How many odd five-digit numbers have distinct digits and no occurrence of the digit 9?
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).
51229
9 x 8 = 72 and pi (this is the correct spelling) out to 512 digits is:\n\n3.1415926535897932384626433832795028841971693993751058209\n749445923078164062862089986280348253421170679821480865132\n823066470938446095505822317253594081284811174502841027019\n385211055596446229489549303819644288109756659334461284756\n482337867831652712019091456485669234603486104543266482133\n936072602491412737245870066063155881748815209209628292540\n917153643678925903600113305305488204665213841469519415116\n094330572703657595919530921861173819326117931051185480744\n623799627495673518857527248912279381830119491298336733624
848063
488,895\n\nWrite every number as a six digit number with leading 0's if necessary.  i.e, 1 becomes 000001, 10 becomes 000010, etc.\nAlso, write 1,000,000 itself as 000000.  (We don't change anything by doing this.) We will subtract out the leading 0's later.\n\nTherefore, there are 1,000,000 6-digit numbers.  Each digit occurs just as many times as any other, so so far there are:\n1,000,000 * 6 / 10 or 600,000 zeros.\nNow subtract out all the leading zeroes.\n\nEvery 1-digit number has 5 too many.  There are 9 of these, so 5*9 = 45 too many zeroes.\nEvery 2-digit number has 4 too many.  There are 90 of these, so 4 * 90 = 360 too many zeroes.\nContinuing this way, there are another 3*900 = 2700 , 2 * 9000 = 18000 and 1 * 90000 = 90000 too many zeroes.\n\nThis leaves 600,000 - 45 - 360 - 2700 - 18000 - 90000\n= 488,895 zeros in total.
324897
Could be 10,14,13. The odd terms differ by 3 (eg. 4-1,7-4), the even terms differ by 4.
1126030
If a five digit number times 9 is another five digit number, then that five digit number must be less than or equal to 11111.  For it to be a five digit number, it must also be >=10000.  So the first digit must be 1. Hence, the last digit must be 9.\n\nSo our number is:\n\n    N = 1XYZ9.\n\nWe also know, since N<=11111, that X=0 or 1.\n\nNote that 9*N + N = 10*N, so you can write this as the addition problem:\n\n1XYZ9 + 9ZYX1 = 1XYZ90.  \n\nLooking at the tens digit, we see we get a carry over of 1 from the 1s digits, so we know that X+Z+1 has last digit 9.  But since X=0 or 1, that means X+Z+1 = 9, or X+Z=8. In the 100s digit, Y+Y has last digit Z.  In the 1000s digit, X+Z we know is 8, but we might get a carry-over from Y+Y of 1.  So either Y=X+Z=8 or Y=X+Z+1=9.  Since Y is always greater than 5, Y+Y always does contribute a 1 to the 1000s digit, so Y=9.  Y+Y has last digit 8, which means Z=8 and X=0.  So our number can only be 10989.
757339
one combination of 5 digits and 120 permutation.\n\none combination 5 digits can be picked in only one way \n\n19367 91367 all are same.\n\nfor permutation\nboth the above are different\n\n1st letter 5 choices , 2nd letter 4 choces so on .\n\nnumber 5*4*3*2*1 = 120 choices
871673
So call the number\nB = {b_0,b_1,b_2,b_3,b_4,b_5,b_6,b_7,b_8,b_9}\n(this reverse indexing makes things easier).\n\nYou must have b_0 != 0, since you were told it was a ten-digit number. So b_0 is between 1 and 9.\nSo b_0 = 0 - (number of the b_i which are non-zero)\n\nAlso note that Sum_i=0..9 (b_i) = 10 since B has ten digits, and each b_i has to have a single value between 0..9\nHence Sum_i=1..9 (b_i) = (10 - b_0) for any particular choice of b_0\n\nTo prove the number is unique, see below:\n\nIt's easy to show 6210001000 is one possible answer.\nb_0 = 6 since there are six 0's in B\nb_1 = 2 since b_2, b_6 =1\nb_6 = 1 since b_0 = 6\nAll other b_i are 0 since there are no other non-zero b_i.\n\nTo prove the number is unique, the easiest way seems to be to show that the other b_0 between 1..5 and 7..9 are impossible by (proof by contradiction).\n\nb_0 = 9 is impossible since then b_9 must be 1, which gives a contradiction giving '9000000001', which only has 8 zeros, also b_1 must be 1, et.c\n\nb_0 = 8 gives '8000000010', but then b_1 must be 1, and it has one less zero, so that becomes '7100000100'...\nbut that has two ones, so b_1++, and b_2 must be 1, and you have again one less zero so b_0=6, giving '6210001000', which finally works.\n\nHarder to show that b_0 = 1..5 are impossible.\n\nLet's start from the opposite extreme, b_0=1.\nThen Sum_i=1..9 (b_i) = (10 - b_0) = 9\nThis constraint means each b_i can only be 0, 1 or 2.\nSince we set b_0=1, this means we have one zero and eight ones e.g. B='11111111110', doesn't matter where you put the 0. But that doesn't work since clearly b_1 must >=8, thus Sum_i=1..9 (b_i) = 8+7>9 which is a contradiction.\n\nIt should be trivial to crank out the cases b_0=2..4, maybe there is a more compact proof.
125542
Let the last digit be n. The first digit is at least 1, so we need to count the number of possible partition of n-1 into 4 digits. There are (n-1+3)C(3) ways.\n\nThere are 4 possibilities for the last digit, 2, 4, 6, and 8. So the answer is (4)C(3) + (6)C(3) + (8)C(3) + (10)C(3) = 4 + 20 + 56 + 120 = 200.
