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
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