During the year 1985, a convenience store, which was open 7 days a week, sold at least one book each day, and a total of 600 books over the entire year. Must there have been a period of consecutive days when exactly 129 books were sold?\nThis task is supposed to have something to do with the euler phi function. Any idea what the connection would be?
130561
One can have only one day with exactly 129 books, because\n600-2*129 = 342 < 365-2\nand each day at least a book was sold.
82985
a) is just sum of an infinite geometric sequence with a=180 and r = (1/3) so thats 180/(1-1/3) = 180*(3/2)=270\nbut it also bounces back so 270*2-180 = 540 - 180 = 360\n\nyou guys remembers the zeno's pardox: a person need to walk from A to B.  But in order to do that he needs to walk to C, where C is midpoint of A and B.  But then he will again need to walk to the midpoint of B and C... etc.  It is inappropriate to say that the person does not ever arrive at B.  This is the same case as part a)
86797
notice that if a remainder occurs twice, then it will immediately repeat.  hence the period is at most 111.  Yes, the pattern is distinctive, thats because 111 = 1 + 10 + 100\ntry for example base 7 with 1/(111) where 111 is in base 7.
373483
The second problem is easier than the first problem, so I'll tackle it first.\n\nLet's imagine that the trip is 120 miles. The distance doesn't matter if the average rates are as given above.\n\nIt takes 3 hours to travel 120 miles at 40 miles per hour.\nIt takes 2 hours to travel 120 miles at 60 miles per hour.\n\nSo it took 5 hours to travel a total of 240 miles.\n\nThis is an average rate of 48 miles per hour.\n\nI'll work on the first problem and hopefully I'll get back to you.\n\nUPDATE\n\nI think I have the first problem solved as well. It involves an infinite series! Call the people involved A and B.\n\nThe probability that A wins on the first toss is 1/2.\n\nIf A doesn't win on the first toss, then the probability that B wins on the second toss is also 1/2. But this only happens in 1/2 of all games. (In the other half, A won on the first toss.)\n\nIf B doesn't win on the second toss, the the probability that A wins on the third toss is also 1/2. But this only happens in 1/4 of all games (In half the other games, A won on the first toss. In half the remaining games - 1/4 of all games, B won on the second toss.\n\nFollowing this logic, A wins:\n\n1/2 + 1/4(1/2) + 1/16(1/2) + 1/64(1/2) + ...\n\nof all games.\n\nB wins:\n\n1/2(1/2) + 1/8(1/2) + 1/32(1/2) + ...\n\nof all games.\n\nLooking again at A, we have:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...)\n\nThe term in parentheses is an infinite series having initial term 1 and ratio 1/4.\n\nThe sum of an infinite series having ratio whose absolute value is less than 1 is equal to:\n\na/(1 - r)\n\nWith a = 1, and r = 1/4, the sum = 4/3. So:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...) = 1/2 * 4/3 = 2/3\n\nLooking again at B, we have:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...)\n\nThis sum should be half of the sum of the infinite series in A.\n\nWith a = 1/2 and r = 1/4, a/(1 -r) = 2/3\n\nSo:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...) = 1/2 * 2/3 = 1/3.\n\nA should win 2/3 of the time, and B should win 1/3 of the time. There is an advantage to going first!
541849
1) If they shared the delivery equally, then they would each deliver p number of papers but Erin delivered 20 of Jennifer's, so \nGilbert = p\nErin = p + 20\nJennifer = p - 20\n\nTotal = p + (p + 20) + (p - 20) ---- (1)\n\nSince erin's delivery is 1/2 of the total papers, we have that \n\np + 20 = [p + (p + 20) + (p - 20)]/2 ----- (2)\n\nFrom (2) we have that \n2(p + 20) = p + (p + 20) + (p - 20)\n2p + 40 = p + p + 20 + p - 20\n\nCollect like terms\n40 + 20 - 20 = p + p + p - 2p\n40 = 3p - 2p\n40 = p\np = 40\n\nSo Gilbert delivered 40 papers\n\n\n2) Considering the last sentence - one-fourth of those remaining left, leaving only 12 people in the audience. Since 1/4 left leaving 12, it means that 12 is 3/4 of the audience\n\nIf 3/4 = 12\n4/4 = (4/4 X 12)/(3/4)\n= 4/4 X 12 X 4/3\n= 16\n\nRemember that 1/3 left remaining 16, so 16 was the 2/3 of the audience.\n\nIf 2/3 = 26\n3/3 = (3/3 X 16)/(2/3)\n= 3/3 X 16 X 3/2\n= 24 \n\nAlso, remember that this fugure was just 1/2 of the original audience\nIf 1/2 = 24\n2/2 = (2/2 X 24)/1/2\n= 2/2 X 24 X 2/1\n= 48 which is the total audience\n\nCheck ...... working backwards\nAt ther beginning, there were 48 people in the audience;\nAfter a few minutes, 1/2 left remaining 24\n\nFive minutes later 1/3 of the remaining people left;\n1/3 of 24 is 8, so if 8 left, we have 24 - 8 = 16 people left.\n\nTen minutes later 1/4 of the remaining people left;\n1/4 of 16 is 4, so if 4 left, we have 16 - 4 = 12 people left!
