# Puzzles

## Archive

Show me a random puzzle**Most recent collections**

#### Advent calendar 2018

#### Sunday Afternoon Maths LXVI

Cryptic crossnumber #2#### Sunday Afternoon Maths LXV

Cryptic crossnumber #1Breaking Chocolate

Square and cube endings

#### Sunday Afternoon Maths LXIV

Equal lengthsDigitless factor

Backwards fours

List of all puzzles

## Tags

time geometry 2d shapes 3d shapes numbers spheres trigonometry complex numbers algebra lines graphs coordinates odd numbers fractions differentiation calculus folding tube maps ellipses triangle numbers money bases triangles squares area square numbers chess probability circles averages speed sport multiples dates factors parabolas functions logic cards games people maths shape prime numbers irreducible numbers probabilty angles proportion dice integration sum to infinity dodecagons hexagons multiplication factorials coins shapes regular shapes colouring grids floors integers rugby crosswords percentages digits sums christmas square roots surds doubling quadratics indices symmetry arrows addition cube numbers star numbers rectangles chocolate cryptic clues cryptic crossnumbers crossnumbers wordplay clocks menace routes taxicab geometry remainders chalkdust crossnumber palindromes sequences means unit fractions division planes volume number partitions ave pascal's triangle mean advent perfect numbers polygons books perimeter## 2 December

Today's number is the area of the largest dodecagon that it's possible to fit inside a circle with area \(\displaystyle\frac{172\pi}3\).

## Cube multiples

Source: Radio 4's Puzzle for Today (set by Daniel Griller)

Six different (strictly) positive integers are written on the faces of a cube. The sum of the numbers on any two adjacent faces is a multiple of 6.

What is the smallest possible sum of the six numbers?

## Polygraph

Draw a regular polygon. Connect all its vertices to every other vertex. For example, if you picked a pentagon or a hexagon, the result would look as follows:

Colour the regions of your shape so that no two regions which share an edge are the same colour. (Regions which only meet at one point can be the same colour.)

What is the least number of colours which this can be done with?