Found problems: 52
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?
Find the last two digits of each of the numbers $3^{1974}$ and $7^{1974}$.
We take $100$ consecutive natural numbers $a_{1},$ $a_{2},$ $...,$ $a_{100}.$ Determine the last two digits of the number $a_{1}^{8}+a_{2}^{8}+...+a_{100}^{8}.$
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 $a_{2021}$.
Show that the number $11^{100}-1$ is both divisible by $6000$ and its last four decimal digits are $6000$.
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 $n> 7$.
Define $a_1 = 1979$ and $a_{n+1} = 9^{a_n}$ for $n = 1,2,3,...$. Determine the last two digits of $a_{1979}$.
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 $(2021, 2021)$?
Prove that the second last digit of each power of three is even .
(V . I . Plachkos)
Prove that none of the digits $2$, $4$, $7$, $9$ can be the last digit of a number
$$
1 + 2 + 3 + \ldots + n,$$
where $n$ is a natural number.
Prove that there exists a positive integer $n$ such that the last digits of $n^3$ are $...201320132013$.
The last three digits of $ N$ are $ x25$. For how many values of $ x$ can $ N$ be the square of an integer?
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?
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] $8$
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 $1\cdot 2\cdot 3 \cdot ... (n - )1 \cdot n$).
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.
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
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 $4$.
Let $n$ be a positive integer. Find as many as possible zeros as last digits the following expression: $1^n + 2^n + 3^n + 4^n$.
Determine the last digit of $5^5+6^6+7^7+8^8+9^9$.
(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?
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 $\overline{abcd}$?
Let $X$ be the sum of all divisors of the number $(3\cdot 2009)^{((2\cdot 2009)^{2009}-1)}$ . Find the last digit of $X$.
The natural number $n$ was multiplied by $3$, resulting in the number $999^{1000}$. Find the unity digit of $n$.
Prove that for any integer $k$ ($k \ge 2$) there exists a power of $2$ that among its last $k$ digits, the nines constitute no less than half. For example, for $k = 2$ and $k = 3$ we have the powers $2^{12} = ... 96$ and $2^{53} = ... 992$.
[hide=original wording]
Probar que para cualquier k entero existe una potencia de 2 que entre sus ultimos k dıgitos, los nueves constituyen no menos de la mitad. [/hide]