Number Theory
# Euler's Theorem

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

Find the last three digits of the above number.

Let \( a \) and \( b \) be positive integers with \(a>b\) such that \[ 7!\Big|\big(x^a-x^b\big) \] 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 \(f(2016)+f(2015)\).

\[\Large 249^3 = 15438\underline{249}\]

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 for all \(m>0\).

Let us define an *almost automorphic number* as a number where \(n\) only appears as the trailing digits of \(n^m\) for **all odd \(m>0\)**. How many almost automorphic numbers are less than 1000?

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