Opublikowany problem matematyczny... za wcześnie // A published math problem ... too early

This is problem S:E22.89 from the GMB exercise supplement, from March 2022 (page 5; proposed for 5th grade).

 

Translated (thanks to DeepL):

               "Let the natural numbers $\;a,\;b,\;c,\;x,\;y,\;z,\;$ be greater than $1$ such that 

$$a^b\;=\;c,\;\;c^x\;=\;y,\;\;y^z\;=\;a^{2025}.$$

             a) Calculate the multiplication $\;b\cdot x\cdot z\;.$

             b) Give an example of numbers that check all the above conditions." 

 

          It is customary in the current year, for many problems, to contain the number 2022, as a kind of parameter, that can be in the problem any number. The author, Mr MUNTEANU Emanuel George, rushed three years earlier to publish this problem. Or is it "because" of the elegant decomposition of the number 2025 into prime factors : $2025=3^4 \cdot 5^2$ ?

 

Answer CiP

 

a) $b \cdot x\cdot z\;=\;2025;$

 b) $a=7,\;b=15,\;c=7^{15},\;x=3,\;y=7^{45},\;z=45.$

           Indeed, $a^b=7^{15}=c,\;c^x=(7^{15})^3=7^{45}=y,\;y^z=(7^{45})^{45}=7^{45\cdot 45}=7^{2025}=a^{2025}.$

                               A more general example is of the form:

$a\geqslant 2\;$ fixed natural number;

 $z=3^{\alpha}\cdot 5^{\beta},\;\;0 \leqslant \alpha \leqslant 4,\;0 \leqslant \beta \leqslant 2$ but $(\alpha,\beta) \neq (0,0),\;(4,2)$ ; a divisor of the number 2025 ;

 $b=3^{\gamma}\cdot 5^{\delta},\;\;0 \leqslant \gamma \leqslant 4-\alpha,\;0 \leqslant \delta \leqslant 2- \beta,$ but not to have both $\gamma =4-\alpha, \delta =2-\beta$; a divisor of the number $\frac{2025}{z};$

$x=3^{4-\alpha-\gamma} \cdot 5^{2-\beta-\delta};$

 $y=a^{3^{4-\alpha} \cdot 5^{2-\beta}}.$

Solution CiP

                

                a) $a^{2025}\underset{def}{=}\;y^z \;\underset{y=c^x}{=}\;\left ( c^x \right )^z=c^{x \cdot z}\;\underset{c=a^b}{=}\left ( a^b \right )^{x \cdot z}=a^{b \cdot x \cdot z}$, and from this it follows

$$b \cdot x \cdot z\;=\;2025. \tag{1}$$

               b) We see from (1) that $b,x,z$ are divisors of 2025. We must have
$$z=3^{\alpha} \cdot 5^{\beta} \;,\tag{2}$$

with $(\alpha,\beta) \neq (0,0)$ since $z \neq 1$, and $(\alpha,\beta) \neq (4,2)$ since $b,x > 1$.

      Beacause $b\cdot x=\frac{2025}{z}=3^{4-\alpha}\cdot 5^{2-\beta}$, we must have

$$b=3^{\gamma}\cdot 5^{\delta}$$

and we choose $\gamma,\; \delta$ so that $b$ and $x$ are greater than 1.

The base $a$ can be any natural number $\geqslant 2$.

         In the concrete example we chose $a=7,\; \alpha =2,\; \beta=1,\;\gamma=\delta=1$.

         We could count all possible cases for a given $a$, but we won't do it here.

$\blacksquare$