A-two-digit-number-AB-10-AB-10-A-B-BA-10-Find-the-number-There-is-a-poem-Having-arrived-at-the-age-of-BA-10-

Question Number 203051 by ajfour last updated on 08/Jan/24

$${A}\:{two}\:{digit}\:{number}\:\:\left({AB}\right)_{\mathrm{10}} \\ $$$$\left({AB}\right)_{\mathrm{10}} −{A}^{{B}} =\left({BA}\right)_{\mathrm{10}} \\ $$$${Find}\:{the}\:{number}.\:{There}\:{is}\:{a}\:{poem}. \\ $$$${Having}\:{arrived}\:{at}\:{the}\:{age}\:{of}\:\left({BA}\right)_{\mathrm{10}} . \\ $$

Answered by Rasheed.Sindhi last updated on 08/Jan/24

(AB)_(10) −A^B =(BA)_(10) ⇒(AB)_(10) −A^B ≥0 10A+B−10B−A=A^B 9A−9B=A^B 9(A−B)=A^B ⇒9∣A^B A=9 or A=3,6 ∧ B≥2 & A≥B 91−9^1 =82 × 92−9^2 =11× 93−9^3 <0 × 32−3^2 =23 ✓ 62−6^2 =26✓ 62−6^3 <0 ×

$$\left({AB}\right)_{\mathrm{10}} −{A}^{{B}} =\left({BA}\right)_{\mathrm{10}} \Rightarrow\left({AB}\right)_{\mathrm{10}} −{A}^{{B}} \geqslant\mathrm{0} \\ $$$$\mathrm{10}{A}+{B}−\mathrm{10}{B}−{A}={A}^{{B}} \\ $$$$\mathrm{9}{A}−\mathrm{9}{B}={A}^{{B}} \\ $$$$\mathrm{9}\left({A}−{B}\right)={A}^{{B}} \\ $$$$\Rightarrow\mathrm{9}\mid{A}^{{B}} \\ $$$${A}=\mathrm{9}\:{or}\:\:{A}=\mathrm{3},\mathrm{6}\:\wedge\:{B}\geqslant\mathrm{2}\:\:\&\:{A}\geqslant{B} \\ $$$$\mathrm{91}−\mathrm{9}^{\mathrm{1}} =\mathrm{82}\:× \\ $$$$\mathrm{92}−\mathrm{9}^{\mathrm{2}} =\mathrm{11}× \\ $$$$\mathrm{93}−\mathrm{9}^{\mathrm{3}} <\mathrm{0}\:× \\ $$$$\mathrm{32}−\mathrm{3}^{\mathrm{2}} =\mathrm{23}\:\checkmark \\ $$$$\mathrm{62}−\mathrm{6}^{\mathrm{2}} =\mathrm{26}\checkmark \\ $$$$\mathrm{62}−\mathrm{6}^{\mathrm{3}} <\mathrm{0}\:× \\ $$

Commented by ajfour last updated on 08/Jan/24

Excellent! i had to subtract 32−23 an hour back when i thought to ask this this way!

$${Excellent}!\:{i}\:{had}\:{to}\:{subtract}\:\mathrm{32}−\mathrm{23} \\ $$$${an}\:{hour}\:{back}\:{when}\:{i}\:{thought}\:{to}\:{ask} \\ $$$${this}\:{this}\:{way}! \\ $$

Commented by MM42 last updated on 08/Jan/24

$$ \\ $$$$\mathrm{62}−\mathrm{6}^{\mathrm{2}} =\mathrm{26}\: \\ $$

Commented by Rasheed.Sindhi last updated on 12/Jan/24

Thanks sir, I′ve edited my solution to cover all answers.

$$\mathbb{T}\mathrm{hanks}\:\mathrm{sir},\:\mathrm{I}'\mathrm{ve}\:\mathrm{edited}\:\mathrm{my}\:\mathrm{solution} \\ $$$$\mathrm{to}\:\:\mathrm{cover}\:\mathrm{all}\:\:\mathrm{answers}.\: \\ $$

Answered by esmaeil last updated on 08/Jan/24

log(10A+B−10B−A)=logA^B → log(9A−9B)=logA^B → log9=logA^B −log(A−B)→ log9=log((A^B /(A−B)))= log((3^2 /(3−2)))→ { ((A=3)),((B=2)) :}

$${log}\left(\mathrm{10}{A}+{B}−\mathrm{10}{B}−{A}\right)={logA}^{{B}} \rightarrow \\ $$$${log}\left(\mathrm{9}{A}−\mathrm{9}{B}\right)={logA}^{{B}} \rightarrow \\ $$$${log}\mathrm{9}={logA}^{{B}} −{log}\left({A}−{B}\right)\rightarrow \\ $$$${log}\mathrm{9}={log}\left(\frac{{A}^{{B}} }{{A}−{B}}\right)= \\ $$$${log}\left(\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{3}−\mathrm{2}}\right)\rightarrow\begin{cases}{{A}=\mathrm{3}}\\{{B}=\mathrm{2}}\end{cases} \\ $$$$ \\ $$

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