For many years, at the start of the year, I posted a large sheet on the noticeboard in my classroom containing a 10 x 10 grid listing the numbers from 1 to 100. I invited anyone using the room to try and complete any of the numbered squares with a calculation using all four digits of the current year, and which equated to the relevant number. The rules were simple: any mathematical operation, such as addition, multiplication and powers could be used, as could any mathematical function, and all four digits must be used every time. The results could be anything but simple, and I especially remember a classics teacher, who enjoyed mathematics, filling in some of the squares with fiendish (and probably unnecessarily complex) calculations whilst passing time taking a detention!

As the 1990s wore on, it became increasingly difficult to fill up every square with a calculation, and once we got to 2000 I moved on to other motivational games. Even now in 2017, it’s not easy. I had a go the other day, and found that the numbers up to 22 could be easily achieved using +, −, ×, ÷, powers and brackets. For example:

1 = 1 + 27 × 0

2 = 2 + 17 × 0

3 = 2 + 1 + 7 × 0 or, alternatively, (21 + 0) ÷ 7

And remembering that *a*^{0} = 1, we can calculate 11 = 12 – 7^{0}.

I’ve puzzled over 23 for a little while, and haven’t found a solution yet (let me know if you crack it), and I’m sure it will get harder as the numbers get bigger; this is because three of the numbers (0, 1 and 2) are small. So although some larger numbers are easy, such as 58 = 70 – 12, and 59 = 7^{2} + 10, I doubt there are solutions for all of them.

So my challenge for you is to try and complete the grid for all numbers 1 to 100 using an easier year: 1968, the year in which IB was founded. There are some nice little tricks, particularly using the 1: for example, remember that 1* ^{n}* = 1. This gives us useful triples such as:

6 × 8 – 1^{9} = 47

6 × 8 × 1^{9}^{ }= 48 (or, more exotically, 6 × 8 + log(1^{9}) = 48)

6 × 8 + 1^{9} = 49

And don’t forget square roots. 68 + 1 + √9 = 72.

The best strategy is not to start at 1 and work your way up, but to begin writing down lots of possible calculations and filling in the results on the grid. This way you will probably complete 30 or 40 random numbers very quickly, then you can start trying to fill in the gaps.

Do let me know if you manage to complete the grid, or if you get completely stuck on any of the numbers.

November 9, 2017

This is an old post at this point, but Numberphile posted a video related to this idea: https://www.youtube.com/watch?v=Noo4lN-vSvw&ab_channel=Numberphile (“The Four 4s”)

This suggests that if you keep extend your set of allowed functions, you will eventually arrive at a set of operators which can calculate any number.