cardinal richelieu wrote:And is there a downloadable version I could practice on?
Not AFAIK.
Nil Satis wrote:Maybe the difficulty varies in different regions but I am certainly getting as close to clearing on Sum Up as I ever do on Word Up,
It is very easy to get "close" to clearing Sum Up, by just playing fast and binning operators, but the reality is that once you've got close (e.g. about 10 numbers left), the chances are in fact still heavily against you. I'd say in Word Up it is slightly harder to get to the endgame (i.e. about 10 decent letters left), but once you're there, it is much easier to clear (or to ascertain if it is possible to do so).
What is your record score? (Overall, and also individual sum?)
The key tip for me in terms of making a clearance possible is to always use as many operators as possible as you go along, rather than
Yes but you should do that anyway to get a bigger score. And you want a maximum of 2 operators when you have about 10 numbers left I think. Not including plus because it's redundant.
One other tip which may seem obvious is that, unlike in Word Up, you don't need to know how your sum ends when you start it. I often do the first half of a sum completely 'blind', e.g. 9 * 8 / 7 ... then see what is possible with the remainder.
This is the crucial strategy to get big sums fast. The counter on the top left tells you the total of what you've entered so far, so it helps you "finish off" the sum correctly. [The top-left counter in Word Up doesn't help you nearly so much, it only tells you the points your partial word is worth, not how to extend it!]
I recommend looking at the initial grid, finding an area with several * and / close together, and making the sort of string with 3 or 4 numbers you suggest, looking at the counter to see what the total is, then trying to use the surrounding numbers to finish it.
So really there is not much arithmetic going on at all, just educated guesswork. It helps to know your thirteen times-table of course.
My favourite sum is still 9x9x9/8/7... most points possible.
unknownpseudonym wrote:Ernest, do you think it is possible to clear in normal circumstances (i.e. without highly improbable order to the given tiles)?
Certainly. If your aim is to clear with a decent score you should play as you normally would (i.e. getting big sums with maximum length), but after a while you need to get to the end-game; start hacking and dealing with areas with too many operators closer together. Also with too many large/small numbers together. By the time you have about 12 numbers left you probably want to have at most 1 or 2 operators and 30 secs or so to try and figure out if it can be done: which involves adding up all the numbers to see what the residue is modulo 13 (i.e. if I subtract the preceding multiple of thirteen, what is the remainder). Then you want to use the last operator(s) you have left to make the remainder zero.
Here's a simple illustrative example:
004X
9433
6785
The numbers total 49, and the preceding multiple of 13 is 13*3 = 39, thus we are 10 over (residue of 10 modulo 13). So we need to create an extra 3, or 3+13, or 3+26 etc (or lose 10, or 10+13, or 10+26, etc) to make it a multiple of 13. This can only be done using the operators (and + wouldn't help).
If the only remaining operator is a multiply (as in this case), then I am looking for two numbers N and M next to the X which can be used to gain 3, i.e. we need N and M such that NxM = N+M+3 (or =N+M+16, or =N+M+29 etc, adding 13 each time). This can be written as:
(N-1)x(M-1)=4 (or 17 or 30 etc)
So possible answers would be N-1 = 1 and M-1 = 4 (i.e. N=2, M=5), or N-1 = 2, M-1 = 2 (i.e. N=3, M=3). I then need to hope that it is possible to make a 2x5 or a 3x3 (it is in this example grid). If not I need to consider (N-1)x(M-1)=30 (17 is no use to us since it is prime). So here we could have N=7, M=6, or N=11 M=4 but N and M must be 9 or lower. Etc etc.
Here it is straightforward since there are two threes next to the X, so it is possible. I can clear with these sums: 3x3+4, 9+4, 6+7, 8+5.
You can follow a similar method if you had a minus or a divide sign instead of X, and if you had more than one operator you could consider combinations. I do not recommend these tactics though as it is quite difficult to do all this in your head in the pub.
So in general it is hard to "make" a clearance, especially if there is a cloud of smoke and jukebox music in your face. It's easier to just wing it and hope for the best. But you can maximise your chance by having not many operators and a good shape near the end. If you play it enough times follow these basic principles to give yourself a better chance, you will get lucky eventually.
I almost never play for the clearance unless I happen to have used almost everything up in an attempt to get as big sums as possible. Then I might think about forcing a clearance, but it hasn't happened yet.