HDY make the numbers 1-20 using 4 four times? You can use () + - * / and ! (factorial).?
358773
1 : (44/44)\n2 : 4-(4+4)/4\n3 : (4+4+4)/4\n4 : ((4-4)*4)+4\n5 : ((4*4)+4)/4\n6 : (4+4)/4+4\n7 : (44/4)-4\n8 : 4+4+4-4\n9 : 4+4+(4/4)\n10: (44-4)/4\n11: ((4+4!)/4)+4\n12: (44+4)/4\n13: 4!-(44/4)\n14: (4!/4)+4+4\n15: (4*4)-(4/4)\n16: 4+4+4+4\n17: (4*4)+(4/4)\n18: (4!+4!+4!)/4\n19: 4!-4-(4/4)\n20: (4/4+4)*4\n\nEDIT : Added remaining numbers (11, 13, 14, 18, 19). Keep in mind that 4! = 4*3*2*1 = 24
632560
There is no simple trick for the addition of two numbers. You will have to calcuate their actual value and then add them. So for you example you would have to go to 4 + 8 = 12.
204914
I only need four 0.\nbinary:0!+0!+0!+0!=100
436598
Remember that a factorial is basically the product of that number, that number - 1, all the way down to 1.\n\n1. (n-1)(n-2)(n-3)(n-4)(n-5)...1\n---------------------------------\n(n-4)(n-5)...1\n\nCancel the (n-4) and everything after.\n\nANSWER : (n-1)(n-2)(n-3)\n\n2. (n+6)(n+5)(n+4)(n+3)...1\n---------------------------\n(n+3)(n+2)...1\n\nCancel the (n+3) and everything after\n\nANSWER : (n+6)(n+5)(n+4)\n\nEDIT : Lattisrectum, keep in mind that (n-4)! is (n-4)(n-5)(n-6)... all the way to 1, not (n-4)(n-3)... :)
1099830
27\n\n(+7 -4 +8 -3 +9 -2 +10)
200114
Philo's answers don't use each operator only once.  Simple fixes to his answers:\n4 = sqrt(4*4)^(4/4)\n7 = sqrt(4)*4-(4/4)\n\nHere's another way to do 2:\n2 = (4*4/4)-sqrt(4)\n\nTo make it even harder, I decided to try requiring them to all use a factorial (and also a power and a square root when possible):\n0 = (4-4)/[4!^sqrt(4)]\n1 = [4!*sqrt(4)]^(4-4)\n2 = sqrt(4)+(4-4)^4!\n3 = sqrt(4)+4!^(4-4)\n4 = 4+[(4-4)^sqrt(4)!]\n5 = 4+[sqrt(4)!^(4-4)]\n6 = (4!*4)/sqrt(4^4)\n6 (also) = 4!*sqrt(4)/(4+4)\n7 = (4!+4)/sqrt(4*4)\n7 (also) = 4*[sqrt(4)!]-4/4\n8 = [4!+4*sqrt(4)]/4\n8 (also) = 4^[4!/(4*4)]\n9 = 4*[sqrt(4)!]+4/4\n\nNote that a lot of the parentheses are implied to not exist if you were to write these out on paper, and therefore shouldn't count as precedence operators.
978426
for example:\n\n-1 < 2\nthen\n1 > -2\n\nanother:\n x < -5\nthen\n-x > 5\n\n-2 x > 9\nx< -9/2\n\ns
277190
1 + 49; 2 + 48; etc. Just match the next highest with the next lowest.
789559
Ten numbers per line, add 'em up and divide by 10.\n\nYou should get these means for the four sets:\n\n0.138, 0.172, 0.184, 0.156\n\nBe careful, since not all the numbers may show up on Yahoo unless you put your mouse pointer on the line.
840708
Remove the trailing zeroes, do the multiplication, and add the removed zeroes to the product.\nFor example,\n530*47000\n=53*47 (add 4 zeroes later)\n=24910000\n=24,910,000
95089
1(1)+0(2)+1(4)+1(8)+1(16)+0(32)+0(64)+0(128)+1(256)=\n1+0+4+8+16+0+0+0+256=\n285
622069
Above answer is ok except for not using the -- operator:\nNeed to start at 10 and to decrement i use the statement:\n\ni--;\n\nThis changes the problem somewhat.  Another problem in the first answer is that fact is not initialized, so it will always equal zero resulting in a division by zero error.
705901
Here is how you do it:\n(2,1) and (3,4) = [(2)(3) + (1)(4)]/(2)(1) - (3)(4)\n                        = (6 + 4)/(2-12)\n                        = 10/-10\n                        = -1\n\nTada
1062229
Technically there are only 4 answers that will work, and they all rely on parentheses to modify the natural precedence rules:\n\n(5+5/5)*5\n(5/5+5)*5\n5*(5+5/5)\n5*(5/5+5)\n\nAs you can see they are all variations on (5 + 1) * 5, where you use 5/5 = 1.  So there is really just one answer.\n\nYou can easily solve problems like this yourself using the link below...\n\nFor example, can you make 24 using 3, 3, 8, 8?  Try it before you cheat... also, note that the solver allows % (mod) as an operator, so you may have to filter out those results.
66293
can you repeat numbers?\nif not\nfor the first digit you get 9 choices\nfor the second one 8\nfor the third 7\nand the forth 6\n\nso the total number is 9*8*7*6= 3024 combination\n\nif you can repeat numbers 9^4=6561\n\n(if you could use the zero, and were allowed to repeat numbers the answer would be 10^4=10000;that numbers from 0001 to 9999 and 0000, a total of 10,000 numbers)
878454
"100 factorial". It means the product of all the whole numbers up to that number multiplied together. For example, 10! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = 3,628,800. Note the two zeros at the end. 100! is a huge number, way too big for my calculator. But I know it ends with an even longer string of zeros. How many zeros are there at the end of 100 factorial? How about 1000!? (Hint: 10 = 2 x 5)\n\nOne hundred factorial (100!) is a huge number of 158 digits, but I know it ends with 24 zeros. We add a zero to the tail of a number whenever we multiply by 10, which is 2 x 5. There are lots of twos as factors of 100!, but not as many fives, so the number of zeros will be the same as the number of fives. The factors 5, 10, 15, 20, ... each contribute a five, and 25, 50, 75, and 100 each add an extra. So the total is 100 / 5 + 100 / 25 = 24. Calculators are handy, but they're not always the best tool for thinking about numbers!\n\nhope that helps!!
237350
Do you mean permutations?\n8P4\n= 8!/(8-4)!\n=8!/4!\n= 8*7*6*5\n= 1680