Puzzles

The vertices of any regular polygon lie on a circle, and so the three vertices we pick must lie on this circle. The triangle will be right-angled if one of the edges of the triangle is a diameter of the circle.

If the regular polygon has \(n=2k\) sides, then there are \(k\) ways to pick two points that make a diameter, and \(2k-2\) ways to pick the third point to make the triangle, and so there are a total of \(k(2k-2)=n(n-2)/2\) right-angled triangles. This means that we need to solve \(n(n-2)/2=14620\). The solution is \(n=172\), and so our polygon has 172 sides.

We can look at how many \(n\)-digit number there are for small values of \(n\) and look for a pattern:

These are the triangle numbers, and there are 231 20-digit numbers.

Why is the pattern the triangle numbers?

One method for putting the five balls into the four boxes is: pick one of the balls (there are 5 ways to do this), then put the other four balls into boxes with one in each box (there are 4! ways to do this), then put the picked ball into a box (there are 4 ways to do this). This method gives exactly two ways of obtaining each possible arrangement, so the number of ways is \(5\times4!\times4\div2\) or 240.

The product of the numbers in the red boxes is 504.

110 is 10×11.

\((a+b+c+d+e+f)^5\) has 252 terms.

Every number except the powers of 2 can be written as the sum of two or more consecutive positive integers. The smallest three-digit power of 2 is 128.

Why can every number that isn't a power of 2 be written as the sum of two or more consecutive positive integers?

The only three digit number whose first digits is 4 and for which the arithmetic and geometric means of its digits are both non-zero integers is 444.

How many three digit numbers are there for which the arithmetic and geometric means of its digits are both non-zero integers? How many four digit numbers are there for which the arithmetic and geometric means of its digits are both non-zero integers?