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 \(AD\).
Let \(ABC\) be an isosceles triangle with \(AB=AC\). Point \(D\) lies on side \(AC\) such that \(BD\) is the angle bisector of \(\angle ABC\). Point \(E\) lies on side \(BC\) between \(B\) and \(C\) such that \(BE=CD\). Prove that \(DE\) is parallel to \(AB\).
Is it possible to place a positive integer in every cell of a \(10\times10\) array in such a way that both the following conditions are satisfied?
Each number (not in the bottom row) is a proper divisor of the number immediately below.
The numbers in each row, rearrange if necessary, form a sequence of 10 consecutive numbers.
A round-robin tournament is one where each team plays every other
team exactly once. Five teams take part in such a tournament getting:
\(3\) points for a win, \(1\) point for a draw and \(0\) points for a loss. At the end of the
tournament the teams are ranked from first to last according to the
number of points.
Is it possible that at the end of the tournament, each team has a
different number of points, and each team except for the team ranked
last has exactly two more points than the next-ranked team?
\(ABCD\) is a rectangle with side lengths \(AB=CD=1\) and \(BC=DA=2\). Let \(M\) be the midpoint of \(AD\). Point \(P\) lies on the opposite side of line \(MB\) to \(A\), such that triangle \(MBP\) is equilateral. Find the value of \(\angle PCB\).