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

[Albert1]

$\overline{ABB},$ 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)

 

[soroban]

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)

Spoiler

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 $\sqrt{ABBCDC} \:=\:\sqrt{399,\!424} \;=\;632$

 

[Albert1]

Hello, Albert!

Spoiler

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 !

 

[Albert1]

$\overline{ABB},$ 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 ?

 

[CellCoree]

Based on the given information, we can create the following equations:

$\overline{ABB} = 100A + 10B + B = 101A + 10B$

$\overline{CDC} = 100C + 10D + C = 101C + 10D$

$CDC - ABB = (101C + 10D) - (101A + 10B) = 101(C-A) + 10(D-B) = 25$

Since $A,B,C,D$ are distinct, we can assume that $C-A = 2$ and $D-B = 5$ or vice versa. This gives us two possible solutions:

Solution 1: $C-A = 2$ and $D-B = 5$
Substituting these values into the equation $CDC - ABB = 25$, we get:
$101(2) + 10(5) = 25$
$202 + 50 = 252$
This means that $CDC = 252$ and $ABB = 227$. Therefore, $\overline{ABBCDC} = 227252$.

Solution 2: $C-A = -2$ and $D-B = -5$
Substituting these values into the equation $CDC - ABB = 25$, we get:
$101(-2) + 10(-5) = 25$
$-202 - 50 = -252$
This means that $CDC = -252$ and $ABB = -227$. However, since $\overline{ABB}$ and $\overline{CDC}$ are 3-digit numbers, we cannot have negative values. Therefore, this solution is not valid.

Now, we need to find the square root of $\overline{ABBCDC}$. Since we have two possible solutions, we will calculate the square root for both and see which one gives us a perfect square.

Solution 1: $\sqrt{227252} \approx 476.885$
Solution 2: $\sqrt{77222} \approx 277.503$

Since $\sqrt{227252}$ is closer to a perfect square (i.e. 476.885 is closer to a whole number than 277.503), we can assume that $\overline{ABBCDC} = 227252$ and $\sqrt{ABBCDC} = 476$.

Therefore, the square root