Find a 9 digit number which has the property that the first digit is equal to number of zeros and ones........
405169
Yeah! At last I have one.\nActually, there are only 3 of them:\n542121000\n533211000\n740110110 (new!)\n\nAlso, I have a proof that there only 3 of them. Email me if you need.
886465
You need the 10 digits 0 through 9 (obviously). You also need another 1 and 2, so you can represent 11 and 22.
797722
AARGHHH!!  I withdraw my earlier attempt.  It must be dementia, I'm afraid.  It's still asking about 1000! (1000 factorial), of course, but a careful accounting is needed, not only of the effect of all numbers ending in: \n\nA. 1, 2, 3, ... 9 (I did this by a variety of ways, always coming up with a final non-zero digit of 6 for the product of all such sets), \n\nbut also of the following two other types of number, which I regrettably neglected the first time around:\n\nB. 10, 20, 30 ...90 (as often as sets of numbers like that occur), and finally, of types:\n\nC. 100, 200, 300, ... 900 (this set occurs but once).  The final, unique 4-digit number (1000) only adds three zeros at the end and doesn't affect the final non-zero digit.\n\nO.K., here goes:\n\nNumbers of Type A.\n\n1 x 2 x 3 x ... x 9 = 362880; final non-zero digit is 8.\n\nThere are 100 such "basic sets" of type A altogether.  \nMultiplying them together, set after set, we get the \nfollowing repeated cycle of non-zero ending digits \n(top row is # of basic sets used so far; bottom row is the corresponding final digit.)\n\n1   2   3   4   5   6   7   8   9 ...\n8   4   2   6   8   4   2   6   8 ...\n\nNotice that after every four basic sets have been used, we end up with 6 as the final non-zero digit.  Since there are exactly 100 such sets (a multiple of 4), we end up with that same last non-zero digit for the product of ALL numbers of type A.  (But we're not done yet.) \n\nNumbers of Type B.\n\nThere are exactly 10 sets of numbers of Type B, i.e. ending with  exactly one (but not two or three) zeros.\n\nMultiplying just these sets together, we'll have the following analogous results:\n\n1   2   3   4   5   6   7   8   9  10  \n8   4   2   6   8   4   2   6   8     4   \n\nSo, the last non-zero digit in these "10's products" is 4 for this lot.\n\nNumbers of Type C.\n\nWe have just one set of "100's".  For this, the last non-zero digit is 8 (it's just our "basic set" with a large number of zeros following it); only the last non-zero digit is of interest.\n\nSo, to summarize, the different sets we've divided our numbers into, when multiplied together, give us final non-zero digits as follows:\n\nType A:   6\nType B:   4\nType C:   8 \n\nFinally, then, the last non-zero digit of the product of everything is the last digit of:\n\n6 x 4 x 8, or 2.  (PHEW!!)\n\nWell, it's now the value of 2 (as reported by zanti3 from the 100 factorial website at:\n\nhttp://www.justinwhite.com/big-calc/1000.html)\n\n(Finish off this URL with 1000.html, instead of 1000... The Yahoo site would not reproduce the full URL from my draft.  Alternatively, just go to Yahoo or Google with 1000 factorial 1000 - this worked for me.) \n\nBoy; what a problem!  Evidently the jury is still out on the question of my impending dementia.  (See zanti3's kind words, below about my earlier fiasco.) \n\nThanks again to zanti3 for bringing the existence of a (1000 factorial) web site to my attention.\n\nLive long and prosper, everyone.
668533
There are 3 pairs:\n\n1. 0 and 0\n2. 2 and 2\n3. 9 and 9\n\nThere won't be any more as the number values get larger. This is because when the no.s are multiplied, there would be more digits than when the no.s are added. So they can't be opposite if they have diff. no. of digits.
1026397
the morse code for 0 = -----\nthe morse code for 1 = .----\n\nso a binary number of 0101 would be\n-----  .----  -----  .----
645487
1 - the initial 1\n11 - all numbers from phrase "previous has 1 digit 1"\n21 - all numbers from phrase "previous has 2 digit 1"\n1211 - all numbers from phrase "previous has 1 digit 2, then 1 digit 1"\n111221 - all numbers from phrase "previous has 1 digit 1, then 1 digit 2, then 2 digits 1"\n312211 - all numbers from phrase "previous has 3 digits 1, then 2 digits 2, then 1 digit 1"\n13112221 - all numbers from phrase "previous has 1 digit 3, then 1 digit 1, then 2 digits 2, then 2 digits 1"\n\nSo, the next will be\n\n1113213211 - all numbers from phrase "previous has 1 digit 1, then 1 digit 3, then 2 digits 1, then 3 digits 2, then 1 digit 1"\n31131211131221\n13211311123113112211\n11131221133112132113212221\nand so on
692386
There are 9 with one digit, 90 with two digits, 900 with 3 digits, and 1 with 4 digits.  So 9(1) + 90(2) + 900(3) + 1(4) = 2893.\n\nSorry about my earlier answer, my brain was off or something!
