MHB Find $\sqrt{ABBCDC}:$ A,B,C,D Distinct, $CDC-ABB=25$

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The problem involves finding the square root of a 6-digit number represented as $\overline{ABBCDC}$, where $ABB$ and $CDC$ are distinct 3-digit numbers satisfying the equation $CDC - ABB = 25$. Additionally, the digits A, B, C, and D must all be distinct. The discussion explores the implications of removing the restriction on the difference between $CDC$ and $ABB$ and questions how many solutions would exist without that condition. Ultimately, the goal is to determine $\sqrt{ABBCDC}$ under the given constraints.
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$\overline{ABB},$ and $\overline{CDC}$
are two 3-digit numbers ,
giving :
(1)$CDC-ABB=25$
(2)$\overline{ABBCDC}$ (6-digit number) is a perfect square
please find :$\sqrt{ABBCDC}$
(here A,B,C,D are distinct)
 
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Hello, Albert!

$ABB,$ and $CDC$ are two 3-digit numbers, such that:

(1) $CDC-ABB\,=\,25$

(2) $\overline{ABBCDC}$ (6-digit number) is a perfect square.

Find: $\sqrt{ABBCDC}$
(where A,B,C,D are distinct digits)
From (1), we have the alphametic: $\;\begin{array}{cccc} & C&D&C \\ -&A&B&B \\ \hline &&2&5\end{array}$

There are only 3 solutions.

$[1]\;\begin{array}{cccc} & 2&0&2 \\ - &1&7&7 \\ \hline &&2&5\end{array}$

$\qquad$But $ABBCDC \,=\,117,\!202$ is not a square.$[2]\;\begin{array}{cccc}&3&1&3 \\ -&2&8&8 \\ \hline && 2&5 \end{array}$

$\qquad$But $ABBCDC \,=\,288,\!313$ is not a square.$[3]\;\begin{array}{cccc}&4&2&4 \\ - & 3&9&9 \\ \hline && 2&5 \end{array}$

$\qquad$And $\sqrt{ABBCDC} \:=\:\sqrt{399,\!424} \;=\;632$
 
soroban said:
Hello, Albert!
From (1), we have the alphametic: $\;\begin{array}{cccc} & C&D&C \\ -&A&B&B \\ \hline &&2&5\end{array}$There are only 3 solutions.$[1]\;\begin{array}{cccc} & 2&0&2 \\ - &1&7&7 \\ \hline &&2&5\end{array}$$\qquad$But $ABBCDC \,=\,117,\!202$ is not a square.$[2]\;\begin{array}{cccc}&3&1&3 \\ -&2&8&8 \\ \hline && 2&5 \end{array}$$\qquad$But $ABBCDC \,=\,288,\!313$ is not a square.$[3]\;\begin{array}{cccc}&4&2&4 \\ - & 3&9&9 \\ \hline && 2&5 \end{array}$$\qquad$And $\sqrt{ABBCDC} \:=\:\sqrt{399,\!424} \;=\;632$
perfect !
 
Albert said:
$\overline{ABB},$ and $\overline{CDC}$
are two 3-digit numbers ,
giving :
(1)$CDC-ABB=25$
(2)$\overline{ABBCDC}$ (6-digit number) is a perfect square
please find :$\sqrt{ABBCDC}$
(here A,B,C,D are distinct)
If restriction (1)CDC-ABB=25 is taken away
how many soluions we can find ?
 

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