Problems

The number $b^2$ is divisible by $8$. Show that it must be divisible by $16$.

Find a number which:

a) It is divisible by $4$ and by $6$, is has a total of 3 prime factors, which may be repeated.

b) It is divisible by $6,9$ and $4$, but not divisible by $27$. It has $4$ prime factors in total, which may be repeated.

c) It is divisible by $5$ and has exactly $3$ positive divisors.

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 $154$?

a) The number $a$ is even. Should $3a$ definitely also be even?

b) The number $5c$ is divisible by $3$. Is it true that $c$ is definitely divisible by $3$?

c) The product $a\times b$ is divisible by $7$. Is it true that one of these numbers is divisible by $7$?

d) The product $c\times d$ is divisible by $26$. Is it true that one of these numbers is divisible by $26$?

a) The number $a^2$ is divisible by $11$. Is $a^2$ necessarily also divisible by $121$?

b) The number $b^2$ is divisible by $12$. Is $b^2$ necessarily also divisible by $144$?

What is the smallest integer $n$ such that $n\times (n-1)\times (n-2)...\times 2$ is divisible by $990$?

Jack believes that he can place $99$ integers in a circle such that for each pair of neighbours the ratio between the larger and smaller number is a prime. Can he be right?

a) Prove that a number is divisible by $8$ if and only if the number formed by its laast three digits is divisible by $8$.

b) Can you find an analogous rule for $16$? What about $32$?

Look at this formula found by Euler: $n^2+n+41$. It has a remarkable property: for every integer number from $1$ to $21$ it always produces prime numbers. For example, for $n=3$ it is $53$, a prime. For $n=20$ it is $461$, also a prime, and for $n=21$ it is $503$, prime as well. Could it be that this formula produces a prime number for any natural $n$?

Denote by $n!$ (called $n$-factorial) the following product $n!=1\cdot 2\cdot 3\cdot 4\cdot ...\cdot n$. Show that if $n!+1$ is divisible by $n+1$, then $n+1$ must be prime. (It is also true that if $n+1$ is prime, then $n!+1$ is divisible by $n+1$, but you don’t need to show that!)