Question

Find two consecutive natural numbers, the sum of whose squares is $181$.

• A. $7,8$
• B. $8,9$
• C. $9,10$
• D. $10,11$

Answer 1

Answer: (C)

$9,10$

Let the two consecutive numbers are $n$ and $n+1$
According to the given condition,
${\left[\text{1st natural number}\right]}^2+{\left[\text{2nd natural number}\right]}^2=181$
$\Rightarrow n^2+{(n+1)}^2=181$
$\Rightarrow n^2+n^2+2n+1=181$
$\Rightarrow {2n}^2+2n-180=0$
$\Rightarrow 2(n^2+n-90)=0$
$\Rightarrow n^2+n-90=0$
$\Rightarrow n^2+10n-9n-90=0$
$\Rightarrow n(n+10)-9(n+10)=0$
$\Rightarrow (n+10)(n-9)=0$
Either, $n+10=0$
$n=-10$ which is not possible being negative
and $n-9=0\Rightarrow n=9$.
Hence, $1^{st}$ natural number $=9$
$2^{nd}$ natural number $=$ $(n+1)=(9+1)=10$.