Waste less time on Facebook — follow Brilliant.
×

Euler's Theorem

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

Level 5

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

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?

×

Problem Loading...

Note Loading...

Set Loading...