Answer the following two questions and justify your answers: (a) What is the last digit of the sum $1^{2012}+2^{2012}+3^{2012}+4^{2012}+5^{2012}$? (b) What is the last digit of the sum $1^{2012}+2^{2012}+3^{2012}+4^{2012}+...+2011^{2012}+2012^{2012}$?

Find the last three digits in the product $1 \cdot 3\cdot 5\cdot 7 \cdot . . . \cdot 2009 \cdot 2011$.

Let $S$ be the set of all positive integers less than $10,000$ whose last four digits in base $2$ are the same as its last four digits in base $5$. What remainder do we get if we divide the sum of all elements of $S$ by $10000$?

Consider positive integers $m$ and $n$ for which $m> n$ and the number $22 220 038^m-22 220 038^n$ has are eight zeros at the end. Show that $n> 7$.

Let $n$ be a positive integer. Find as many as possible zeros as last digits the following expression: $1^n + 2^n + 3^n + 4^n$.

An infinite sequence $p_1, p_2, p_3, \ldots$ of natural numbers in the decimal system has the following property: For every $i \in \mathbb{N}$ the last digit of $p_{i+1}$ is different from $9$, and by omitting this digit one obtains number $p_i$. Prove that this sequence contains infinitely many composite numbers.

Prove that the square of any integer cannot end with two fives.

Does there exist a $6$-digit number $A$ such that none of its $500 000$ multiples $A$, $2A$, $3A$, ..., $500 000A$ ends in $6$ identical digits? (S Tokarev)

Let $N$ be a positive integer such that $N$ and $N^2$ both end in the same four digits $\overline{abcd}$, where $a \ne 0$. What is the four-digit number $\overline{abcd}$?

Find the last two digits of each of the numbers $3^{1974}$ and $7^{1974}$.

Let $a_1=1$, $a_2=2^{a_1}$, $a_3=3^{a_2}$, $a_4=4^{a_3}$, $\dots$, $a_9 = 9^{a_8}$. Find the last two digits of $a_9$.

Find the last four digits of $5^{5555}$.

Determine all sets $(a, b, c)$ of positive integers where one obtains $b^2$ by removing the last digit in $c^2$ and one obtains $a^2$ by removing the last digit in $b^2$. .

The sequences $a_n$ and $b_n$ members are the last digits of $[\sqrt{10}^n]$ and $[\sqrt{2}^n]$ respectively (here $[ ...]$ denotes the whole part of a number). Are those sequences periodical?

Juan writes the list of pairs $(n, 3^n)$, with $n=1, 2, 3,...$ on a chalkboard. As he writes the list, he underlines the pairs $(n, 3^n)$ when $n$ and $3^n$ have the same units digit. What is the $2013^{th}$ underlined pair?

There are many sets of two different positive integers $a$ and $b$, both less than $50$, such that $a^2$ and $b^2$ end in the same last two digits. For example, $35^2 = 1225$ and $45^2 = 2025$ both end in $25$. What are all possible values for the average of $a$ and $b$? For the purposes of this problem, single-digit squares are considered to have a leading zero, so for example we consider $2^2$ to end with the digits 04, not $4$.

(a) Is there a natural number $n$ such that the number $2^n$ has last digit $6$ and the sum of the other digits is $2$? b) Are there natural numbers $a$ and $m\ge 3$ such that the number $a^m$ has last digit $6$ and the sum of the other digits is 3?

The positive integer $a$ is relatively prime with $10$. Prove that for any positive integer $n$, there exists a power of $a$ whose last $n$ digits are $\underbrace{0...0}_\text{n-1}1$.

Determine the last digit of $5^5+6^6+7^7+8^8+9^9$.

The natural number $n$ was multiplied by $3$, resulting in the number $999^{1000}$. Find the unity digit of $n$.

If there is a natural number $n$ such that the number $n!$ has exactly $11$ zeros at the end? (With $n!$ is denoted the number $1\cdot 2\cdot 3 \cdot ... (n - )1 \cdot n$).

Denote by $a_n$ the last digit of the number $n^{(n^n)}$ (let $n\ne 0$ be a natural number ). Prove that the numbers $a_n$ form a periodic sequence and state this period!

Write using with the floor function: the last, the second last, and the first digit of the number $n$ written in the decimal system.

The cells of a $2021\times 2021$ table are filled with numbers using the following rule. The bottom left cell, which we label with coordinate $(1, 1)$, contains the number $0$. For every other cell $C$, we consider a route from $(1, 1)$ to $C$, where at each step we can only go one cell to the right or one cell up (not diagonally). If we take the number of steps in the route and add the numbers from the cells along the route, we obtain the number in cell $C$. For example, the cell with coordinate $(2, 1)$ contains $1 = 1 + 0$, the cell with coordinate $(3, 1)$ contains $3 = 2 + 0 + 1$, and the cell with coordinate $(3, 2)$ contains $7 = 3 + 0 + 1 + 3$. What is the last digit of the number in the cell $(2021, 2021)$?

Is there such a natural number that the product of all its natural divisors (including $1$ and the number itself) ends exactly in $2001$ zeros?