\(\textbf{Solution}\)

Let the \(\textit{complementary}\) divisors be \((x-10)\) ,\((x+10)\) and \((y-\frac{23}{2})\) and \(y+\frac{23}{2})\).

So, \(N=(x-10)(x+10)=x^2-10^2=x^2-100\)

and, \(N=(y-\frac{23}{2})(y+\frac{23}{2})=y^2-(\frac{23}{2})^2=y^2-\frac{529}{4}\) .

\(\therefore\) \(x^2-100=y^2-\frac{529}{4}\)

\(\Rightarrow y^2-x^2=\frac{529}{4}-100\)

\(\Rightarrow (y+x)(y-x)=\frac{129}{4}\)

\(\Rightarrow 4(y+x)(y-x)=129\)

\(\Rightarrow (2y+2x)(2y-2x)=129\)

\(\Rightarrow (m+n)(m-n)=129\) [Assuming \(2y=m\) and \(2x=n\)]

So, either \((m+n)(m-n)=43\times 3\) or \((m+n)(m-n)=129\times 1\)

• Case I

\((m+n)(m-n)=43\times 3\)

So, \((m+n)=43\) and \((m-n)=3\) because if  \((m+n)=3\) and \((m-n)=43\) then we are having negative solutions of \(x\) and \(y\), which cannot happen.

Now, by adding the equations we have \((m+n)+(m-n)=43+3 \Rightarrow 2m=46\Rightarrow m=23 \Rightarrow 2x=23 \Rightarrow x=11.5\)

\(x\) cannot have a non-integral or a negative solution, so this is not the required solution.

• Case II

Either \(m+n=129\) and \(m-n=1\) or \(m+n=1\) and \(m-n=129\).

But the in the former case we will get the negative solutions of \(n\), so that cannot be a solution.

So, adding the equations we get, \(m+n+m-n=129+1\Rightarrow 2m=130\Rightarrow m=65 \Rightarrow \boxed{y=32.5}\)

Again, \(n=64\Rightarrow \boxed{x=32}\)

\(\therefore\) This is the only possible solution.

So, \(N=(32-10)(32+10)=42\cdot 22 = 924\)

\(\therefore\) The sum of the digits of \(N=\boxed{9+2+4=15}\)

• This reply was modified 1 year, 6 months ago by Deepan Dutta.
• This reply was modified 1 year, 6 months ago by Deepan Dutta.