitute $x=p+1$ , $y=p+2$ :

x(2p+5)-y^2=(p+1)(2p+5)-(p+2)^2.

Compute:

(p+1)(2p+5)=2p^2+7p+5,

(p+2)^2=p^2+4p+4.

x(2p+5)-y^2=p^2+3p+1.

Thus

D=2(p+1)^2(p+2)(p^2+3p+1).

Hence

\Delta=2(p!)^3(p+1)^2(p+2)(p^2+3p+1).

Use the condition that $p$ and $p+2$ are both prime

If $p$ and $p+2$ are both prime, they are twin primes.

The only even prime is $2$ . If $p>2$ , then $p$ is odd and $p+2$ is also odd. So in general twin primes are possible for odd $p$ .

We need the highest powers of $p$ and $p+2$ dividing $\Delta$ .

Find the exponent $\alpha$ of $p$ in $\Delta$

From

\Delta=2(p!)^3(p+1)^2(p+2)(p^2+3p+1),

count powers of $p$ .

In $(p!)^3$ , since $p!$ contains exactly one factor of $p$ , we get contribution $3$ .
$(p+1)^2$ contributes no factor of $p$ .
$(p+2)$ contributes no factor of $p$ .
$p^2+3p+1 \equiv 1 \pmod p$ , so no factor of $p$ .
The factor $2$ contributes no factor of $p$ for odd prime $p$ .

So for odd twin-prime $p$ ,

\alpha=3.

(If $p=2$ , then $p+2=4$ is not prime, so this case is impossible.)

Find the exponent $\beta$ of $p+2$ in $\Delta$

Now count powers of $p+2$ .

In $p!$ , there is no factor $p+2$ since $p+2>p$ . So $(p!)^3$ contributes $0$ .
The explicit factor $(p+2)$ contributes $1$ .
$(p+1)^2$ contributes $0$ .
Check whether $p+2$ divides $p^2+3p+1$ .

Modulo $p+2$ , we have $p\equiv -2$ , so

p^2+3p+1 \equiv (-2)^2+3(-2)+1=4-6+1=-1 \not\equiv 0 \pmod{p+2}.

So no extra factor of $p+2$ comes from this term.

Also $2$ is not divisible by $p+2$ since $p+2\ge 5$ for twin primes.

Hence

\beta=1.

Required sum

Therefore,

\alpha+\beta=3+1=4.

So the required integer is

\boxed{4}.

Comparison with stored answer

Stored correct answer = $4$ .

Our derived answer is also $4$ , so they agree.

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