Imagine you see a really huge party bus pulling out, an infinite bus with no seats. Instead everyone on board is identified by their unique name, which is an infinite sequence of \(0\)s and \(1\)s. The bus has every person named with every possible infinite sequence of \(0\)s and \(1\)s, someone is named \(00010000..00...\), someone else \(0101010101...\), and so on. Prove that this time you will not be able to accommodate all the new guests no matter how hard you try.
Prove the triangle inequality: in any triangle \(ABC\) the side \(AB < AC+ BC\).
In certain kingdom there are a lot of cities, it is known that all the distances between the cities are distinct. One morning one plane flew out of each city to the nearest city. Could it happen that in one city landed more than \(5\) planes?
Prove that the set of all finite subsets of natural numbers \(\mathbb{N}\) is countable. Then prove that the set of all subsets of natural numbers is not countable.
How many subsets are there of \(\{1,2,...,n\}\) (the integers from \(1\) to \(n\) inclusive) containing no consecutive
digits? That is, we do count \(\{1,3,6,8\}\) but do not count \(\{1,3,6,7\}\).
For example, when \(n=3\), we have
\(8\) subsets overall but only \(5\) contain no consecutive integers. The
\(8\) subsets are \(\varnothing\) (the empty set), \(\{1\}\), \(\{2\}\), \(\{3\}\), \(\{1,3\}\), \(\{1,2\}\), \(\{2,3\}\) and \(\{1,2,3\}\), but we exclude the final three
of these.
Remainder is the number that is “left over" from division. Even if a number is not divisible by another number fully, we can still divide, but leaving a remainder. The remainder is less than the number we’re dividing by. For example, a remainder of \(44\) in division by \(7\) is \(2\), because \(44 = 6 \times 7 + 2\). More generally, we can write \(n=qk+r\), where \(0\leq r<k\). We say that \(k\) goes into \(n\) \(q\) times, and a little bit (\(r\)) is left. If that little bit was larger than \(k\), it could “go into" \(n\) once more.
The general rule is that a remainder of a sum, difference or a product of two remainders is equal to the remainder of a sum, difference or a product of the original numbers. What that means is if we want to find a remainder of a product of two numbers, we need to look at the individual remainders, multiply them, and then take a remainder.
For example, \(10\) has remainder \(3\) when dividing by \(7\) and \(11\) has remainder \(4\) when dividing by \(7\). The product \(10\times11=110\) will have the same remainder as the product of the individual remainders. We first multiply \(3\times4=12\) and then take a remainder upon division by \(7\), which is \(5\) because \(12=7+5\). That means that \(110\) gives a remainder \(5\) in division by \(7\) - and it does, because \(110=15\times7+5\). If a number is divisible by a number we are dividing it, nothing remains and we say the remainder is \(0\).
Let’s have a look on some examples:
How many integers less than \(2025\) are divisible by \(18\) or \(21\), but not both?
Determine all prime numbers \(p\) such that \(p^2-6\) and \(p^2+6\) are both prime numbers.
Let \(ABCD\) be a square and let \(X\) be any point on side \(BC\) between \(B\) and \(C\). Let \(Y\) be the point on line \(CD\) such that \(BX=YD\) and \(D\) is between \(C\) and \(Y\). Prove that the midpoint of \(XY\) lies on diagonal \(BD\).
Let \(ABCD\) be a trapezium such that \(AB\) is parallel to \(CD\). Let \(E\) be the intersection of diagonals \(AC\) and \(BD\). Suppose that \(AB=BE\) and \(AC=DE\). Prove that the internal angle bisector of \(\angle BAC\) is perpendicular to \(AD\).