I’m thinking of a 6-digit number. The sum of the digits is 43. It’s a square number and it's under 500000.
199459
499849\n\nI found the answer by taking the square root of 500,000.  This was a bit over 707.  When 707 is squared it gives the above answer whose digits happen to add up to 43.  If this wasn't the answer, I could've used trial and error or found a more systematic method.  (For that matter, there may be more answers.)
318166
5050\n\nIf you add the numbers like this\n (1 + 100) + (2 + 99) + (3 + 98) + . . . . + (50 + 51) = ?\nthen each set adds up to 101 and there are 50 sets of numbers, so 101*50=5050
78450
26*26*26*10*10*10 = 17,576,000\n\nRemember there are 26 letters in alphabet, so you have 26 choices for your first 3 digits. And you have 10 digits to choose from (0-9) for your last 3 digits.  Just multiply them together.
327331
272\n\nThe digits 2,7 must be present in all numbers and the average number length equals 5 and you want symmetry in the 3rd digit(i.e 2,7,8,7,2). Also you want the first 3 digits to increase for each number; therefore the number is 272
437964
1296 = 6^4\n\nThe first digit can be chosen in 6 ways. the second digit can be chosen in 6 ways. similarly for the third and fourth digits. that makes it 6 * 6 * 6 * 6 = 1296.
239833
Here are 10,000 digits... I could do many millions with ease, but Yahoo probably does not like huge posts.\n\nBy the way, the sqrt(Pi) has an infinite number of digits.  This can be proved using a "diagonalization proof" that proves that the number is irrational.\n\n1.77245385090551602729816748334114518279754945612238712821380778\n9852911284591032181374950656738544665416226823624282570666\n2361528657244226025250937096027870684620376986531051228499\n2517302895082622893209537926796280017463901535147972051670\n0190185234018585446974494912640313921775525906216405419332\n5009063984076137334774751534336679897893658518364087954511\n6516173876005906739343179133280985484624818490205465485219\n5613251561647467515042738761056107996127107210060372044483\n6723652966137080943234988316684242138457096091204204277857\n7806869476657000521830568512541339663694465418151071669388\n3321942929357062268865224420542149948049920756486398874838\n5059306402182140292858112330649789452036211490789622873894\n0324597819851313487126651250629326004465638210967502681249\n6930595420461560761952217391525070207792758099054332900662\n2230676144696612481887430699788352050614644438541853079735\n7425717918563595974995995226384924220388910396640644729397\n2841345043002140564233433039261756134176336320017037654163\n4763206692765418128357624903269045084853201341924359897308\n7119379948293873011126256165881888478597787596376136321863\n4246546641333954355703201522654193952186030497310513829498\n4396591656142459554212266151024785360980955103956007894021\n8809961338285402501680074580272911936642519282051000193635\n0073914643295493433951928853735459200563766502880540575532\n1231890091263228191509149808366956244831008522219239736463\n2484286326114576693242537157737789441409054457359535122562\n6391080239236909732127905807617134603914574791879794124850\n2184451458113418888804132209553321846467097274910285652627\n0784545326222784880098238583630075495095476406237708338835\n7225436621567481327668384244972420874516161833205077991480\n1846668142366936519028454638576148278570377743883762974799\n8273770543158368241099868322850380552635536972229313380526\n4428410372312043967004307612454138311792278275363715598398\n3768845370278429857070905112238405367790133854145853162080\n7304313806973998743669316601381707927205604195488285806309\n3111636297047867814026963272962701226135985897754505289483\n1130166840015320748519824024633375558517135680193412289759\n8071956874025057150214178379254364303036592821125092588061\n8903117074543127903953553660682611001188965742048727593919\n9769955383521150866962555964413705038292449535903106362345\n3056471711621685872545868744002961175792172319055405719868\n1727588419089649657906696515601728351482903856551169807210\n7953309161308435985243894654406821655003275379960238665037\n9888648152118657999585718656377511331597475359604341377664\n5119143460134292508116324806409073773212629335747472967679\n3412716029665120809898090577996603051021816255579789074870\n7621107623802626785429701502710915350498535149239083248428\n0828987595575678848892608420885521269510357370208661259115\n5413204403735608643388371239620642939023786663116326167888\n4192279819499524039424578422044303042043042071096927339294\n6085104969289739161855607837870336428902342932718872968029\n7215816594261297283663959050411303747457435097490580163269\n1653757690981097485625377850342879942192237718584431832793\n7312980065870368619230102174545052030845533038619506602364\n3341153205974122426426535421655815003647269194638254412455\n7259473900719227187097960549897246028226024291050774020812\n3756748076276377437063450463147737694803316835707620587495\n7817335099676287338739670296881906614261445737132282280393\n5444608377601058646738943855776887203971767059273166300903\n1126081448843086241142922479654540261637551586587351984372\n1755055272950641504357687583814935940105565912150377044197\n3880370568097143463163605434136507669997576211981184258742\n252081404364062117230919743945524596778174219146433350671
