# (10 + 98 – 7 × 6 – 5) × (4 × 3 + 21)

Yesterday’s post was inspired by Peter Norvig’s new year’s greeting, which was:

Happy 10/9!*8!*7!-6!*5!/4!+3!*2!+1!

I thought his greeting was wonderful, but it bothered me that it relied so heavily on factorials, which many people don’t understand. I wondered whether it would be possible to use only add, subtract and multiply, plus concatenation (eg: 1 and 2 can become 12), so that more people would be able to read the message. For variety, I decided to put the digits in forward order.

My strategy for finding a solution was to start with 2013 and work backwards, trying to whittle things down to zero. First I tried 2013 – 1234, but that gave 779, and I couldn’t see how to get from 779 down to zero using just the remaining digits 5,6,7,8,9.

So I then backtracked and tried 2013 – 123. That gave me 1890, and I knew I was in luck, since 1890 is a multiple of 90, so it must also be a multiple of 45. That gave (2013 – 123) / 45 = 42, which is just 7 × 6 — two more digits down.

At this point it was done, except for incorporating 8 and 9. Fortunately (8 – 9) is -1, which is easy to slip in, giving the final result: 0 + 123 – 45 × 6 × 7 × (8 – 9) = 2013.

Still, there was something very cool about the fact that Peter Norvig’s sequence was an actual countdown from 10 to 1 — just like the countdown to midnight at Times Square on New Years Eve. So this morning I figured I’d try such a countdown, again using only “easy to read” arithmetic.

Since 2013 = 61 × 33, this time I took a different approach:

61    10 + 56 – 5    10 + (98 – 42) – 5    10 + 98 – 7 × 6 – 5

33    12 + 21    4 × 3 + 21

And there you have it — the title of today’s post.

## 3 thoughts on “(10 + 98 – 7 × 6 – 5) × (4 × 3 + 21)”

1. Very interesting writeup of your search process, Ken; thanks! I wasn’t nearly as clever; I just stole my formula; I believe it was from Bill Gosper, but now I can’t find it …

2. Sharon Perl says:

I think there’s a bigger lesson in all this. We tend to find things more readily when we’re clear about what we want and we look for it. This wasn’t the type of result that you’d happen to observe by accident.

(Given that the set of digits and operators is finite, the number of numbers to be produced by this type of manipulation is clearly finite, but how many of them are there? I haven’t tried to do the counting. Ken?)