System of equations in digits

Algebra Level 2

Let A , B , C , D A,B,C,D be distinct non-zero decimal digits that satisfy the following system of equations:

{ A B C = ( C D ) 2 D B C = ( A C ) 2 C B D = ( C A ) 2 . \begin{cases}ABC=(CD)^2\\ DBC=(AC)^2\\ CBD=(CA)^2. \end{cases}

What is the sum A + B + C + D ? A+B+C+D?


The answer is 19.

David Ingerman by David Ingerman , 48, USA

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2 solutions

X X
Sep 18, 2018

Notice the D B C = ( A C ) 2 , C B D = ( C A ) 2 DBC=(AC)^2,CBD=(CA)^2 .

If the digit reverse, then the digit of their square reverse, too, then the only two possibilities are 169 and 961, and 144 and 441( but B B can not be equal to C C .

So, C A CA must be 13, C B D CBD must be 169. Hence, 3 + 6 + 1 + 9 = 19 3+6+1+9=19

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David Ingerman
Sep 17, 2018

{ 361 = 1 9 2 , 961 = 3 1 2 , 169 = 1 3 2 . \begin{cases} 361=19^2,\\ 961=31^2,\\ 169=13^2. \end{cases} A+B+C+D=3+6+1+9=19.

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