# Numerical puzzle

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

## 5 Answers

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 $4$

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

$A=1,$

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 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$(4,1,6)$- 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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Well I've written following piece of code in C++ MSVC 2012

#include <iostream>

using namespace std;

int main()
{
int A;
int B;
int C;
for( A=0 ; A<=9 ; A++ )
{
for( B=0 ; B<=9 ; B++ )
{
for( C=0 ; C<=9 ; C++ )
{
if(((A*A)/10==B)&&((A*A)%10==C))
{
if(((A*A*A)/10==C)&&((A*A*A)%10==A))
{
cout<<" A is "<<A<<endl;
cout<<" B is "<<B<<endl;
cout<<" C is "<<C<<endl;
cout<<endl;
}
}
}
}
}
cout<<"Done!";
return 0;
}


It yields following Results

A = 0 , B = 0 , C = 0.

and

A = 4 , B = 1 , C = 6.

I hope it helps...

Have a nice day.

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