Each Number Has a Rule.\nFigure out the rule to get the next number\n\nJanuary  - 20\nAugust   - 10\nMarch    - 5\nNovember - 15\nJuly     - ?\n\nI have been working on this all day, I think the months have something to do with the answer, please help so I can put my mind at ease.\n\nThanks
120568
10.\nYou want to figure out the answer yourself, so I'll give you some other months as a clue:\n\nFebruary - 20\nApril - 10\nMay - 5\nJune - 5
604774
The explanation is really simple, really!\n\nOkay, take a look at the mystic ball chart right now. Take a look at the numbers 0, 9, 18, 27, 36, 45, 54, 63, 72, and 81. They are all the same symbol, right? And they are all multiples of 9! Now, before doing anything, just click on the mystic ball. You'll notice that the symbol that pops up is the same symbol that the numbers mentioned above show. Click on the blue link to try it all again. You should get another sheet with different symbols. Now, look at the very same numbers that you looked at before. They all share the same symbol, but now it is a different one, right?\n\nThis is the trick. No matter what number you choose from 0 to 99, the final answer will always be a multiple of 9. Try it out yourself! Go through each number and figure out what the final number of each number would be. Figure out what the final symbol would be, too. You will see that no matter what number you choose, the same symbol will keep popping up.\n\nIf you want the equation, it would be 10a + b - (a+b). 'a' represents the first digit of the number you chose, and 'b' represents the second digit of the number you chose. '10a + b' represents the number as a whole (for example, let's choose the number 32. a = 3, b = 2. 10(3) + 2 = 32. Make sense?). the '- (a+b)' part represents the time when you add the digits together and subtract it from the original number to get your final answer. After reducing the equation, you get 9a, which is why the answers all come out to be a multiple of 9.\n\nHopefully this makes sense!
542292
OMG a math riddle I cant seem to answer!!!! :-)  Awesome. I cant sleep because now i want to knwo the anwer. And so i continue to try and solve it.\n\nBest i can come up with is 113  or 240 or 50-something..\nAlso maybe another number in the 100-n-teens\n\nI did find a 1-3-4 pattern aasuming the next number in 1-3-4-1 was a 3 but was unsure of how to continue it. (its1230 in the moring here)\n\nAnd again another possible patter 1-3-4-8 next number would be 16.\n\nI know some may not even know how I came up with these, but I love math and I love a challenge. I have been trying to find any possible correlation in the numbers that I can. \n\nHope someone figures it out. I know I cant wait to see the answer.
148610
Ok, I got it.  I recommend writing the numbers 1-12 and crossing them out as you use them.\n\nYou've got your 5,6,1,and 8 on the inside.  So on the upper row of four, 5+6=11, so you need 15 more to make 26.  The only numbers that are left where 2 add up to 15 are 12&3 or 11&4.  I tried 11&4 first and will just tell you it didn't work, so it's 12&3.  Now to decide which goes where.  If you put the 3 on the left, it lines up with the 1 making 4, and there's no other numbers that will add up to the required 22 - so 12 must go on the left and 3 on the right.\n\nNow you can see the line  from the bottom point up to the 3.  8+3=11, so you need 15 again, and the only choices left are the 11&4.  Put the 11 at the bottom point and the 4 between the 8&3.\n\nNow looking at the line that goes from 12 down to the 11 you have 12+1+11=24, so you must use the 2 here.  Now you have 7,9 & 10 left.\n\nLooking at the line with the 5&2, that makes 7, so you need 19.  The only way to get that is with the 10 & 9, but you don't know which goes where.  You DO know, though, that the only other point that is left gets the 7.  That's the bottom right point.\n\nThis made a line with 7,6,4 and the very top point.  7+6+4=17, so you need 9 at the top to make 26, which leaves the space for the 10 at the bottom left.\n\nIn order, from top to bottom/left to right you have\n\n9\n12,5,6,3\n2,4\n10,1,8,7\n11
