How many 3 digit numbrs are divisible by 9?
920247
There are exactly 100\n108 is the smallest and is 12*9\n999 is the largest and is 111*9\n\nTherefore there are 111 - 12 + 1 or 100 such numbers.\n(the number of numbers between a and b inclusive is a-b+1)\n\nNice try heartsensei but that wasnt your original answer.  copying others is low.
744222
Let the number be X\nThen X = 1000a + 100b + 10c + d, where a,b,c,d are the digits of X.\nLet the 'flipped' number be Y.\nThen Y = 1000d + 100c + 10b + a\nSo, X - Y = 1000a + 100b + 10c + d - (1000d + 100c + 10b + a)\nor\nX - Y = 999a + 90b - 90c - 999d\n= 9 * (111a + 10b - 10c - 111d) .\nwhich is divisible by 9.\nThe sum of the digits of any number divisible by 9 is itself divisible by 9.*  So if you sum the digits of X-Y, you will get a number divisible by 9.  Then if you sum the digits of that number, you will get another (smaller) number divisible by 9.  If you keep doing that, you will eventually get to 9. [I assume that's what you meant.  If you sum the digits just once you wont necessarily get 9, as in your example. 6354-4536 = 1818 which sums to 18.]\n\n* You can prove that the sum of the digits of any number divisible by 9 is itself divisible by 9 by induction.\nThe first 2-digit multiple of 9 is 18, and 1+8 = 9.\nAssume it is true for k*9 where k is a natural number (1,2,3,...)\nProof for (k+1)*9:\n(k+1)*9 = k*9 + 9\nIf the last digit of k*9 is 0, then adding 9 just adds 9 to the sum of the digits of k*9.  Since the sum of the digits of k*9 is divisible by 9, so is this sum + 9.  We are done.\nIf the last digit is anything else and the tens digit is <9, adding 9 is the same as adding 1 to the tens digit and subtracting one from the one's digit.  So the sum is the same as the sum of k*9 + 1 - 1 = k*9 and we are done\nIf the last digit is anything else and the tens digit=9 then adding 1 to the tens digit means subtracting 9 from the tens digit and adding 1 to the hundred digit, which means the sum is the same as the sum of k*9 + 1 - 9 + 1 = the sum of k*9 - 9 which is divisible by 9.  (This can be extended to carrying a 1 to the other places as well.)
403642
just remeber, if you want to know if any number is divisible by 4 just check to see if the last two digets are divisible by 4... that should greatly help in solving this problem\n\nNow if you would please, stop asking all of your homework questions
210139
the only way to make 1991_1991 divisible by nine is to choose _ the missing number by 5\n\n199151991 is divisible by nine\n\nso 199151991 / nine =221 2 7999 \n\nSo there is 3 numbers not 4 in the first so i say\n0221 2 7999 x nine = 1991 5 1991
1022090
99.\n\n2-55 can be thought of a 2 and then a matching tens and ones place 3 digits higher.\n\n6-xx then, can be thought of as the x being 3 tens and 3 ones higher than 6, ant that' 6-99, making 99 the answer.
1098310
There are lots of them.  The easy way is to multiply 5 times 6 times 7.  This number is divisible by 5, 6, and 7.  So -- just add two to it.\n\nThis is 212.\n\nYou want another?  How about 2102?  210000000002?\n422? 420002?\n\nI got a million of 'em.
178354
if there are finitely many digits after the decimal point, count them, if there's n of them, write what you had behind the decimal point over 10^n, that is a 1 with n zeroes.\nIf there are infinitely many, find the moment when the digits begin repeating.  put the repeating sequence over as many nines as there are repeating digits.For a complex case, see the example (when there's a repeating and not repeating part.\nExample\n0.34356356356...=0.34+0.00356356356.... [notation is sometimes 0.00(356)]\n0.34=34/100=17/50\n0.00(356)=1/1000*0.(356)=1/1000*(356/999)=356/999000
1169470
792\n\n1 thru 9 = 9 digits\n10 thru 99 = 180 digits (90 numbers with 2 digits each)\n100 thru 300 = 603 digits (201 numbers with 3 digits each)\nTOTAL = 792
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).
