How many even 3 digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7 if no digit is repeated?
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.)
152234
1, 3, 5, 7, 11, 17....\n\n\nNumbers that can only be divided by that number or 1 with the result of a whole number.
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
688342
There are 24 possibilities to have 3 consecutive letters.\nThe total possible ways of typing any 3 letters is 26^3.\nAnswer: 24/26^3 = 24/17576 = 3/2197 = .001365498407
789066
Firstly, it has to start with either a 1 or a 2, and end with a 2 or a 6. So we just take each case one at a time.\n1..2: that leaves 23556, so there are 5!/2! ways, ie 60 ways.\n1..6: that leaves 22355, so there are 5!/(2!2!) ways, ie 30 ways.\n2..2: that leaves 13556, so there are 5!/2! ways, ie 60 ways.\n2..6: that leaves 12355, so there are 5!/2! ways, ie 60 ways.\n\nNote that in each case there are 5! ways of arranging the other 5 numbers, but we need to divide by 2 if there is a pair of repeated numbers.\nThat gives a total of 210.\n\n(note that both of the other answers given so far are wrong - its not 360, because that person has forgotten they have to be even; and its not 120. They counted 2x2 possibilities for the first and last digits which is correct, but one of those cases already uses both 2s, so you *don't* want to divide by 2 in the end for that particular case (you're not counting them twice are you!)\nTheres really no other way of doing it than splitting them into cases like I did.
324897
Could be 10,14,13. The odd terms differ by 3 (eg. 4-1,7-4), the even terms differ by 4.
674175
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
112973
103 people:\n\n00 to 99 is 100 unique two digit plus 3 duplicate\n(zero is a valid digit)
817181
The first digit has nine possibilities: 1 thru 9.\nThe second digit can then also be any of nine possibilities:\nanything but whatever the first digit was.\nThen the third digit can be any of the remaining eight\ndigit possibilities, and this leaves seven possibilities for\nthe fourth digit.\nSo the total number of IDs is 9x9x8x7 = 4536.
798343
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.
146975
If you know the area code and nothing else, then it's a 7-digit number and so 10 million combinations.  If you know the area code and the first three numbers then there are 4 digits or 10,000 numbers.
947933
The answer is 4.\n11 is eleven, which has 6 letters, so the next number is 6.\n6 is six, which has 3 letters, so the next number is 3.\n3 is three, which has 5 letters, so the next number is 5.\n5 is five, which has 4 letters, so the next number is 4.\n4 is four, which has 4 letters, so the next number is 4. It's all 4's from there!\n\nBy the way, 4 is the only number that has the same number of letters as its value, and if you start this sequence at any whole number, you will end up at four! Just a tidbit I knew! ^_^
1170732
6 SIX\n\nLAST THREE ARE CONSECUTIVE
252901
In the experiment of choosing the 3 digits\n\nTotal no. of possible choices \n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n==> n = 9 * 10 * 10\n==> n = 900\n\n[Since the first digit cannot be zero, there would be 9 possibilities for choosing it, 10 for second and third]\n\nLet "A" be the event that all the 3 digits are different and "B' the event that the 3rd digit is different from the first 2 digits.\n\nFor Event "A"\n\nNo.of favorable choices\n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n= 9 * 9 * 8\n==> m(a) = 648\n\n[Since the three digits to be chosen should be different, there would be 9 possibilities for choosing the first digit (since 0 is excluded), 9 possibilities for the second (excluding the one chosen for the first and including 0) and 8 for the third (excluding the ones chosen for the first and the second and including 0)\n\nProbability that all the digits are different\n==> Probabilty of occurance of event "A" = m(a)/n\n==> P(A) = 648/900\n==> P(A) = 18/25\n\nFor Event "B"\n\nNo.of favorable choices\n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n= 9 * 10 * 8\n==> m(a) = 720\n\n[Since the first two digits can be any of the ten and the last should be different from the first two, there would be 9 possibilities for choosing the first digit (since 0 is excluded), 10 possibilities for the second (any digit can be chosen including 0) and 8 for the third (excluding the ones chosen for the first and the second)\n\nProbability that the 3rd digit is different from the first two\n==> Probabilty of occurance of event "B" = m(b)/n\n==> P(A) = 720/900\n==> P(A) = 4/5\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
906203
Assuming you cannot reuse a digit unless there is more than one copy in the list...\n\nYou could do it this way...\n5 one-digit numbers: 2, 5, 6, 7, 9\n22 two-digit numbers: 22, 25, 26, 27, 29, 52, 56, 57, 59, 62, 65, 66, 67, 69, 72, 75, 76, 79, 92, 95, 96, 97\netc.\n\nBut it would take a while. The answer is: at least a few.
427494
The number of combinations is determined by multiplying the number of possible numbers of the first by the number of possible numbers of the second, third and last. If the any of the numbers could be used more than once, the answer would be 4 X 4 X 4 X 4 = 256. Since this problem says that the number can only be used once, then the number of possible choices for the first is 4, the second is only 3, the third is 2, and the last is only 1. So the way to arrive at 24 is 4 X 3 X 2 X 1. Maybe an example to make sure it is clear. If the first digit was 3, then the second could only be one of 5, 8, or 9. If the second was chosen as 5, then the third could only be one of 8 or 9. The last then could only be the one not selected as the third.
842276
1,2,3,4,6,8,12,16,24,48 \n\nlolz hope this helps :)