×
Back to all chapters

# Euler's Theorem

Euler's theorem relate to the remainder of various powers and has applications ranging from modern cryptography to recreational problem-solving.

# Euler's Theorem: Level 5 Challenges

$\Large 9^{8^{7^{6^{5^{4^{3^{2^{1}}}}}}}}$

Find the last three digits of the above number.

Let $$a$$ and $$b$$ be positive integers with $$a>b$$ such that $7!|(x^a-x^b)$ for all integers $$x.$$

Find the smallest possible value of $$a+b.$$

Clarification:
$$!$$ denotes the factorial notation. For example, $$8! = 1\times2\times3\times\cdots\times8$$.

$\large 2,2^2,2^{2^2},2^{2^{2^2}},\ldots$ The sequence above can be recursively defined by $a_1=2, a_{k+1}=2^{a_{k}}$

Given a positive integer $$n$$, a new sequence is created by dividing each term $$\{a_k\}$$ by $$n$$ and writing down the remainder of the division. We denote this new sequence by $$\{b_{k}\}$$.

For a given $$n$$, it is known that the terms of $$\{b_{k}\}$$ eventually become constant. Let $$f(n)$$ denote the index at which the constant values begin, i.e. $$f(n)$$ is the smallest number $$i$$ for which $$b_{i-1} \neq b_i = b_{i+1} = b_{i+2} = \cdots$$.

Find $$f(2016)+f(2015)$$.

###### Proving that the terms of $$\{b_{n}\}$$ become constant is an old USAMO problem, which inspired this problem.

How many integers $$1\leq{a}\leq{2015}$$ are there such that $$a^{a^a}$$ and $$a^a$$ end with the same digit?

$\Large 249^3 = 15438\underline{249}$

An automorphic number is defined as a positive integer $$n$$ such that the trailing digits of $$n^m$$, where $$m$$ is a positive integer, is $$n$$ itself.

Let us define an almost automorphic number as a number where $$n$$ only appears as the trailing digits of $$n^m$$ when $$m$$ is odd. How many almost automorphic numbers are less than 1000?

×