All Questions
1 question
2
votes
1
answer
513
views
Prove ${^{b}a} \equiv {^{b+1}} a \pmod {10^{\lfloor{\log_{10} (^{b}a) }\rfloor + 1}} \Rightarrow a=5$ as $a$ and $b$ are two integers greater than $1$
$\DeclareMathOperator\len{len}$Let $a, b \in \mathbb{N} -\{ 0, 1 \}$ and define ${^{b}a}$ to be $a^a$ if $b = 2$ and $a^{\left(^{b-1}a \right)}$ if $b \geq 3$ (e.g., ${^{3}5} = 5^{\left( 5^5 \right)} =...
- 1,451