First puzzle:\n\n8549176320 \n"Gee, Mom, this homework puzzle is too tough!"\n"No isn't, Timmy. All you have to do is to discover the order they're written in!"\n\nSecond puzzle:\n\n99+9=9\nAt the turn of the century, Professor otto Doppelganger was one of Europe's leading puzzle experts.  Here we see him dictating his memoirs into an early Edison recording machine.  Let's listen in:\n"And theen, in the summer of 1899, invented my most famous puzzle, the Nifty Nines Impossibility.  I proved that ninety-nine plus nine is equal to nine if you add a certain five-letter word to the equation.  That word was ****"\nCan you figure out what that nifty five-letter word was?
204697
dozen... 99+9 = 108... 9 dozen = 9 * 12 = 108
143403
The correct answer is "NO"... (That's why kindergarten students can answer it.)\n\nThe last line of the riddle should read, "Can you guess the correct answer?"
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.
408290
Firstly, it has to start with either a 1 or a 2, and end with a 2 or a 6. So we just take each case one at a time.\n1..2: that leaves 23556, so there are 5!/2! ways, ie 60 ways.\n1..6: that leaves 22355, so there are 5!/(2!2!) ways, ie 30 ways.\n2..2: that leaves 13556, so there are 5!/2! ways, ie 60 ways.\n2..6: that leaves 12355, so there are 5!/2! ways, ie 60 ways.\n\nNote that in each case there are 5! ways of arranging the other 5 numbers, but we need to divide by 2 if there is a pair of repeated numbers.\nThat gives a total of 210.\n\n(note that both of the other answers given so far are wrong - its not 360, because that person has forgotten they have to be even; and its not 120. They counted 2x2 possibilities for the first and last digits which is correct, but one of those cases already uses both 2s, so you *don't* want to divide by 2 in the end for that particular case (you're not counting them twice are you!)\nTheres really no other way of doing it than splitting them into cases like I did.
261466
I saw this before..a long time ago. The answer is 2, 2 and 9.\n\n1) Break down 36 into prime numbers - 36 = 2*2*3*3*1\n\n2) Now, we must combine these into ages. Three of them. Any of the ages can be 1, but the others must still come to 36 when multiplied. We also know that when added up, they must be equal to the bus number.\n\nPOSSIBLE COMBINATIONS (Product of ages, Sum of ages)\n\n1, 1, 36  = 38\n1, 2, 18 = 21\n1, 3, 12 = 16\n1, 4, 9 = 14\n1, 6, 6 = 13\n2, 2, 9 = 13\n2, 3, 6 = 11\n3, 3, 4 = 10\n\n3) The number that they add up to is a number the friend knows - the bus number.YET he still says asks for more data! That means that there must be more than one occurrence of the sum in the addition answers. Otherwise, he would already know their ages. Therefore, the only sum that appears more than once is 13, with the sets {1, 6, 6} and {2, 2, 9}. The answer must be one of these.\n\n4) Your clue said "The older one has blue eyes". This means that must be ONLY ONE older child. Therefore, leading us to the answer of the kids' ages to be 2, 2 and 9.
185480
Yes.  Sounds like a crazy, crazy puzzle that even Blaine himself never imagined someone would solve this quickly...\n\nFor the answer:\n\nhttp://thefoolsparadise.com/db/solution.htm\n\nAnd the story of the winner:\n\nhttp://thefoolsparadise.com/db/winners-tale.htm
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.