521941
the odd numbers between 50 and 60 are:\n51 53 55 57 59\nand the only number that add up to  10,\n5+5=10\nthus: 55 is your answer.
592829
A 9 digit number has exactly 1,000,000,000 combinations if you include the code 000-00-0000 through 999-99-9999.  So we are about 1/3 of the way to running out of numbers if we are recycling.
88776
Are you looking for a set of 8 consecutive whole numbers that when you add 7 of them together in different combinations you receive those 3-digit responses?\n\nWell you've only listed 6 possibilities and only one of those is odd.\n\nStarting at 29 the lowest combination is 224 and the highest is 231.  Starting at 28 the lowest is 217.
953820
There are 6 correct numbers.\nThat means 49-6 = 43 numbers are incorrect.\nYou need to choose 4 correct numbers and 2 incorrect numbers.\nC(6,4) = 15\nC(43,2) = 903\n15*903 = 13545
930433
The answer is 393.\n\n Start with 333.\nIt's an easy 3-digit number that is divisible by 3\n Tens digit id 3 times more than hundreds.\nSo 393. \n Does it add up to 15? \nYes it does!\n\n:) :) :)
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).
1116625
In the UK we tend to say Nought rather than Zero so 8 5 4 9 0 1 7 6 3 2 would be in alphabetical order this side of the pond.\n\nIn France it would be cinq deux huit neuf quatre sept six trois un 5 2 8 9 4 7 6 3 1\n\nIn German it would be acht drei eins funf neun sechs sieben vier zwei 8 2 1 5 9 6 7 4 2\n\nRe the last-but-one contributor\n\nV100 = 10 and 10 + 0 = 10 \nV9801 = 99 and 98 + 01 = 99\nV998001 = 999 and 998 + 001 = 999\nV99980001 = 9999 and 9998 + 0001 = 9999\nV9999800001 = 99999 and 99998 + 00001 = 99999 etc\n\nso not unique but one member of an infinite set of numbers with this property.
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
1127227
N = 285714 and 1.5*N = 428571\n\nHere's the analysis:\n\nLet's assume that it's an n digit number with digits x_1, x_2, ...  x_n. Then\n  N = Sum{i=1..n}(10^(n-i)*x_i)\nand the number you get by moving the last digit to the front will be:\n  10^(n-1)*x_n + Sum{i=1..n-1}(10^(n-i-1)*x_i)\nWhat you are asking for is:\n  1.5*Sum{i=1..n}(10^(n-i)*x_i) \n    = 10^(n-1)*x_n + Sum{i=1..n-1}(10^(n-i-1)*x_i)\nor:\n  1.5*x_n + 1.5*Sum{i=1..n-1}(10^(n-i)*x_i) \n    = 10^(n-1)*x_n + Sum{i=1..n-1}(10^(n-i-1)*x_i)\nor by transposing to the other side:\n  1.5*Sum{i=1..n-1}(10^(n-i)*x_i) - Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 10^(n-1)*x_n - 1.5*x_n\nor: \n  Sum{i=1..n-1}(1.5*10^(n-i)*x_i) - Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\nor:\n  Sum{i=1..n-1}(15*10^(n-i-1)*x_i) - Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\nor:\n  Sum{i=1..n-1}(14*10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\nor:\n  14 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\n\nNow the left hand side is a positive integer and on the right hand side 10^(n-1)-1.5 will be a fraction ending in 0.5. So x_n has to be an even digit to make (10^(n-1)-1.5)*x_n a positive integer. \n\nNow, if x_n were 2 or 6, multiplying it with something-point-five will give you an odd number. But the left hand side is an even number (because you are multiplying by 14). So x_n cannot be 2 or 6. That leaves 4 and 8 as the two possible choices for x_n i.e., the number N has to end in either 4 or 8.\n\nIf x_n is 4, then we are looking for:\n  14 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*4\nor:\n  14 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 4*10^(n-1)-6\nor: \n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 2*10^(n-1)-3\nor:\n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 2*10^(n-1)-3+10-10\nor:\n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 2*10^(n-1)-10+7\nor:\n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 20*10^(n-2)-10 + 7\nor:\n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 10*(2*10^(n-2)-1) + 7\nor:\n  Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 10/7*(2*10^(n-2)-1) + 1\n\nAgain, the left hand side is a positive integer. The right hand side can be a positive integer only if 2*10^(n-2)-1 is a multiple of 7. \n\nFor values of n, 2*10^(n-2)-1 is 1(n=2), 19(n=3), 199(n=4), 1999(n=5), 19999(n=6), etc. n=6 is the first number in this series (19999) that's a multiple of 7. Putting n=6, we get:\n  Sum{i=1..5}(10^(5-i)*x_i)\n    = 10/7*(2*10^(6-2)-1) + 1 = 28571\ni.e., a five digit number = 28571.\n\nThat gives us our first answer: N = 285714 and 1.5N = 428571.\n\nNow I have only done the analysis so far for x_n being 4. If you do similar analysis for x_n=8, you get\n  14 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\n    = (10^(n-1)-1.5)*8\n    = 8*10^(n-1)-12\n    = 8*10^(n-1)+2-14\nor:\n  Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (4*10^(n-1)+1)/7 - 1\n\ni.e., 4*10^(n-1)+1 has to be a multiple of 7. For various values of n, 4*10^(n-1)+1 gievs you numbers like 41, 401, 4001, 40001, etc. and the first number in this series that's a multiple of 7 is 400001 for n=6. So we end up with:\n\n  Sum{i=1..5}(10^(5-i)*x_i)\n    = (4*10^(6-1)+1)/7 - 1 = 57142.\n  \nThat gives us our second answer: N = 571428 and 1.5N = 857142.\n\nWhile this is also correct, the first number is smaller. So the smallest number that satisfies your constraint is 285714.