281955
1) The sum of the positive 3-digit odd integers = 101 + 103 + 105 + ... + 995 + 997 + 999.  Adding the first of these to the last, we get 101 + 999 = 1100.  Adding the second of these to the next to last, we get 103 + 997 = 1100.  This pattern keeps repeating until 549 + 551 = 1100.  The question then becomes, how many of these 1100's do we have?  We have one for 101, 103, 105, ..., 549 (the later numbers share their 1100's with these).  How many numbers are in the set 101, 103, 105, ..., 549?  Well, there are 449 numbers in 101, 102, 103, ..., 549.  Since there is one more odd number than even number, we must have 225 odd numbers and 224 even numbers.  Thus, there are 225 odd numbers in the set 101, 103, 105, ..., 549.  Thus, we have 225 1100's to add together.\n\n225 * 1100 = 247500.\n\n\n\n2) This time we wish to add the numbers of the set 10, 11, 13, 14, ..., 94, 95, 97, 98.  Adding the first to the last we get 10 + 98 = 108.  Adding the second to the second to last we get 11 + 97 = 108.  The pattern keeps repeating until we get 53 + 55 = 108.  How many 108's do we have?  Well, we have one for 10, 11, 13, 14, 16, 17, ..., 52, 53 (again, the later numbers share their 108's with these).  How many numbers are in this set?  In the set 10, 11, 12, 13, ..., 52, 53, there are 44 numbers.  The multiples of three range from 12 to 51--from 3*4 to 3*17.  Thus, there are 14 multiples of three in this range.  This means that there are 30 numbers in the set 10, 11, 13, 14, 16, 17, ..., 52, 53.  Thus, we have 30 108's to add together.\n\n30 * 108 = 3240.\n\n\n\n3) Here, I am assuming that the raise of 10% also applies to the previous year's raise and not just to the "principal" of $18000.  Thus, I am assuming a compound raise instead of 5 simple raises.\n\nDuring the first year, you make $18000.  During the second year, you make 18000 * 1.1 = $19800.  During the third year, you make 19800 * 1.1 = $21780.  During the fourth year, you make 21780 * 1.1 = $23958.  Finally, during the fifth year, you make 23958 * 1.1 = $26353.80.\n\nGood luck.
540057
Start by picking the units digit, you have 4 choices {1, 3, 5, 7} since 9 is not a choice.\n\nNext pick the ten-thousands digit (one of the digits is removed already, but you can't have zero or nine as the first digit either.)  So there are 7 choices for the ten-thousands digit.\n\nFor the thousands digit you have 7 choices (take away the two choices from before, and ignore the 9, but now you can use zero).\n\nContinuing down, you have 6 choices for the hundreds digit and 5 choices for the tens digit.\n\nThat comes out to 7 x 7 x 6 x 5 x 4 ways or 5,880 ways for an odd five-digit number with distinct digits (no nine).\n\nFor the decreasing digits, start with:\n87654321 (it can't include 9 nor can it end in 0).\n\nYou need to remove 3 digits leaving the last digit as a 1 or 3 (to be odd)\n\nIn the first case (ending in 3), you will have:\n876543 and one more digit to remove (5 choose 1, or 5 choices)\nThey are:\n87653, 87643, 87543, 86543, 76543\n\nIn the second case (ending in 1), you start with:\n87654321\n\nYou must leave the 1, so you have 7 digits choose 3, or 35 ways to remove 3 digits.\n87651, 87641, 87631, 87621, 87541,\n87531, 87521, 87431, 87421, 87321,\n86541, 86531, 86521, 86431, 86421,\n86321, 85431, 85421, 85321, 84321,\n76541, 76531, 76521, 76431, 76421,\n76321, 75431, 75421, 75321, 74321,\n65431, 65421, 65321, 64321, 54321\n\nAdding it up there are only 40 ways to create a 5 digit odd number, with distinct digits in descending order (no nine).
452461
Try this...pick a number (say 7)\nsubtract 1 (6)\nsquare (36)\nsubtract 1 (35)\ndivide by original number (5)\nadd 2 (7!)\n\nSteve ;-)
156601
2, 9, 16, 23, 30\n\nU(n+1) = Un + 7\n\nMoving through weeks on a calendar.\n\n\n\n50, 100, 200, 400\n\nU(n+1) = 2Un\n\nDoubling ingredients in a recipe.
