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)


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 203051 by ajfour last updated on 08/Jan/24

$A t w o d i g i t n u m b e r {(A B)}_{10}$ ${(A B)}_{10} - A^{B} = {(B A)}_{10}$ $F i n d t h e n u m b e r . T h e r e i s a p o e m .$ $H a v i n g a r r i v e d a t t h e a g e o f {(B A)}_{10} .$
	Answered by Rasheed.Sindhi last updated on 08/Jan/24

	${(A B)}_{10} - A^{B} = {(B A)}_{10} \Rightarrow {(A B)}_{10} - A^{B} ⩾ 0$ $10 A + B - 10 B - A = A^{B}$ $9 A - 9 B = A^{B}$ $9 (A - B) = A^{B}$ $\Rightarrow 9 ∣ A^{B}$ $A = 9 o r A = 3, 6 \land B ⩾ 2 & A ⩾ B$ $91 - 9^{1} = 82 \times$ $92 - 9^{2} = 11 \times$ $93 - 9^{3} < 0 \times$ $32 - 3^{2} = 23 ✓$ $62 - 6^{2} = 26 ✓$ $62 - 6^{3} < 0 \times$
	Commented by ajfour last updated on 08/Jan/24

	$E x c e l l e n t! i h a d t o s u b t r a c t 32 - 23$ $a n h o u r b a c k w h e n i t h o u g h t t o a s k$ $t h i s t h i s w a y!$
	Commented by MM42 last updated on 08/Jan/24

	$62 - 6^{2} = 26$
	Commented by Rasheed.Sindhi last updated on 12/Jan/24

	$T hanks sir, I^{'} ve edited my solution$ $to cover all answers .$
	Answered by esmaeil last updated on 08/Jan/24
	$log(10A+B−10B−A)=logA^B → log(9A−9B)=logA^B → log9=logA^B −log(A−B)→ log9=log((A^B /(A−B)))= log((3^2 /(3−2)))→ { ((A=3)),((B=2)) :}$
	$l o g (10 A + B - 10 B - A) = {l o g A}^{B} \to$ $l o g (9 A - 9 B) = {l o g A}^{B} \to$ $l o g 9 = {l o g A}^{B} - l o g (A - B) \to$ $l o g 9 = l o g (\frac{A^{B}}{A - B}) =$ $l o g (\frac{3^{2}}{3 - 2}) \to {\begin{cases} A = 3 \\ B = 2 \end{cases}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum