Question

The largest integer for which \[\dfrac{{{{\left( {n + 1} \right)}^2}}}{{n + 23}}\]is also an integer is
A. Greater than 470
B. Less than 440
C. Between 455 and 470
D. Between 440 and 455

Answer 1

Hint: We need to find the value of n for which the value of the expression is an integer. So we can first equate the expression to a variable k and form a quadratic equation in n, as n is an integer, the discriminant of the equation will be a perfect square. So we can equate it to another variable t. It also forms a quadratic equation. Its discriminant will be a perfect square and can be equated to another variable a. Then we can solve for a and t using factorization. Then we can solve for k using t and solve for n using k. Then we can compare the value of n with the options to get the correct option.

Complete step by step Answer:

We need to find the value of integer n where \[\dfrac{{{{\left( {n + 1} \right)}^2}}}{{n + 23}}\]is also an integer.
So, we can assume \[\dfrac{{{{\left( {n + 1} \right)}^2}}}{{n + 23}} = k\] where k is an integer.
We can multiply the above equation with $n + 23$on both sides.
\[ \Rightarrow {\left( {n + 1} \right)^2} = k\left( {n + 23} \right)\]
We can expand the square using the identity, ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.
\[ \Rightarrow {n^2} + 2n + 1 = kn + 23k\]
We rearrange it to a quadratic equation.
\[ \Rightarrow {n^2} + \left( {2 - k} \right)n + 1 - 23k = 0\]
We can find the value of n using the quadratic formula.
$ \Rightarrow n = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
On substituting the values, we get,
$ \Rightarrow n = \dfrac{{ - \left( {2 - k} \right) \pm \sqrt {{{\left( {2 - k} \right)}^2} - 4 \times 1 \times \left( {1 - 23k} \right)} }}{2}$
On simplification, we get,
$ \Rightarrow n = \dfrac{{k - 2 \pm \sqrt {4 + {k^2} - 4k - 4 + 92k} }}{2}$
$ \Rightarrow n = \dfrac{{k - 2 \pm \sqrt {{k^2} + 88k} }}{2}$
As k is a positive integer, n will become an integer only if ${k^2} + 88k$ is a perfect square.
So we can write,
${k^2} + 88k = {t^2}$where t is an integer.
${k^2} + 88k - {t^2} = 0$
By using quadratic formula, we get,
$ \Rightarrow k = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
On substituting the values, we get,
$ \Rightarrow k = \dfrac{{ - 88 \pm \sqrt {{{\left( {88} \right)}^2} - 4 \times 1 \times - {t^2}} }}{2}$
We can divide by 4 inside the radicle and by 2 on other terms,
$ \Rightarrow k = - 44 \pm \sqrt {{{\left( {44} \right)}^2} + {t^2}} $
As, k is positive, the root will be positive,
$ \Rightarrow k = - 44 + \sqrt {{{\left( {44} \right)}^2} + {t^2}} $
As k is an integer ${\left( {44} \right)^2} + {t^2}$will be perfect square.
So we can write,
${\left( {44} \right)^2} + {t^2} = {a^2}$
\[ \Rightarrow \]${\left( {44} \right)^2} = {a^2} - {t^2}$
On factorising using the relation ${a^2} - {t^2} = \left( {a + t} \right)\left( {a - t} \right)$, we get,
\[ \Rightarrow \]$44 \times 44 = \left( {a + t} \right)\left( {a - t} \right)$
We can factorise$44 \times 44$,
$44 \times 44 = 2 \times 2 \times 2 \times 2 \times 11 \times 11$
\[ \Rightarrow \]$44 \times 44 = \left( {2 \times 2 \times 2 \times 11 \times 11} \right) \times \left( 2 \right)$
\[ \Rightarrow \]$44 \times 44 = 968 \times 2$
Thus $\left( {a + t} \right)\left( {a - t} \right) = 968 \times 2$
\[ \Rightarrow \]$\left( {a + t} \right) = 968$ and $\left( {a - t} \right) = 2$
By solving, we get,
\[
  \,\,\,\,\,\,\,\,\left( {a + t} \right) = 968 \\
  \left( - \right)\underline {\left( {a - t} \right) = \,\,\,\,\,\,\,\,2} \\
  \,\,\,\,\,\,\,\,\,\,0 + 2t = 966 \\
 \]
$ \Rightarrow t = \dfrac{{966}}{2} = 483$
On substituting the value of t in $k = - 44 + \sqrt {{{\left( {44} \right)}^2} + {t^2}} $, we get,
$ \Rightarrow k = - 44 + \sqrt {{{\left( {44} \right)}^2} + {{\left( {483} \right)}^2}} $
On squaring we get,
$ \Rightarrow k = - 44 + \sqrt {1936 + 233289} $
On adding we get,
$ \Rightarrow k = - 44 + \sqrt {235225} $
On taking root we get,
$ \Rightarrow k = - 44 + 485$
On simplification we get,
$ \Rightarrow k = 441$
Now, on substituting the value of k in $n = \dfrac{{k - 2 \pm \sqrt {{k^2} + 88k} }}{2}$, we get,
$ \Rightarrow n = \dfrac{{441 - 2 \pm \sqrt {{{441}^2} + 88 \times 441} }}{2}$
On simplification we get,
$ \Rightarrow n = \dfrac{{439 + \sqrt {19481 + 38808} }}{2}$
On adding we get,
$ \Rightarrow n = \dfrac{{439 \pm \sqrt {233289} }}{2}$
On taking positive square root we get,
$ \Rightarrow n = \dfrac{{439 + 483}}{2}$
On solving further we get,
$ \Rightarrow n = \dfrac{{922}}{2} = 461$
So the required integer is 461.
Therefore the correct answer is option C.

Note: We use the concept of a quadratic equation and the quadratic formula to solve this problem. For the solution of a quadratic equation to be a rational number, its discriminant must be a perfect square. For a quadratic equation of the form $a{x^2} + bx + c = 0$, its discriminant is given by the equation, $D = {b^2} - 4ac$ and its solution is given by$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. In this question, we only took the positive roots as the largest integer will be positive and terms inside the square root must be positive. After finding the factors of $44 \times 44$, we must only select the factors which give integer values for all the variables. For that, we can try each of the factors starting with the highest factor.