1117885
23\n\nAny number exactly divisible by three can be represented just using the three. Any number that yields a remainder of 1 when divided by three and is at least 13 can be represented as 13 plus some number of 3s. Any number that yields a remainder of two when divided by three and is at least 26 can be represented as 13+13+some number of threes. Because all numbers must have a remainder of either 0, 1, or 2 when divided by 3, all numbers at least 26 can be represented with one of these schemes. It is easily seen that adding the ability to include more than two 13s, or any number of sixes, adds no additional power to the system. Therefore, the greatest number not representable would be the largest number less than 26 and having a remainder of 2 when divided by three. This number is 23.
838732
000000 to 999999 --> 1 million combinations\n\n10 choices for the first digit\n10 choices for the second digit\n10 choices for the third digit\n10 choices for the fourth digit\n10 choices for the fifth digit\n10 choices for the sixth digit\n\n10 x 10 x 10 x 10 x 10 x 10 = 1 million\n\nHowever, if you disallow 0 as a starting digit then you have the following:\n\n*9* choices for the first digit (1-9)\n10 choices for the second digit (0-9)\n10 choices for the third digit (0-9)\n10 choices for the fourth digit (0-9)\n10 choices for the fifth digit (0-9)\n10 choices for the sixth digit (0-9)\n\n9 x 10 x 10 x 10 x 10 x 10 = 900,000
146728
no, it is divisible by 1, 3, 23, 29, and itself (with the numbers being reduced as far as possible)
845925
If you problem, or part of it, is:\n\nHow many 8 digit numbers can be formed from this list of digits?\n8! / (3! * 2!)\nThe number combinations of digits has tobe reduced by 2! because of the 2 "6"s.\nThe number combinations of digits has tobe reduced by 3! because of the 3 "2"s\n.\nBut since the number cannot exceed 65,500; all larger numbers must be eliminated. Subtract the number of 8 digit numbers that can be produced from the list that start with "9". Likewise "7". Likewise "69", "67", "66", "659", ""657", and "656".\n\n\n\n(Try to state your problem clearly and completely.)
700622
you take any number which is divisible by nine\nsum of its digits will be nine or any multiple of nine\ne.g.\n99    9+9=18    1+8=9\nkeep on suming the digits if the sum comes to be more than one digit it will at last become nine\ne.g.   \n123456789        1+2+3+4+5+6+7+8+9=45=4+5=9\nhence any number you multiply by nine you will get the sum 9 finally
779216
Each digit is even\nEach digit is different\nThe number has 4 digits\nIt is divisible by 4.\nIt is not divisible by 8.\nIt is a square.\n\nYour answer must be: 6084\nIt is the square of 78.\nAll digits are even and different.\nIt is divisible by 4.\nIt is not divisible by 8
112973
103 people:\n\n00 to 99 is 100 unique two digit plus 3 duplicate\n(zero is a valid digit)
90506
I'll take it as far as I can.\n\n3 - Add all the numbers and if the result is divisible by 3 then the entire number is 9+1+2+0+3+1+2=18 which is divisible by 3.\n4 - The last 2 numbers are divisible by 4. 12 is divisible by 4 so the entire number is.\n6 - It's divisible by 2 and 3.\n8 - If the last 3 numbers are divisible by 8, then the entire number is. 312/8=39. \n9 - Like for 3, except it has to be divisible by 9. The result, from above, is 18 which is divisible by 9.\n12 - If divisible by 3 and 4.\n18 - If divisible by 2 and 9.\n24 - If divisible by 8 and 3.\n36 - If divisible by 4 and 9.\n72 - If divisible by 8 and 9.