54139
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.
206207
Okay, this is not going to be pretty, but it works, but only for years from A.D. 1582 and later.\n\nFirst, let M equal the number corresponding to the chosen month according to the following list\n\nJan=11\nFeb=12\nMar=1\nApr=2\nMay=3\nJune=4\nJuly=5\nAug=6\nSep=7\nOct=8\nNov=9\nDec=10\n\nNext, let N equal the day of the month for which you are inquiring.  So if you want to look for August 15, M would be 6 and N would be 15 (M=6, N=15).\n\nNow, take the last two digits from the year you are searching for, call that number Y.  For 1947, Y=47.\n\nThe leftover digits of the year make a new number, call it C.  For 1947, after removing 47 we see that C=19.\n\nFinally, if the year is a leap year, let L=1, otherwise, let L=0.\n\nNow we're ready to get down to business.  The following formula makes use of the greatest integer function (ipart) which essentially means that if you get some number with a decimal, chop off the decimal.  3698.2461 becomes 3698.  OK?  Good.\n\nNow we must calculate the following formula:\nN + ipart(2.6M - 0.2) + Y + ipart(Y/4) + ipart(C/4) - 2C - (1 + L)*ipart(M/11)\n\nAfter calculating this number, divide it by seven and look at the remainder.  The remainder corresponds to the day of the week as follows:\n\n0=Sunday\n1=Monday\n2=Tuesday\n3=Wednesday\n4=Thursday\n5=Friday\n6=Saturday\n\nLet's test this with your August 15, 1947.\n\nSo, M=6, N=15, Y=47, C=19, and since 1947 was not a leap year, L=0.\n\nPlugging these values into the formula we get:\n15 + ipart(2.6*6 - 0.2) + 47 + ipart(47/4) + ipart(19/4) - 2*19 - (1 + 0)*ipart(6/11)\n\n=15 + ipart(15.4) + 47 + ipart(11.75) + ipart(4.75) - 38 - 1*ipart(.5454545454)\n\n=15 + 15 + 47 + 11 + 4 - 38 - 0\n\n=54\n\nDividing this by 7, we get a remainder of 5.\n\nThis says that August 15, 1947, was a Friday.
372043
1. 101\n2.  There are 2 ones.  The first and last entries of the row.\n3.  Each entry is the sum of the 2 entries above and to     each side of it.\n4.  If the 2'nd entry in a row is a prime, it divides each of the other entries in the row (except for the last entry in the row which is always a 1)\n\nCheck out http://ptri1.tripod.com/ for all manner of kewl sh|t about Pascal's Triangle.  Also Fibonacci Series and other fun things ☺\n\n\nDoug\n\nEDIT:  Give Peter D the points.  He needs 'em worse than I do ☺
546155
D)6\n\nNote the pattern:\nIf f(1) = a, f(2) = b, then\nf(3) = b-a\nf(4) = -a\nf(5) = -b\nf(6) = -b+a\nf(7) = a\nf(8) = b\n\nAt f(7), the pattern starts repeating.  Therefore, the six values are f(1) through f(6) above.\nThis really isn't a permutation/combination problem, but you can often find good answers here on Yahoo Answers!  If you have specific problems with which you have difficulty, why dont you post a few related ones in one question here?  I'm sure I or others should be able to help.\n\nYOUR FOLLOW-UP:\nThis is two small problems in one.\nFirst, find the # of ways to choose the order of the backgrounds.  We have 3 choices for the first one, then 2 choices for the second, then 1 left for the third.  Therefore, there are 3*2*1=6 ways to choose the order of the backgrounds.\n{ (b, e, r), (b, r, e), (e, b, r), (e, r, b), (r, b, e), ( r, e, b) }\nSecond, there would be 3*2*1 = 6 ways to interview the business people.  There would be 2*1 = 2 ways to interview the education people, and just 1 way to interview the recreation person.  So that would make 6*2*1 = 12 ways to interview the people after the background order is set.\n\nThe total number of ways is then 6*12=72 ways to interview all 6.
