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 \(A\))
Find the last two digits of the number \[33333333333333333347^4 - 11111111111111111147^4\]
Replace all stars with ”+” or ”\(\times\)” signs so the equation holds: \[1*2*3*4*5*6=100\] Extra brackets may be added if necessary. Please write down the expression into the answer box.
In how many ways can one change \(\pounds 2\) into coins worth \(50\)p, \(20\)p and \(10\)p? One does not necessarily need to use all available coin types, i.e. having \(5\) coins of \(20\)p and \(10\) coins of \(10\)p is allowed.
Show that for any natural number \(n\geq 1\), the number \[n^4+4n^2+9\] is not a perfect square.
Convert the binary number \(10011\) into decimal, and convert the decimal number \(28\) into binary. Multiply by \(2\) as binary numbers both \(10011\) and the result of conversion of \(28\) into binary numbers.
The ternary numeral system has only \(3\) digits: \(0,\) \(1\) and \(2\). Therefore the number \(3\) is written in ternary as \(10\). Write down the numbers \(23\) and \(156\) in ternary and add them as ternary.
Write down the first fifteen numbers in binary system.