26155
You should use scientific notation to describe this number. Correctly expressed, this number is 1*10^n where n is the number of zeros you typed. (I don't know how many zeros follow the ...)
26758
Why would it be 11, 10, 12?  It's 0, 1, 2, not 1, 0, 2.\n\nThe second digit (the 1 in 10) is the tens place, and the first digit (the 0 in 10) is the ones place, so after 9 you have no more digits to put a higher number in the ones place, so you put a 1 in the tens place and a 0 in the ones place and get 10.  To add one to that you put a 1 in the ones place and get 11.  Thus 11 is one more than 10 so it comes after it when you arrange the numbers from smallest to largest (which is essentially what you're doing when you're counting).
613818
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.
1014042
The range of consecutive prime numbers for a "strictly" 9-digit product begins at 463 (* 467 * 479 = 103569859) and ends with 991 (* 997 * 1009 = 996919243).  Within this numerical range, there are NO consecutive primes whose product results in unique digits.\n\nEven if you expand the definition of a 9-digit product to include those with a leading zero as a digit, you can also check all 8-digit products and consider the first digit to be a zero.  The range then begins at 211 (* 223 * 227 = 010681031).  Unfortunately, even with this expanded definition, there are no products with unique digits.  (The solution 233*239*241=13420567 is incorrect because if expressed as nine digits, the product becomes 013420567 which repeats the 0 digit.)\n\nFor the morbidly curious, the full solution set follows:\n\n211 x 223 x 227 = 10681031 -- 0x3 1x3 repeat\n223 x 227 x 229 = 11592209 -- 0x2 1x2 2x2 9x2 repeat\n227 x 229 x 233 = 12112039 -- 0x2 1x3 2x2 repeat\n229 x 233 x 239 = 12752323 -- 2x3 3x2 repeat\n233 x 239 x 241 = 13420567 -- 0x2 repeat\n239 x 241 x 251 = 14457349 -- 4x3 repeat\n241 x 251 x 257 = 15546187 -- 1x2 5x2 repeat\n251 x 257 x 263 = 16965341 -- 1x2 6x2 repeat\n257 x 263 x 269 = 18181979 -- 1x3 8x2 9x2 repeat\n263 x 269 x 271 = 19172437 -- 1x2 7x2 repeat\n269 x 271 x 277 = 20193023 -- 0x3 2x2 3x2 repeat\n271 x 277 x 281 = 21093827 -- 0x2 2x2 repeat\n277 x 281 x 283 = 22027871 -- 0x2 2x3 7x2 repeat\n281 x 283 x 293 = 23300239 -- 0x3 2x2 3x3 repeat\n283 x 293 x 307 = 25456133 -- 3x2 5x2 repeat\n293 x 307 x 311 = 27974761 -- 7x3 repeat\n307 x 311 x 313 = 29884301 -- 0x2 8x2 repeat\n311 x 313 x 317 = 30857731 -- 0x2 3x2 7x2 repeat\n313 x 317 x 331 = 32842151 -- 1x2 2x2 repeat\n317 x 331 x 337 = 35360399 -- 0x2 3x3 9x2 repeat\n331 x 337 x 347 = 38706809 -- 0x3 8x2 repeat\n337 x 347 x 349 = 40811711 -- 0x2 1x4 repeat\n347 x 349 x 353 = 42749359 -- 4x2 9x2 repeat\n349 x 353 x 359 = 44227723 -- 2x3 4x2 7x2 repeat\n353 x 359 x 367 = 46508809 -- 0x3 8x2 repeat\n359 x 367 x 373 = 49143869 -- 4x2 9x2 repeat\n367 x 373 x 379 = 51881689 -- 1x2 8x3 repeat\n373 x 379 x 383 = 54143561 -- 1x2 4x2 5x2 repeat\n379 x 383 x 389 = 56466073 -- 0x2 6x3 repeat\n383 x 389 x 397 = 59147839 -- 9x2 repeat\n389 