how many 6 digit numbers have 3 odd and 3 even digits?
340631
First you must consider that the first digit can not be zero, otherwise it would not be a 6 digit number.\nThen you must consider how you can arrange odd or even within the digits. (O=ODD, E-EVEN)\n1.OOOEEE\n2.OOEOEE\n3.OOEEOE\n4.OOEEEO\n5.OEOOEE\n6.OEOEOE\n7.OEOEEO\n8.OEEOOE\n9.OEEOEO\n10.OEEEOO\n11.EOOOEE\n12.EOOEOE\n13.EOOEEO\n14.EOEOOE\n15.EOEOEO\n16.EOEEOO\n17.EEOOOE\n18.EEOOEO\n19.EEOEOO\n20.EEEOOO\nFor each of the #1-10 above we can consider 5 possible digits for the odds and 5 for the evens\nso for each there are 5*5*5*5*5*5=15625 different digit combinations.\nFor #11-20 we must consider that the first digit can not be zero, so there are 4*5*5*5*5*5=12500 different digit combinations for each.\nSo there is a total of 10*15625 + 10*12500=281250 different numbers with 3 odd and 3 even digits.
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
938348
There are many combinations. One of them is 1,2,3,4,5,16. Is there some specific property you wanted these six numbers to have?
924460
It depends on whether you allow a leading zero or not. \nIf you are allowed a leading 0, its easy. The last digit can be 0,2,4, or 6 (4 possibilities), and after you've chosen that, theres 7 possibilities for the first digit and 6 possibilities for the second digit, which makes 4*7*6 = 168.\nHowever, usually 3 digit numbers can't have leading zeroes. Now we have to split it up into two cases:\nCase 1:\n_ _ 0\nThere are 7 choices for the first digit, and 6 choices for the second - 42 possibilities.\nCase 2:\nIt ends with a 2, 4, or 6.\nThere are now 3 choices for the last digit, 6 choices for the first digit (anything *but* 0), and 6 choices for the second digit - 3*6*6 = 108.\nSo the total is 150.\n\nNote that we dealt with the 0 case specially, because it made a difference whether we had a 0 to work with for the first digit or not.\n\n(Note - again, in response to Wal C below, you *have* to split it up into cases. He said 4*6*6; however, if you happened to choose the 0 for the last digit as part of that '4', there will be *7* possibilities for the first digit, not 6. So you can't do it all together like that.)
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.
681251
1. If any number has an even number at its end, it is divisible by two. So 23000 is divisible.\n\n2. If the sum of digits of a number equal a number divisible by three, then the first number is divisible by three. In 23000, 2+3+0+0+0=5, which is not divisible by three. So 23000 is not divisible.\n\n3. Sorry, but I dont remember the divisiblity test for 4.\n\n4. If any number has 5 or 0 as its last digit, it is divisible by 5. So 23000 is.
901509
2, i think....8*6 = 48\n9*7 = 63\n7*2+1 = 15\n9*3+1 = 28
129045
Even numbers are divisible by 2 (2,4,6,8,10, etc) whereas odd are not (1,3,5,7,9, etc)
136789
That's an extremely good question. \n\nThe answer is simple. Before that, i will explain what your question means to those who don't get. Well, it means that we must get 24 using 1,3,4 and 6 any number of times however involving operations +,-,* and / only once. \n\n  Then, the answer is very simple :\n\n  {[(6*6)/(3-1)]+[6]}  =  24\n\n---------------------------------------------------------\nWOKRING :\n\n {[(6*6)/(3-1)]+[6]}   =  {[(36)/(2)]+[6]}\n                       =  18 + 6\n                       =  24\n\n----------------------------------------------------------\n\nI hope you post more questions like these. They evoke a lot of thought. I hope the solution serves th purpose.
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!
226140
I'm not sure I understand your question - do you mean adding odd numbers only to odd numbers and even numbers only to even numbers?  For odd I think it would be the numbers that end in (3 and 7) (1 and 9) and (5 and 5) where they add up to 50, and for the evens the numbers ending in (2 and 8) (4 and 6) that add up to 50.
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.
499304
For each of the 4 positions in the password, there are 26+26+10-3=59 different choices.  Since there are no other restrictions, the number of permutations is:\n\n59x59x59x59=12,117,361
536540
1^2, 2^2, 3^2, 4^2, 5^2, 6^2...\n\nSo 6^2 is the next number = 36!
955173
First you can only define even and odd for integers, so it only applies to {... -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5...}\n\nAn integer is even if you divide it by 2 and have *no* remainder.  \nAn integer is odd if you divide it by 2 and have a remainder.\n\nThe set of even numbers:\n{... -6, -4, -2, 0, 2, 4, 6 ...}\n\nThe set of odd numbers:\n{... -5, -3, -1, 1, 3, 5, 7 ...}\n\nAn alternate definition is if the last digit of a number is 0, 2, 4, 6, 8 the number is even.\nIf the last digit of a number is 1, 3, 5, 7, 9, the number is odd.
889893
2+7\n3+6\n9+0\n\nThat's it. They all add up to 9... I think that's what the question is asking.
133768
Assuming each digit is used no more than once in any given integer, this one would be fairly simple (and perhaps time-consuming) to draw out on paper:\n\n1. All integers have to end with the 3. For single-digit integers you of course have:\n\n3\n\n2. For two-digit integers, you have:\n\n43\n63\n83\n\n3. Three-digit integers:\n\n403\n603\n803\n463\n643\n483\n843\n683\n863\n\n4. Four-digit integers:\n\n4063\n4083\n4603\n4803\n6043\n6083\n6403\n6803\n8043\n8063\n8403\n8603\n8463\n8643\n4683\n4863\n6483\n6843\n\nI count 31.