Did you solve it? The world’s most fascinating number – revealed! | Mathematics

Earlier today I set three puzzles, and also explained why 108 is possibly the most fascinating number in the universe.

Here are some more reasons:

108 = 6² + 6² + 6²

108 is the smallest number that can be written as the sum of a square and a cube in two different ways. (108 = 2³ + 10² = 3³ + 9²)

Many ratios in Moon-Sun-Earth astronomy seem to be around 108: the distance between the Earth and the Sun is about 108 times the diameter of the Sun; The distance between the Earth and the Moon is about 108 times the diameter of the Moon.

The upper frequency of FM radio is 108Mhz.

(Examples taken from the book Exploring the Beauty of Fascinating Numbers by Shyam Sunder Gupta. For more reasons click here.)

Here are the puzzles again with solutions.

1. Brilliant billions

You have ten cards. On each of the cards is one of the digits 0 to 9. When you arrange the cards in a line you get a number between 0123456789 and 9876543210.

i) How many of these numbers are divisible by 2?

ii) How many are divisible by 3?

Solution. i) half of them, ii) all of them!

The sum of the digits 0-9 is 45, which is divisible by three, hence all numbers made from these ten digits are divisible by three.

2. How low can you go?

What is the smallest even number between 1000 and 9999 written with four different digits?

Solution 1024

Most people will try to use the three lowest digits, i.e 1032. I hope you didn’t fall into that trap.

3. All about me

An autobiographical number is one where the first digit describes how many 0s it has, the second digit describes how many 1s it has, and so on, so that the (n + 1)th digit describes how many n’s it has. For example, 1210 is an autobiographical number because it has 1 zero, 2 ones, 1 two and 0 threes.

Find the only ten digit autobiographical number.

Solution 6210001000

Let the solution be ABCDEFGHIJ.

Each digit n + 1 describes how many times digit n appears. Since there are only ten possible positions for digits, we can deduce that A + B + C + D + E + F + G + H + I + J = 10.

Let’s proceed by trial and error. Image A = 9. Then J = 1, since there is a single 9 in the number. But that means A < 9, so we have a contradiction.

Let A = 8. Then I is 1, which means B = 1, which means A <8, so this doesnt work either.

Let A = 7. Then H = 1, so B must be either 1 or 2. (Since the digits must add up to ten.) If B = 1, then another digit must be 1, but this would mean B = 3, (since thee are three 1s) which is a contradiction. If B = 2, then C = 1 and we have another contridiction.

Following this logic, we finally hit a solution that works when A = 6.

I’ve been setting a puzzle here on alternate Mondays since 2015. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.

Sources of today’s puzzles: 1) Leon Gelkoff, 2) SmartFriends, a daily IQ challenge, smartfriends.app.link 3) An old classic.