361031
Oh Dear! lots of dodgy maths out there!\nWelcome to Mr Cantor's infinity paradise hotel!\nWe got infinite rooms, but they're all full just now.\nDon't worry!.. we'll squeeze you in!..I'll just get every guest to move one room along\nNow you can get in room1\nHey there's an infinite stretch limo arrived with an infinite number of guests..no probs just get all our guests to move from room n to 2n!\nYes infinity is an uncountable number, but you can still do mathematical ops with it\nHope that helps, you can work out the rest foor yourselves. PS always agree with your wife- she's right even when she's wrong-now that's a true paradox
292904
Nice problem. Here's the solution: 1000! is a multiple of 9 which is the same thing with saying that: S(1) = the sum of its digits is a multiple of 9 (Using the following known fact: a number is divisible by 9 if and only if the sum of its digits is divisible by 9). \n   Clearly S(1) < 1000! and 9 | S(1). Summing the digits of S(1) one obtains a number S(2) which is strictly smaller than S(1) and is a multiple of 9. And so on, until finally one arrives at a one digit number, say S(r) for some r  which is divisible by 9. But this can only be 0 or 9. Since S(r) is the sum of the digits of the previous number obtained by this algorithm, namely S(r-1), it can not be 0. So it must be 9! \n   Note that this type of proof works for any number divisible by 9, not only factorials of numbers, and in fact, with exactly the arguments displayed above it can be prooved that the single digit after adding the digits of an arbitrarly chosen number N is the digit d > 0 such that N-d is a number divisible by 9. (If you don't belive it, simply work out a computer programme that verifies this for all the numbers smaller than 1 000 000 for example).
453447
This is the third time that this question has been asked!?!?!\n\nIt's a tough one, too.\n\nI'm taking it to work with me and asking my colleges to help me out on it.\n\nBut someone "did" crack this CHALLENGE Question.\n\nSo here's the answer. . .\n\nhttp://answers.yahoo.com/question/index;_ylt=AvnsMpyNhtwrHV1JVMvXrE4jzKIX?qid=20061130045858AAB57A7\n\nANSWER:  ABCDEFGHJ = 25,693,148\n\nA=2\nB=5 \nC=6\nD=9\nE=7\nF=3\nG=1\nH=4\nJ=8
489918
4 and 1.\n\nThe first scientist knows the product is 4, so the possibilities are 2X2and 4x1.  So he doesn't know, but he knows the sums are either 4 or 5.  \n\nIf the sum were 4, then the two combinations would be 2+2 (product=4) and 3+1 (product=3, only combination); if the sum were 5 then the two possibilities would be 3+2 (product=6) and 4+1 (product =5).  \n\nIf the sum were four, the second scientist would know it was 3+1 if the first scientist said he knew, and he would know it was 2+2 if the scientist said he did not know - but the second scientist still does not know.  \n\nThus the first scientist realizes that the sum must be 5, which means the ages are 4 and 1.  The 2nd scientist, realizing that the 1st scientist would not have be able to figure this out if the product was 6, also realizes that the ages are 4+1.
170982
I believe it is 9599...because the digits of the right side add up to the left side.\n\n3+9+8+3 = 23\n9+5+9+9 = 32
513529
I see too many combinations to make anything out easily.\nI thought maybe the first might be PR, could also be AXIO...(didn't quite make AXIS POWERS, too bad!)  All those dots in a row have me stumped.  Letters using only dots: e (1 dot), i (2 dots), s (3 dots) and h (4 dots), as well as the number 5 (5 dots).\nAny clues?
114322
The answer is:\nIn its proper, original form, the first two sentences have absolutely nothing to do with the question: "Think of words ending in -GRY. Angry and hungry are two of them." Ignore those two sentences. They are there only to throw you off course. (And it worked, didn't it?) What's left is the actual riddle itself: "There are only three words in the English language. What is the third word? The word is something that everyone uses every day. If you have listened carefully, I have already told you what it is."\nThe key is the phrase "the English language." In this three-word phrase, the third word is simply the word "language." Get it? "Language" is definitely something that "everyone uses every day"! Without that quirky little twist, the puzzle would be just another trivia question, not a riddle.
478226
I understand your confusion, and suggest that you focus on how the two problems are grouped.\n\nIn mathematics, it's customary to group added terms together using parentheses.  Therefore, the first sum is merely 6.\n\nOn to the next problem:  With nine 1's being added, the sum is 9.\n\nNever would you string these together to form a two-digit answer.\n\n\nHowever, as the first Answerer suggests, perhaps your problem was a single multiplication problem.  In that case (and assuming the problem was written all on one line), you would work it out as he/she did:\n\n (6)(9) = 6 * 9 = 54\n\nHope this helps!
165790
Look up Gauss. Story is he added up 1..100 in grade school and invented the n*(N+1)/2.
290759
This is more of a word problem or a riddle than a math problem.\n\nThe answer is 7 (SEVEN).  Remove its first letter and it becomes EVEN!