x 397 x 401 = 61927633 -- 3x2 6x2 repeat\n397 x 401 x 409 = 65111573 -- 1x3 5x2 repeat\n401 x 409 x 419 = 68719771 -- 1x2 7x3 repeat\n409 x 419 x 421 = 72147191 -- 1x3 7x2 repeat\n419 x 421 x 431 = 76027969 -- 0x2 6x2 7x2 9x2 repeat\n421 x 431 x 433 = 78568283 -- 8x3 repeat\n431 x 433 x 439 = 81927497 -- 7x2 9x2 repeat\n433 x 439 x 443 = 84208541 -- 0x2 4x2 8x2 repeat\n439 x 443 x 449 = 87320173 -- 0x2 3x2 7x2 repeat\n443 x 449 x 457 = 90900499 -- 0x4 9x4 repeat\n449 x 457 x 461 = 94593973 -- 3x2 9x3 repeat\n457 x 461 x 463 = 97543451 -- 4x2 5x2 repeat\n461 x 463 x 467 = 99677881 -- 7x2 8x2 9x2 repeat\n463 x 467 x 479 = 103569859 -- 5x2 9x2 repeat\n467 x 479 x 487 = 108938491 -- 1x2 8x2 9x2 repeat\n479 x 487 x 491 = 114537043 -- 1x2 3x2 4x2 repeat\n487 x 491 x 499 = 119319383 -- 1x3 3x3 9x2 repeat\n491 x 499 x 503 = 123239527 -- 2x3 3x2 repeat\n499 x 503 x 509 = 127757473 -- 7x4 repeat\n503 x 509 x 521 = 133390067 -- 0x2 3x3 repeat\n509 x 521 x 523 = 138693847 -- 3x2 8x2 repeat\n521 x 523 x 541 = 147413303 -- 1x2 3x3 4x2 repeat\n523 x 541 x 547 = 154769821 -- 1x2 repeat\n541 x 547 x 557 = 164831339 -- 1x2 3x3 repeat\n547 x 557 x 563 = 171534277 -- 1x2 7x3 repeat\n557 x 563 x 569 = 178433279 -- 3x2 7x2 repeat\n563 x 569 x 571 = 182918137 -- 1x3 8x2 repeat\n569 x 571 x 577 = 187466723 -- 6x2 7x2 repeat\n571 x 577 x 587 = 193397129 -- 1x2 3x2 9x3 repeat\n577 x 587 x 593 = 200848507 -- 0x3 8x2 repeat\n587 x 593 x 599 = 208506509 -- 0x3 5x2 repeat\n593 x 599 x 601 = 213479407 -- 4x2 7x2 repeat\n599 x 601 x 607 = 218519393 -- 1x2 3x2 9x2 repeat\n601 x 607 x 613 = 223626691 -- 2x3 6x3 repeat\n607 x 613 x 617 = 229580147 -- 2x2 repeat\n613 x 617 x 619 = 234118799 -- 1x2 9x2 repeat\n617 x 619 x 631 = 240993413 -- 3x2 4x2 9x2 repeat\n619 x 631 x 641
806909
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)
121493
We are used to ten digits in our numbering system, but binary only has two (zero and one, or on and off) In our system when we get to nine, our ones spot is at the top and we have to use two columns for higher numbers. Same applies in binary. Column starts at zero and is full at one.\nOn our regular system our first two columns fill at nine nine - in binary the first columns fill at one one. nine nine in our system equals 99. one one in binary equals 3.\n00000000 = 0\n00000001 = 1\nnow that column is full, so to count to 2 you have to use the next column.\n00000010 = 2\n00000011 = 3\nnow that column is full, so to count to 4 you have to use the next column.\n00000100 = 4\nand so on.\n\nInsted of having the "ones" "tens" "hundreds" etc. columns as our regular system does - \nbinary has the "ones" "twos" "fours" "eights" "sixteens" thirty-twos" etc colmns\n\nA byte is 8 columns.
454049
dozen... 99+9 = 108... 9 dozen = 9 * 12 = 108
1110740
I -1 digit\nLOVE - 4 digits\nYOU - 3 digits
966093
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).
1145815
6\n\nYou are putting the single digit numbers in alphebetical order:\neight, five, four, nine, one, seven, six, three, two, zero