Problem E:15726 GMB 6-7-8/2020, pag 365

 Proposed for the fifth grade. In translation:

 

          "Find the natural numbers $\overline{ab}$ for which $a^{2}+b^{b}=\overline{ab}$."

 

ANSWER CiP

$43$ and $63$

Solution CiP

          In writing $\overline{ab}$ is usually assumed $a \neq 0$. We also assume that the numbering base is ten (so $\overline{ab}=10\cdot a+b$). Because of the term $b^{b}$ we also require that $b \neq 0$. So

$a \in\{1,2,3,4,5,6,7,8,9\}\;\;\; b\in\{1,2,3,4,5,6,7,8,9\}$.

     From the given equation we obtain

$(1)$                    $b^{b}=\overline{ab}-a^{2}$

and because $4^{4}=256>\overline{ab}$ we can only have cases $b=1$, $b=2$ or $b=3$.

     If $b=1$ we must have $a^{2}+1=\overline{a1}$ or $ a^{2}+1=10\cdot a+1$ or $ a^{2}=10a$. But none of the values that verify this ($a=0\; and \;a=10$) are admissible.

     If $b=2$ we must have $a^{2}+2^{2}=\overline{a2}$ $\Leftrightarrow$ $a^{2}+2=10\cdot a$ and the last equation is not verified by any of the allowable values of $a$. (Another reason is that for $a^{2}+2$ to be a multiple of $10$, the last digit of $a^{2}$ must be $8$, which cannot be !).

     If $b=3$ we must have $a^{2}+3^{3}=\overline{a3}$ $\Leftrightarrow$ $a^{2}+24=10\cdot a$, and the equation

$a^{2}-10a+24=0$ has roots $a_{1}=4$ and $a_{2}=6$. Hence the answer.

     To be rigorous, we must also check the values we found:

$4^{2}+3^{3}=43$ and $6^{2}+3^{3}=63$

which is easy to do. 

$\blacksquare$

  

     REMARK  

          The problem being proposed for the fifth grade, the use of a second degree equation should be avoided in solution (case $b=3$). We can write the equation $a^{2}+24=10a$ in form

$(2)$                $a\cdot (10-a)=24$.

     - We can check each of the possible values of $a$ (by brute force !)

or

     - We check only the values 1, 2, 3, 4, and 5, because $a$ and $10-a$ play symmetrical roles

or

     - Seeing that in equation (2) , $a$ must be an even number, so we check only the values 2, 4, 6, 8

or  ALL of THIS