Answer this puzzle:\n\n98 33 79\n23 __ 55\n\n12 26 13\n11 __ 6\n\n125 21 20\n70 __ 35\n\nHINT:\nThe code for the puzzle is different for each section of the puzzle\n\nTHANK YOU!!!
202223
98 33 79\n23 __ 55\n\n9+7  = 16\n8+7  = 17\n16+17= 33\n\n2+5= 7\n3+5= 8\n7+8= 15\n\nso Answer for the first puzzle is 15\n\n********************\n12 26 13\n11 __ 6\n\n12 26 13 12+13=25+1= 26\n11 _  6  11+6= 17+1= 18\n\nso Answer for the second puzzle is 18\n\n********************\n125 21 20\n70 __ 35\n\n125/5 = 25 20/5 = 4 25-4= 21\n70/5 = 14 35/5 = 7  14-7= 7\n\nso Answer for the third puzzle is 7\n\nI hope these are the required answers.\n\nPls post the correct answers.
532858
something is wrong w/ your puzzle, because if you look in the fourth column, there are two 2s.\n\n* thanks for fixing it,  below are the number in the shape of the sudoku puzzle \n8 6 9 5 4 1 2 3 7\n2 3 1 6 7 8 4 5 9\n4 5 7 2 9 3 6 1 8\n6 1 4 8 3 2 9 7 5\n7 9 5 4 1 6 8 2 3\n3 2 8 9 5 7 1 4 6\n1 8 2 7 6 5 3 9 4\n5 4 6 3 2 9 7 8 1\n9 7 3 1 8 4 5 6 2
398103
the following is the total permutation of\nthe four numbers:-\n\n1234\n1243\n1324\n1342\n1423\n1432\n2134\n2143\n2314\n2341\n2413\n2431\n3124\n3142\n3214\n3241\n3412\n3421\n4123\n4132\n4231\n4213\n4321\n4312\n\ntotal amount of permutations=4!\n4*3*2*1=24\n\nnotice that 4 is grouped with 123 six times\nnotice that 3 is grouped with 124 six times\nnotice that 2 is grouped with 134 six times\nnotice that 1 is grouped with 234 six times\n\nthe six permutations of the three other\nnumbers =3!=3*2*1=6\n\nthere are two permutations of the other\ntwo numbers =2!=1*2\n\nhence,total for four numbers= 4*3*2*1\n=24\n\nthey are symmetric since each number\nhas to appear the same number of\ntimes in the permutation\n\ni hope that this helps
520135
No, but you can work it out on paper. I know because I did it once. You have five letters, so there should be 25 combinations, Of course, as two letters are the same that would make it 20. I think. Just give it a go. It's always more fun to work these tings out yourself.
52724
Hey asker, this is a great puzzle.  Could you provide the correct answer after a couple of days?  I am really curious.  Thanks.
428803
You have to look for the repeating pattern of units digits for each problem.\n\n1998^1 ends in 8\n1998^2 ends in 4  (8X8=64)\n1998^3 ends in 2\n1998^4 ends in 6\n1998^5 ends in 8 so it's starting to repeat.  It repeats every 4 so 1998^1999 is the same as 1998^3, ends in a 2.\n\nNow for 1999^1998.\n1999^1 ends in a 9\n1999^2 ends in a 1 9X9 = 81\n1999^3 ends in 9  It's already repeating\n\nSo 1999^1998 ends just like 1999^2 did, with a 1\n\nFinally subtract these two end digits since that's all you care about: 2 - 1 = 1\n\nSo it ends in a 1
187316
Please visit the reference link to know the rules of the game!\n\nA Brief Overview:\nEach row and column of the WHOLE BIG SQUARE(containing nine smaller squares, which in turn contain nine number placeholders each holding one number) should contain all the numbers from 1-9. The individual smaller squares should contain all the numbers from 1-9 too with no repetition whatsoever! Get it? Or follow the links!
