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

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Discussion Overview

The discussion revolves around 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 $CDC - ABB = 25$. Participants are exploring the implications of this equation and the conditions on the digits A, B, C, and D.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Some participants reiterate the conditions that $CDC - ABB = 25$ and that $\overline{ABBCDC}$ must be a perfect square.
  • There is a suggestion to consider the scenario where the restriction $CDC - ABB = 25$ is removed, prompting a question about the number of possible solutions in that case.

Areas of Agreement / Disagreement

Participants generally agree on the initial conditions of the problem, but the discussion remains unresolved regarding the implications of removing the restriction and the potential number of solutions.

Contextual Notes

The discussion does not resolve the mathematical steps involved in finding the perfect square or the implications of the distinct digits condition.

Albert1
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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)
 
Last edited:
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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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