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 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

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}

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