Puzzles

Today's number is the largest possible remainder which can be obtained when dividing one of the answers in this advent calendar by another answer smaller than it (not including today's answer!).

This number is a prime number. If you treble it and add 16, the result is also prime. Repeating this will give 11 prime numbers in total (including the number itself).

What is the largest number which cannot be written as the sum of distinct squares?

This year, I posted instructions for making a dodecahedron and a stellated rhombicuboctahedron.

To get today's number, multiply the number of modules needed to make a dodecahedron by half the number of tube maps used to make a stellated rhombicuboctahedron.

Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums reading across and down are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, 4+3×2 is 14, not 10.

The answer is the product of the digits in the red boxes.

1 = 1/2 + 1/4 + 1/8 + 1/16 + 1/16. This is the sum of 5 unit fractions (the numerators are 1).

In how many different ways can 1 be written as the sum of 5 unit fractions? (the same fractions in a different order are considered the same sum.)

Put the digits 1 to 9 (using each digit once) in the boxes so that the three digit numbers formed (reading left to right and top to bottom) have the desired properties written by their rows and columns.

Today's number is the multiple of 6 formed in the left hand column.

In March, I posted the puzzle One Hundred Factorial, which asked how many zeros 100! ends with.

What is the smallest number, n, such that n! ends with 50 zeros?