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 ?
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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