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If the positive integer \(n\) has positive integer divisors \(a\) and \(b\) with \(n=ab\) , then \(a\) and \(b\) are said to be \(\textit{complementary}\) divisors of \(n\) . Suppose that \(N\) is a positive integer that has one complementary pair of divisors that differ by \(20\) and another pair of complementary divisors that differ by \(23\) . What is the sum of the digits of \(N\)?
\(\textbf{Solution}\)
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 \(N=\boxed{9+2+4=15}\)
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