149503
1. The only square digits are 1, 4, and 9, so the hundreds place could be 1 or 9.  It is 9 (See#3 for reason)\n2. The sqrt(16) is 4, so one more is 5.\n3.  If the hundreds place is 1, then 1/10 of 1 is 1/10.  Adding 1/10 you get 2/10 which is not a digit.  Thus, the hundreds digit must be 9.  1/10 of 9 is 9/10.  9/10 + 1/10 is 1.\n4. 30/5 = 6\n5. 2\nThus, the number is\n952.216
219512
Without resorting to a search of the web I have memorized the following digits. (I hope I don't mistype any):\n\n3.1415926535897932384626433832\n795028841971693993751058209749\n445923078164062862089986280348\n253421170679821480865132823066\n470938446095505822317253594081\n284811174502841027019385211055\n5964462294895493038196
149451
By adding a single line to the first '+' you can make it into a '4'. So...\n\n545 + 5 = 550
325426
We don't know exactly how long the number is but generally the form will be:\n\n2...cba\nx 3\n--------\n...cba2\n\nFirst what multiples of 3 end in 2?\n\nA quick enumeration will show you that only 4 will work (4 x 3 = 12) so the last digit (a) = 4\n\n2...cb4\nx 3\n--------\n...cb42\n\nSince we are carrying a 1 (from the 12) and the total has to be 4 (from our last answer), we need 3*b to end in 3.  The only choice is 1:\n\n2...c14\nx 3\n--------\n...c142\n\nNow we have no carry, but 3*c must end in 1.  The only thing that works is 3*7 = 21\n\n2..d714\nx 3\n--------\n..d7142\n\nSo we have to carry a 2 (from 21) and the sum must equal 7, so the last digit in d*3 must be 5.  The only choice is d=5 (3 x 5 = 15)\n\n2..e5714\nx 3\n---------\n..e57142\n\nSo we are carrying a 1 (from 15) which will add to 5, so 3*e must end in 4.  The only choice is 3*8 = 24.\n\n2..f85714\nx 3\n----------\n..f857142\n\nCarrying the 2 (from 24) we need to make 8, so 3*f must end in 6.  Only f=2 makes 2x3 = 6\n\nAnd we have found our last digit because we are back to 2.\n\nThe answer is the first number was:\n285714\n\n(And the second number was 857142 which is 3 x 285714)
417645
If the number begins with 5, there are still three of the odd digits available to fill the last place.  If a 3 is used for the last place, the others are filled using 1, 2, 3, 4, and so there are 4! arrangements.\nIf 1 is used for the last place, that leaves 2, 3, 3, 4 for the others, and so the number of arrangements is  4!/2! because of the repeated 3.  So that gives \n4! + 4!/2! = 36\n\nIf it begins with a 3, the last place can be filled with 1, 3, or 5, and  the numbers left to fill the remaining four places are all different and so the number of possibilities is\n3*4! =72\n\nIf it begins with a 2 or 4 ( two choices), and the last with 1 or 5 (two choices), there are 4!/2! ways of filling the other places because 3 is repeated.\n2*2*4!/2! = 48\n\nIf it begins with 2 or 4 and ends with 3, then the other four digits are all different, and the number of possibilities is\n2*4! = 48\n\nAdding all these together gives 204.  Yes.
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).
461405
I'm going to start by assuming that numbers starting with 0, other than zero itself, don't count.  Normally, since there are two choices for any individual digit, the number of numbers possible for any number of digits (d) is 2^d.\n\nHowever, since 0 can't be the starting digit for any but the one-digit numbers, your choices are cut in half, which would be 2^(d-1).\n\nThus, your answer would be the sum of:\n2 + 2^1 + 2^2 +...+ 2^9\n\nInterestingly enough, you'll notice that this answer ends up being 2^10.
309200
69..\n\nand try to paraphrase your friggin homework...
421609
I POSTED THE FIRST CORRECT ANSWER HERE AND WAS WRONGLY DELETED FOR A BOGUS VIOLATION REPORT\n\nI posted last night around 12pm, here was my (correct) answer:\n\nOk so let the tens digit be x, then the ones digit is (12-x). The value of the number is 10*x+(12-x) = 9x+12 which you are told = 11x 9x+12=11x 12=2x x=6 12-x=6 also So the number is 66. (Check x = 6*11 as claimed.)\n\n(There is some severely dishonest point-gaming going on around here)
415198
The first 3 digits cannot be 800, or 900, or all the numbers from 000 to 099.  That leaves a total of 1000 - 102 = 898 possibilities left.   The next 3 digits can be anything from 000 to 999, or 1000 possibilities.  The last 4 digits must be 0000, or just one possibility.  The answer is the product: 898 x 1000 x 1, or 898,000.
353401
Here is one possibility\n\n{2, 4, 4, 5, 5, 6, 6, 8, 8, 8, 12, 16}