75103
I will answer this entirely in 1337.\n\n7h3 7h1n6 15, 4nd m057 1np0r74n7 0f 4|| 7h1n65 7h47 y0u n33d 70 r34|1z3, 15 7h47 y0ur w1f3 |0v35 7h3 (0n(3p7 0f w0r|d d0m1n4710n 50 v3ry, v3ry mu(h. 5u853qu3n7|y, y0u 5h0u|d d0 3v3ry7h1n6 1n y0ur p0w3r 70 455157 h3r 1n 4(h13v1n6 h3r dr34m5 4|| 7h3 wh1|3 3n5ur1n6 7h47 n07h1n6 84d h4pp3n5 70 h3r; 7h47 15 y0ur du7y 45 h3r hu584nd.\n\n|&#92;/|057|y, y0ur r35p0n5181|17y 70 y0ur w1f3 15 70 pr073(7 h3r, (4r3 f0r h3r, 4nd u|71m473|y fu3| h3r r463-f1||3d p|4n5 4nd d3v1(35 70 7h3 p01n7 0f &#92;/&#92;/()RL|) |)()|&#92;/|1|&#92;|471()|&#92;|! Y()|_| 4R3 73|-| R()XX()RZ!!!!!!111!!!!
654838
They change the symbols to confuse you but the pattern emerges if you write it out.\n\n11 - 2 = 9\n12 - 3 = 9\n13 - 4 = 9\n14 - 5 = 9\n...\n21 - 3 = 18\n22 - 4 = 18\n23 - 5 = 18\n\nAny number selected within a given decade (like 11 to 20, 21 to 30) will resolve to the same number.  The designers place the same symbol into each decade in this position.
228889
For each of the 4 positions in the password, there are 26+26+10-3=59 different choices.  Since there are no other restrictions, the number of permutations is:\n\n59x59x59x59=12,117,361
397207
21,\n52,\n40,\n80,\n104\n\neach pattern, you are adding a number in the one before the number\n1 + 0 = 1, 1 + 1= 2, 2 + 1 = 3, 3 + 2 =5, 5 + 3 = 8, and so on and so forth\n\nsorry if my answer is wrong im just a kid trying to answer correctly your question....\n\nand im very sorry if my words are illiterate!!
293289
0123456789\n6210001000\n\nHere you can see there are 6 zeros, 2 ones, 1 two, and 1 six.\n\nI suppose you were looking for something more like this though.\n1234567890\n2100010006\n\nFor those of you with the 1000.0008 solution, where is the 1 8?  I suspect that solution is incorrect.
512971
s=9\ne=5\nn=6\nd=7\nm=1\no=0\nr=8\nn=6\ny=2\n\nSEND=9567\nMORE=1085\n--------\nMONEY=10652\n\nBut I've assumed that you can't use the same number to be represented by different letters.
224164
6670903752021072936960\n\nThey wrote  an article that can be found here:\nhttp://www.afjarvis.staff.shef.ac.uk/sudoku/sudoku.pdf\n\nThere is also a whole dicussion on wheather symetrical Soduko puzzles are actually the same puzzle.
378090
The answer is 119\n\nTake 2 at a time end up with 1 (59 times=118 one left)\nTake 3 at a time end up with 2 (39 times=117 two left)\nTake 4 at a time end up with 3 (29times =116 three left)\nTake 5 at a time end up with 4 (23times=115 four left) \nTake 6 at a time end up with 5 (19times=114 five left)\nTake 7 at a time end up with 0 (17times=119 o left)\n\nI am dead sure this time = ]\n\nThe trick is look at 5 at a time. there is 4 left this means the last digit must be 9. So look at all possible answer of 7s that ends with 9 and work it out from that!!!!!!!!!!
190549
Oh man, that was way harder than I thought. This is what I got:\n\nONE NINETY ONE ZERO NINETEEN EIGHT SIXTY TWO ZERO SIXTY ONE\n\nI have no idea what it means or was it even the one you were looking for, because there might be more "correct" solutions.
516885
C is at the end, so its just the other 8 positions to play with.\n\nCount how many permutations of those 8 there are.\n\nThen count how many there are where A and B stand side by side.\n\nThen subtract those out.\n\nEdit:  Poster below me is way off.  without any restrictions, 9 men could be arranged in over 360000 ways.   Not 81.\n\nBasically, you can take 'C' out of it.  There are 8 spots left, and 8 men.  Spot 1 can hold any of the 8.  spot 2 any of the remaining 7 and so on, giving 8X7X6X5X4X3X2 which is 8! possibilities.\n\nThat's 40320.\n\nOf those 40320, some involve A and B being side by side.  How many?  Well, they could be in slot 1&2, 2& 3 .... up to 7&8.  And they could be A-B or B-A.  That's 14 possibles.  For each of those 14, the other 6 spots could be filled 6! ways.\n\nSo there are 14 X 6! = 14 X 720 = 10800 combos with C in the 9 spot and A & B side by side.\n\nTherefor, of the 40320 with C in the 9 spot, 40320-10800 = 30240 possible arrangements have C at the end, but A and B NOT side by side.