Number Theory

Euler's Theorem

Euler's Theorem: Level 5 Challenges

         

987654321 \Large 9^{8^{7^{6^{5^{4^{3^{2^{1}}}}}}}}

Find the last three digits of the above number.

Let a a and b b be positive integers with a>ba>b such that 7!(xaxb) 7!\Big|\big(x^a-x^b\big) for all integers x. x.

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

Clarification: !! denotes the factorial notation. For example, 8!=1×2×3××88! = 1\times2\times3\times\cdots\times8 .

2,22,222,2222,\large 2,2^2,2^{2^2},2^{2^{2^2}},\ldots The sequence above can be recursively defined by a1=2,ak+1=2aka_1=2, a_{k+1}=2^{a_{k}}

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

For a given nn, it is known that the terms of {bk}\{b_{k}\} eventually become constant. Let f(n)f(n) denote the index at which the constant values begin, i.e. f(n)f(n) is the smallest number ii for which bi1bi=bi+1=bi+2= b_{i-1} \neq b_i = b_{i+1} = b_{i+2} = \cdots .

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


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

How many integers 1a20151\leq{a}\leq{2015} are there such that aaaa^{a^a} and aaa^a end with the same digit?

2493=15438249\Large 249^3 = 15438\underline{249}

An automorphic number is defined as a positive integer nn such that the trailing digits of nmn^m, where mm is a positive integer, is nn itself for all m>0m>0.

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

×

Problem Loading...

Note Loading...

Set Loading...