We take $100$ consecutive natural numbers $a_{1},$ $a_{2},$ $...,$ $a_{100}.$ Determine the last two digits of the number $a_{1}^{8}+a_{2}^{8}+...+a_{100}^{8}.$

Define alternate sum of a set of real numbers $A =\{a_1,a_2,...,a_k\}$ with $a_1 < a_2 <...< a_k$, the number $S(A) = a_k - a_{k-1} + a_{k-2} - ... + (-1)^{k-1}a_1$ (for example if $A = \{1,2,5, 7\}$ then $S(A) = 7 - 5 + 2 - 1$) Consider the alternate sums, of every subsets of $A = \{1, 2, 3, 4, 5, 6, 7, 8,9, 10\}$ and sum them. What is the last digit of the sum obtained?

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

Show that the number $11^{100}-1$ is both divisible by $6000$ and its last four decimal digits are $6000$.

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?

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 $1^{2012}+2^{2012}+3^{2012}+4^{2012}+...+2011^{2012}+2012^{2012}$?

How many zeros does $101^{100} - 1$ end with?

Define $a_1 = 1979$ and $a_{n+1} = 9^{a_n}$ for $n = 1,2,3,...$. Determine the last two digits of $a_{1979}$.

Let $X$ be the sum of all divisors of the number $(3\cdot 2009)^{((2\cdot 2009)^{2009}-1)}$ . Find the last digit of $X$.

Consider the sequence of numbers defined by $a_1 = 7$, $a_2 = 7^7$ , $ ...$ , $a_n = 7^{a_{n-1}}$ for $n \ge 2$. Determine the last digit of the decimal representation of $a_{2021}$.

What is the last digit of $4^{3^{2019}}$? [b]A.[/b] $0$ [b]B.[/b] $2$ [b]C.[/b] $4$ [b]D.[/b] $6$ [b]E.[/b] $8$

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!

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

Determine the last two digits of the number $2^5 + 2^{5^{2}} + 2^{5^{3}} +... + 2^{5^{1991}}$ , written in decimal notation.

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 $10000$?

In the sequence of digits $2,0,2,9,3,...$ any digit it equal to the last digit in the decimal representation of the sum of four previous digits. Do the four numbers $2,0,1,5$ in that order occur in the sequence? Folklore

We take $100$ consecutive natural numbers $a_{1},$ $a_{2},$ $...,$ $a_{100}.$ Determine the last two digits of the number $a_{1}^{8}+a_{2}^{8}+...+a_{100}^{8}.$

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?

Regarding a natural number $n$, it is stated that the number $n^2$ has $7$ as the second to last digit. What is the last digit of $n^2$?

Prove that there exists a positive integer $n$ such that the last digits of $n^3$ are $...201320132013$.

Call a positive integer $x$ $n$-[i]cube-invariant[/i] if the last $n$ digits of $x$ are equal to the last $n$ digits of $x^3$. For example, $1$ is $n$-cube invariant for any integer $n$. How many $2015$-cube-invariant numbers $x$ are there such that $x < 10^{2015}$?

How maay zeros are there at the end of the number $$1000! = 1 \cdot 2 \cdot 3 \cdot ... \cdot 999 \cdot 1000?$$

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 $\overline{abcd}$?

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 $4$.

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 $(2021, 2021)$?