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I'm stuck here with some numerical rebus -

Given: $A^2=BC, A^3=CA$

Find: $A+B+C$

  1. $13$
  2. $12$
  3. $11$
  4. $10$

(only one correct solution)

Note that letters represent digits.

I can't think of any idea to solve this one, and according to the book from which this question was taken, it is possible to solve in one minute. I will appreciate any idea.

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Taking apart the numbers, as lab bhattacharjee did, is a useful way to make good equations out of expressions of digits. When given a number like $CA$, this should be an early try. – Ross Millikan Nov 26 '12 at 19:03
$A^2$ has $2$ digits, so $A\ge4$, and $A^3$ has two digits, so $A\le4$. That kind of narrows things down.... – Gerry Myerson Feb 13 '13 at 12:18
up vote 6 down vote accepted

Assuming $A^2=10B+C$ and $A^3=10C+A$

We get $A^3<100\implies A<5$

As the last digits of $A$ and $A^3$ are same, $A$ must be $0,1$ or $4$

$A=0\implies C=B=0$


from the 1st equation $1^2=10B+C\implies B=0,C=1$

from the 2nd equation $1^3=10C+1\implies C=0$ so $A\ne 1$ as it would make the system inconsistent.

$A=4\implies C=6,B=1 $

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Thank you very much sir. An elegant and short solution indeed. I will accept your answer as soon as the system will allow me. – artyh Nov 26 '12 at 18:37
@artyh, my pleasure. Hope I could clear the matter. – lab bhattacharjee Nov 26 '12 at 18:38

From $A^3 = CA$, we know $A = 3$ or $A = 4$ (since $A^3$ must be two digits). Testing both, we see only $A = 4$ works, so $A^3 = 64 \rightarrow C = 6$

From $A^2 = BC$ and $A = 4$, $C = 6$, we see that $B = 1$. Thus, $A+B+C = 11$.

For digits, it's common to use overline: $\overline{abc}_k = ak^2 + bk + c$

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From the second equation, so $c = A^2$. Apply the $C$ value in the first equation, so $B = 1$. So $A + B + C = (A \cdot A) + A + 1$. If $A = 3$, then $(3 \cdot 3 ) + 3 + 1 = 13$

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In this question, $A^3=CA$ means that A cubed equals the number formed by writing the digits $C$ and $A$ next to each other, i.e. $10C+A$. – Peter Phipps Feb 13 '13 at 11:10

We have the equations $$A^2-10B-C=0$$ $$A^3-10C-A=0$$ that lab bhattacharjee established. also we know that $$\prod_{i=0}^9(A-i)=0$$ because $A \in \{0,1,2,3,4,5,6,7,8,9\}$. Also we have $$\prod_{i=0}^9(B-i)=0$$ $$\prod_{i=0}^9(C-i)=0$$ Now I calculate the grobner basis ${6 C-C^2,3 A-2 C,C-6 B}$ of this system using Maxima

(%i1) load(grobner)$
(%i2) poly_reduced_grobner([A^2-10*B-C,A^3-10*C-A,prod(A-i,i,0,9),prod(B-i,i,0,9),prod(C-i,i,0,9)],[A,B,C]);
(%o2) [6*C-C^2,3*A-2*C,C-6*B]

Solving the system of equations

$$ \begin{eqnarray} 6C-C^2&=&0 \\ 3A-2C&=&0 \\ C-6B&=&0 \end{eqnarray} $$

we get two solution triples$(A,B,C)$: $(0,0,0)$ and $(4,1,6)$

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