Find all natural numbers $n$ with the following property: Given the decimal writing of $n$, adding a few digits one can obtain the decimal writing of $1996n$.

Let $a, b, c$ be two-digit, three-digit, and four-digit numbers, respectively. Assume that the sum of all digits of number $a+b$, and the sum of all digits of $b + c$ are all equal to $2$. The largest value of $a + b + c$ is (A): $1099$ (B): $2099$ (C): $1199$ (D): $2199$ (E): None of the above.

Find the smallest positive integer $x$ such that $x^2$ ends with the four digits $9009$.

It is possible to write the numbers $111$, $112$, $121$, $122$, $211$, $212$, $221$ and $222$ at the vertices of a cube, so that the numbers written in adjacent vertices match at most in one digit?

A natural number $N$ is [i]good [/i] if its digits are $1, 2$, or $3$ and all $2$-digit numbers are made up of digits located in consecutive positions of $N$ are distinct numbers. Is there a good number of $10$ digits? Of $11$ digits?

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 $(2021, 2021)$?

Find all natural numbers $ n$ for which every natural number whose decimal representation has $ n \minus{} 1$ digits $ 1$ and one digit $ 7$ is prime.

Does there exist a ten-digit number such that all its digits are different and after removing any six digits we get a composite four-digit number?

Natural number $k$ has $n$ digits in its decimal notation. It was rounded up to tens, then the obtained number was rounded up to hundreds, and so on $(n-1)$ times. Prove that the obtained number $m$ satisfies inequality $m < \frac{18k}{13}$. (Examples of rounding: $191\to190\to 200, 135\to140\to 100$.)

For each integer $n \ge 100$ we define $T(n)$ to be the number obtained from $n$ by moving the two leading digits to the end. For example, $T(12345) = 34512$ and $T(100) = 10$. Find all integers $n \ge 100$ for which $n + T(n) = 10n$.

The combination to open a safe is a five-digit number. different, randomly selected from $2$ to $9$. To open the box strong, you also need a key that is labeled with the number $410639104$, which is the sum of all combinations that do not open the box. What is the combination that opens the safe?

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?

Determine the number of ten-digit positive integers with the following properties: $\bullet$ Each of the digits $0, 1, 2, . . . , 8$ and $9$ is contained exactly once. $\bullet$ Each digit, except $9$, has a neighbouring digit that is larger than it. (Note. For example, in the number $1230$, the digits $1$ and $3$ are the neighbouring digits of $2$ while $2$ and $0$ are the neighbouring digits of $3$. The digits $1$ and $0$ have only one neighbouring digit.) [i](Karl Czakler)[/i]

Find the maximal integer $ M$ with nonzero last digit (in its decimal representation) such that after crossing out one of its digits (not the first one) we can get an integer that divides $M$. (A Galochkin)

For a positive integer $n$, we consider all its divisors (including $1$ and itself). Suppose that $p\%$ of these divisors have their unit digit equal to $3$. (For example $n = 117$, has six divisors, namely $1,3,9,13,39,117$. Two of these divisors namely $3$ and $13$, have unit digits equal to $3$. Hence for $n = 117$, $p =33.33...$). Find, when $n$ is any positive integer, the maximum possible value of $p$.

Suppose for some positive integer $n$, the numbers $2^n$ and $5^n$ have equal first digit. What are the possible values of this first digit? Note: solved [url=https://artofproblemsolving.com/community/c6h312638p1685546]here[/url]

With three different digits, all greater than $0$, six different three-digit numbers are formed. If we add these six numbers together the result is $4.218$. The sum of the three largest numbers minus the sum of the three smallest numbers equals $792$. Find the three digits.

Let $A$ be an even number but not divisible by $10$. The last two digits of $A^{20}$ are: (A): $46$, (B): $56$, (C): $66$, (D): $76$, (E): None of the above.

For every three-digit number, we take the product of its three digits. Then we add all of these products together. What is the result? (G Galperin)

How many numbers $\overline{abcd}$ with different digits satisfy the following property: if we replace the largest digit with the digit $1$ results in a multiple of $30$?

Find, with proof, all positive integers $n$ such that neither $n$ nor $n^2$ contain a $1$ when written in base $3$.

Let $k$ be a positive integer, and let $s(n)$ denote the sum of the digits of $n$. Show that among the positive integers with $k$ digits, there are as many numbers $n$ satisfying $s(n) < s(2n)$ as there are numbers $n$ satisfying $s(n) > s(2n)$.

Given a positive integer $A$ written in decimal expansion: $(a_{n},a_{n-1}, \ldots, a_{0})$ and let $f(A)$ denote $\sum^{n}_{k=0} 2^{n-k}\cdot a_k$. Define $A_1=f(A), A_2=f(A_1)$. Prove that: [b]I.[/b] There exists positive integer $k$ for which $A_{k+1}=A_k$. [b]II.[/b] Find such $A_k$ for $19^{86}.$

Two natural numbers have the same decimal digits in different order and have the sum $999\cdots 999$. Is this possible when each of the numbers consists of $2000$ digits?

When $4444^{4444}$ is written in decimal notation, the sum of its digits is $ A.$ Let $B$ be the sum of the digits of $A.$ Find the sum of the digits of $ B.$ ($A$ and $B$ are written in decimal notation.)