Courses
Courses for Kids
Free study material
Offline Centres
More

Three consecutive positive integers are such that the sum of the square of the first and the product of other two is $46$, the smallest integer is:

seo-qna
Last updated date: 01st Mar 2024
Total views: 408.6k
Views today: 8.08k
IVSAT 2024
Answer
VerifiedVerified
408.6k+ views
Hint: Proper conversion of wordings to equation form is must. It is easy to solve such problems with variables, as they can be interrelated easily like $b=a+1$ to denote they are consecutive in nature.

Complete step-by-step answer:
Here, three consecutive positive integers are such that the sum of the first and product of the other two is $46$.
Let us consider these integers as $a,b$ and $c$ such that
$\begin{align}
  & \Rightarrow b=a+1 \\
 & \Rightarrow c=b+1 \\
\end{align}$
Or, $\Rightarrow c=\left( a+1 \right)+1=a+2...\text{ }\left( 1 \right)$
Now, as per the given conditions, we have
$\Rightarrow {{a}^{2}}+bc=46$, where $a$ is first integer, $b$ is second integer and $c$ is third integer.
Substituting the values of $b$ and $c$ from equation (1), we get
$\begin{align}
  & \Rightarrow {{a}^{2}}+bc=46 \\
 & \Rightarrow {{a}^{2}}+\left( a+1 \right)\left( a+2 \right)=46 \\
 & \Rightarrow {{a}^{2}}+\left( a\left( a+2 \right)+1\left( a+2 \right) \right)=46 \\
 & \Rightarrow {{a}^{2}}+\left( {{a}^{2}}+2a+a+2 \right)=46 \\
 & \Rightarrow {{a}^{2}}+{{a}^{2}}+3a+2=46 \\
 & \Rightarrow 2{{a}^{2}}+3a+2=46 \\
\end{align}$
By transposing values, we get
$\begin{align}
  & \Rightarrow 2{{a}^{2}}+3a+2=46 \\
 & \Rightarrow 2{{a}^{2}}+3a+2-46=0 \\
 & \Rightarrow 2{{a}^{2}}+3a-44=0 \\
\end{align}$
Applying middle-term split method, we get
$\begin{align}
  & \Rightarrow 2{{a}^{2}}+3a-44=0 \\
 & \Rightarrow 2{{a}^{2}}+11a-8a-44=0 \\
 & \Rightarrow a\left( 2a+11 \right)-4\left( 2a+11 \right)=0 \\
 & \Rightarrow \left( a-4 \right)\left( 2a+11 \right)=0 \\
\end{align}$
Thus, solving the above equation, we get
\[\Rightarrow \left( a-4 \right)\left( 2a+11 \right)=0,\] i.e.,
\[\begin{align}
  & \Rightarrow \left( a-4 \right)=0 \\
 & \Rightarrow a=4 \\
\end{align}\]
And,
$\begin{align}
  & \Rightarrow \left( 2a+11 \right)=0 \\
 & \Rightarrow 2a=-11 \\
\end{align}$
On cross-multiplication, we get
$\Rightarrow a=\dfrac{-11}{2}$
But as per the given conditions, $a$ is a positive integer, so $a=\dfrac{-11}{2}$ is not applicable.
Thus, $a=4$ and from equation (1), we can calculate other two integers, i.e.,
$\begin{align}
  & \Rightarrow b=a+1 \\
 & \Rightarrow b=4+1 \\
 & \Rightarrow b=5
\end{align}$
Similarly,
\[\begin{align}
  & \Rightarrow c=a+2 \\
 & \Rightarrow c=4+2 \\
 & \Rightarrow c=6 \\
\end{align}\]
Hence, the smallest integer out of three consecutive integers, we have $a=4$.

Note: This type of problem can also be solved with the hit and trial method. Considering values of those three consecutive integers one by one, trying to fit it in given conditions.
Recently Updated Pages