Integer numbers \(a,b\) and \(c\) are such that the sum of digits of a number \(a+b\) is less than \(5\), the sum of digits of a number \(b+c\) is less than \(5\), the sum of digits of a number \(a+c\) is less than \(5\), but the sum of digits of a number \(a+b+c\) is greater than \(50\). Can you find such three numbers \(a,b\) and \(c\)?
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.
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!)