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Four Fours unblocked – after forty years January 1, 2008

Posted by Peter Hornby in Uncategorized.
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The New Year seems to have started pretty auspiciously for me. I was fooling around this morning with a little maths problem which has been bugging me every now and then for, well, essentially my whole life. Every couple of years, I’d take it out, give it some mostly aimless thought, get nowhere, and put it away again. Today I played around with it for about twenty minutes, and found the solution. It’s not quite Andrew Wiles proving Fermat’s Last Theorem, but I was totally astonished and delighted that I’d found what what I’d been looking for, after having been stuck since I was maybe 10 years old.

Allow me to explain. Don’t laugh too much.

When I was about eight years old, sometime around 1963-1964, my teacher, a wonderfully inspiring man called Dr Fred Everett, presented the class with a challenge, which he called “Four Fours”. He asked us to take each integer in turn, and represent it using the digit ‘4’, used four times. Each use of ‘4’ could be modified in various ways. You could raise it to a power, you could take its square root, and raise that to a power, you could use the factorial – and there were several other possibilities which we asked about at the time, and which were ruled legal or illegal. The idea was to see how far you could get.

Let’s try a few easy examples.

16 = 4 + 4 + 4 + 4
17 = \sqrt{4}^3 + 4 + 4 + 4^0 \hspace{20 mm}(8+4+4+1)
179 =  \sqrt{4}^7 + (\sqrt{4}^3 - 4^0)^2 + \sqrt{4}\hspace{10 mm}(128+49+2)

Clearly, the first few hundred are pretty easy, although a few caused us some thought. Once you get into the high hundreds, the values you can construct from the basic rules start becoming a little wider spaced, and you have to start getting inventive. We were stuck on 811 for some time.

811 =  (4^!)^2 + (4-4^0)^5 - \sqrt{4}^3\hspace{10 mm}(576+243-8)

With that out of the way things moved along fairly easily for a while. We asked for a ruling on

888 = 444 \times \sqrt{4}

and were told that it was acceptable, not that it would have been too hard to come up with another answer. So time went by, and after a few months only my friend William Irving and myself were pushing on. We arrived eventually at

1190 =  (\sqrt{4}^5+\sqrt{4})^2 + (\sqrt{4})^5 + \sqrt{4}\hspace{10 mm}(1156+32+2)

and then ran into a brick wall. Neither of us could solve 1191, and after a while, we both gave up. This would be around 1965-1966.

All of my results were written in a notebook which moved with me from place to place, which is a miracle in itself. Whenever I’d clear out a cupboard in preparation for a move, I’d run across the book. get it out, think about 1191 for a while, and put it away again. The book came out again yesterday as I was freeing up some space in one of our storage closets, and I started thinking about 1191 again, with no expectation of success. This morning, the book was still on my desk, so I sat down, made some notes, and, within twenty minutes, had come up with

1191 =  \sqrt{4}^{11} - (4-4^0)^6 - \sqrt{4}^7\hspace{10 mm}(2048-729-128)

I walked around for a while, not quite believing that the solution had turned out to be so simple. I even broke out a calculator to make sure that my arithmetic was correct.

So there we are. It’s New Year’s Day 2008, and I can start writing in the Four Fours notebook again, after a gap of over forty years. It’s a very strange feeling.

Happy New Year!

Update: Bugger. Now I’m stuck on 1194. Check in here again on my 90th birthday.

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Comments»

1. sam - February 4, 2008

that did not help me but thanks


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