421596
This sounds like 4th grade everyday math (which I teach, so I know this stuff).\n\nNames for 17 - there are an infinite number of them, but here are some:\n\n2 + 15\n20 - 3\n(2 * 8) + 1\netc - basically any number sentence that equals 17\n\n\nBroken calculator problem - just make a number sentence, but don't use a 2\n\n22 = 11 + 11\n22 = 19+1+1+1\n22 = 30-8
514130
1275 may be the correct answer but not for that reason. i don't agree with the logic in finding the pattern. i have been trying to find the answer for the last 20 mins and will keep trying to find the answer but i want to know two things:\n\ndoes ab represent just the unknown number or is it a two digit number\n\nand are the numbers correct? \n\none more thing, how was this question written i.e. horizontally, vertically?\n\ncheers\n\nthe differences between each number starting from the bottom is \n4, 8, 200, 330 and the differences between those numbers are 4, 192, 132. so i can't get a guesstimate for you other than it should be around 1000 and probably ends in five.
65170
The pattern works this way: \nSubtract 6, add 11, divide by 5, add 5\n20-6=14. 14+11=25. 25 divided by 5=5.\nHere’s what you need: 5+5=10.\n10-6=4.  4+11=15. 15 divided by 5=3. 3+5=8.\n\nFill in the blank with 10.
79346
this one is pretty easy\nafter 46 i will give u the next no.\n\n54,55,62,63,69,70,75,76\n\nc it goes like that 80,81,84,85,....
296485
7 will be after five\nand eight will be after six.\n\nReason: The upper numbers in 'numerals' contains straight lines, the lower one contains curves.\nWell...
674426
Okay I got the answer 3121 from Scott, but I had a mathematical strategy.\n\nI narrowed down the possibilities to all multiples of 5 times a number with numbers digit of 4 or 9 plus 1.\n\nThis is because to keep the chain of divide by 5 subtract 1 remainder going, we would start with a number x ending in either 4 or 6. We subtract x/5 and 1, so we are basically subtracting the whatever factor times 5 yields the closest muliple. The result from x-(x/5)-1 has to end in either 1 or 6, so that we have a multiple of 5 with remainder 1 and can keep the sequence going. Now if it ends in 1, the factor was even. Ends in 6, factor was odd. If factor was even, we need to subtract the factor plus one, an odd number, to get another number ending in either 1 or 6. If we subtract a number ending in 5, we get a number ending in 6. Therefore, if the factor is even it has to end in 4. And for numbers ending in 6, we have to subtract an even number (odd factor + 1)subtract a number ending in 0, so only factors ending in 9 work.\n\nI was then testing every factor ending in 4 or 9. Take the factor, multiply by 5, add one, subtract (factor + 1), see if the new number minus one divided by 5 yields a factor ending in either 4 or 9.\n\nI was up to 149 when I decided to refresh the page and saw the answer.  Jeez I would've had to go up to 624. -.-;;\n\nAs it is, I only got a chain of 3 which didn't end in a multiple of 5.\n\nSo if I had continued testing throughout the night I would've gotten it. :)
541278
Take each number, multiply by two, and then subract two.\n\n3*2 - 2 = 4\n4*2 - 2 = 6\n6*2 - 2 = 10\n10*2 - 2 = 18\n\nedit: I'm sure that the two possible answers are related.  I would also get 34, then 66, then 130 as the next numbers in the sequence.  It has to do with the fact that it started with three.  If we had started at 2, then my system would always yield 2 every time, but you guys wouldn't.  I think both are acceptable, although I will admit that the other method is more interesting.
654356
The first one is (-5, 3). I don't remember how to do the others.
514903
Hello,\n\nThe symbols in relationship to the numbers change each time. \nThe symbol you choose is the correct one  for all answers. No matter the number you choose there is one correct symbol for all answers.