# 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 at 12:18

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 at 11:10