In how many ways can this be done so that the result is less than 600?
917858
Only 1, 3, and 4 are allowable for the hundreds digit.  Anything goes for the rest.  So there are 3 choices for the hundreds digit, 5 choices for the tens digit since you can't use whatever you cose for the hundreds, and 4 choices left for the ones digit.\n\n3*5*4 = 60
967058
It depends on what you are wanting to prove. Generally speaking, when you substitute numbers, you are restricting to a specific example. You cannot prove a general result by only looking at specific examples (unless you somehow show those specific cases give *all* the possibilities). However, you *can* show a result is false, if it is false for some specific example, so to show a general result is false, it is enough to 'plug in numbers'.
472542
Take each one and put in over 1000 and then reduce.\nYou use 1000 because there are 3 decimal places.
422347
2\n\n60 minutes to a hour\n2x60 = 120
974447
minimum of 1 to maximum of 8
409160
This is possible only if at least half of those 256 values will not be used at all, and that you know which ones won't be used beforehand. If that's the case, you can map the remaining values that will be used with 7 bits.\n\nSee Shannon's entropy equation...
217638
simple answer, no.\ncomplex answer, yes.\n\nbut it is not simple, too many hoops, and you dont know if you are over-priting.\n\nbest to find another solution besides user tagging.
713027
Just try out small numbers. 1, 2, 3, 4, 6, 8 all work. Once you start doing anything higher, the second factor becomes smaller again, so the others are 96/8, 96/6, 96/4, 96/3, 96/2, 96/1.\nIn all: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96.
845925
If you problem, or part of it, is:\n\nHow many 8 digit numbers can be formed from this list of digits?\n8! / (3! * 2!)\nThe number combinations of digits has tobe reduced by 2! because of the 2 "6"s.\nThe number combinations of digits has tobe reduced by 3! because of the 3 "2"s\n.\nBut since the number cannot exceed 65,500; all larger numbers must be eliminated. Subtract the number of 8 digit numbers that can be produced from the list that start with "9". Likewise "7". Likewise "69", "67", "66", "659", ""657", and "656".\n\n\n\n(Try to state your problem clearly and completely.)
247429
there has been alot of work on showing that at least a certain percent of zeros must lie on the line. I think we're up to around 90%. It is unclear to me as to whether or not a proof can be found in this way. My guess is no
846187
if you are using a "while" or "do while" loop\n\nwhile ((x1 - x0) > 0.001)\n{\n//your code here\n}\n\nor you can do this\n\ndo\n{\n//your code here\n}\nwhile ((x1 - x0) > 0.001);
185940
We puzzled for years to find the result in the 1960's having been promised sincerely that it could be done.  I am convinced that per se it cannot.  However, if you have a small piece of paper, you can take the last line along the back of the page.
57233
4100\n\nbut most people while adding they calculate 90 as 900 and give the result as 5000!!!!
621145
With just ! (Factorial), +, -, * and /, it is impossible to get 150 using only 3 5s without the x^2 (Squared) operation (Which means x*x, in case you are in Fifth Grade), or using a '^2' as you say. If you say there is an alternate way, and there really is, then I give up and admit defeat. (I'm sure most others do.) If you are so sure, why don't you post the answer yourself?
861788
There is no accurate way to do this because too many variables are involved. You can do like I do though and run a series of tests and then average them.\n\nhttp://www.toast.net/
805759
Something less than 10 microns x 10 microns x 10 microns.\n\nor a list of "My Successes" by George W Bush.
779807
Yiu would first decide how many decimal places you are going to consider.  The usual is 3.  Then work it from there...;