Problems

List the first $10$ prime numbers and write the prime decomposition of $2910$.

Find three different natural numbers, larger than $100$ such that each of them is divisible by the difference of the other two numbers? The values of differences also have to be different from each other.

There are four numbers written in a row. The first number is $100$. It is known that if we divide the first number by the second number we will get a prime number as a result, if we the second number by the third number we will get a prime number, and if we divide the third number by the fourth number we will also get a prime number. Can all the resulting prime numbers be distinct?

We know that the product $c\times d$ is divisible by a prime $p$. Show that either $c$ or $d$ must be divisible by $p$.

The number $a$ has a prime factorization $2^3\times 3^2\times 7^2\times 11$. Is it divisible by $54$? Is it divisible by $154$?

Two numbers are given in terms of their prime factorizations: $a=2^3\times 3^2\times 5$ and $b=2\times 5^2\times 7^2$.

a) What is the greatest common divisor $\gcd (a,b)$ of these numbers?

b) What is their least common multiple $\mathrm{lcm}(a,b)$?

c) Write down the prime factorization of $\gcd (a,b)\times \mathrm{lcm}(a,b)$. Then write the prime factorization of $a\times b$. What do you notice?

A five-digit number is called indecomposable if it is not decomposed into the product of two three-digit numbers. What is the largest number of indecomposable five-digit numbers that can come in a row?

Let’s "prove" that the number $1$ is a multiple of $3$. We will use the symbol $\equiv$ to denote "congruent modulo $3$". Thus, what we need to prove is that $1\equiv 0$ modulo $3$. Let’s see: $1\equiv 4$ modulo $3$ means that $2^1\equiv 2^4$ modulo $3$, thus $2\equiv 16$ modulo $3$, however $16$ gives the remainder $1$ after division by $3$, thus we get $2\equiv 1$ modulo $3$, next $2-1\equiv 1-1$ modulo $3$, and thus $1\equiv 0$ modulo $3$. Which means that $1$ is divisible by $3$.

The numbers $a$ and $b$ are integers and the number $p\ge 3$ is prime. Suppose that $a+b$ and $a^2+b^2$ are divisible by $p$. Show that $a^2+b^2$ is divisible by $p^2$.

Find all possible non-zero digits $A$ for which the following holds $(AA+AA+1)\times A=AAA$. (Recall $AA$ means the two-digit number whose first and second digits are $A$)