573129
two times 9 factorial\n(since the two books are always together,assume it is a single book,and the number of books is 9)\nyou multiply by two,because u can keep them as A-B or B-A
167949
As the terms you give are clearly an arithmetic progression, with the difference between successive terms incrementing by 2 each time\n\n7 - 4 = 3\n12 - 7 = 5\n19 - 12 = 7\n28 - 19 = 9\n\nIt is clear that the next few terms are 39, 52, 67, 84, 103, 124 etc.\n\nI recognise this as rather similar to 36, 64, 81, 100, 121 because I am familiar with the squares of the positive integers having the property of being in Arithmetic Progression, with the difference incrementing by 2 (adding a constant, c, won't alter that property), so I can see where this is heading, but let's work the simultaneous equations through to prove what my observation and intuition is telling me ...\n\nThere are values of a b c to be found for which, with n increasing by 1 each time\n\na + b + c = 4 (1)\n4a + 2b + c = 7 (2)\n9a + 3b + c =12 (3)\n\nsubtracting (1) from (2)\n\n3a + b = 3 (4)\n\nsubtracting (2) from (3)\n\n5a + b = 5 (5)\n\nSubtracting (4) from (5)\n\n2a = 2\na = 1\n\nThen, substituting in (4),  b = 0 and substituting in (1) c = 3\n\nAnd it can be seen that this is indeed a sequence of the squares of the positive integers, with 3 added to them, each time.\n\nT(n) = n^2 + 3
128677
that math formula that proves it is SO WRONG! i just talked about this with someone, if you look at the formula it gets messed up like halfway through.\n\ndo we need to break this down?\none fish + one fish = 2 fish
604333
Let the numbers be n and n+1 (consecutive)\n\nthen the answer is 2n(n+1)\n\nYou know, I can tell you're really trying, (not just fishing for homework answers) because you didnt ask the whole question, just the part you needed help with setting up properly.\nGood Luck!
151627
The Fibonacci series has a number of interesting properties.\n\nThe ratio of successive terms (term divided by next term) converges on 0.618 (the Golden Ratio). Get a calculator and fill in the gaps below\n\n5/8 = 0.625\n8/13 =\n13/21 =\n21/34 =\n34/55 =\n55/89 =\n\nAND\n\nThe ratio of successive terms (term divided by previous term) converges to 1.618 = the inverse of the Golden Ratio.\n\n8/5 = 1.6\n13/8 = 1.625\n21/13 =\n34/21 =\n55/34 =\n89/55 =\n\nThe numbers appear in nature eg in the arrangement of leaves on a stalk, sunflower heads, pine cones etc.\n\nThe generating formula, each term is the sum of the last two, can be used to generate other series e.g.\n\n1, 3, 4. 7, 11, 18, 29, 47, 76,\n\nCompare this to\n\n1, 2, 3, 5, 8, 13, 21, 34, 55\n\nand notice how the difference between 55 and 65 is 21, the differemce between 34 and 47 is 13, the difference between 21 and 29 is 8 etc.\n\nThe series starting 1, 4 and 1, 5 and 1, 6  show similar properties\n\n1, 6, 7, 13, 20, 33, 53, 86, 139\n1, 5, 6, 11, 17. 28, 45, 73, 118\n1, 4, 5, 9. 14, 23, 37, 60, 97, \n1, 3, 4. 7, 11, 18, 29, 47, 76,\n1. 2. 3, 5, 8, 13, 21, 34, 55\n\nand they all converge on the golden ratio too.\n\nOne can experiment with other generating formulae\n\neg each term = the sum of the 3 previous terms\n\n0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81\n\nor each term is twice the previous term plus the one prior to that\n\n0, 1, 2, 5, 12, 29, 70, 169\n\nand explore the properties of such series. Have fun!
281955
1) The sum of the positive 3-digit odd integers = 101 + 103 + 105 + ... + 995 + 997 + 999.  Adding the first of these to the last, we get 101 + 999 = 1100.  Adding the second of these to the next to last, we get 103 + 997 = 1100.  This pattern keeps repeating until 549 + 551 = 1100.  The question then becomes, how many of these 1100's do we have?  We have one for 101, 103, 105, ..., 549 (the later numbers share their 1100's with these).  How many numbers are in the set 101, 103, 105, ..., 549?  Well, there are 449 numbers in 101, 102, 103, ..., 549.  Since there is one more odd number than even number, we must have 225 odd numbers and 224 even numbers.  Thus, there are 225 odd numbers in the set 101, 103, 105, ..., 549.  Thus, we have 225 1100's to add together.\n\n225 * 1100 = 247500.\n\n\n\n2) This time we wish to add the numbers of the set 10, 11, 13, 14, ..., 94, 95, 97, 98.  Adding the first to the last we get 10 + 98 = 108.  Adding the second to the second to last we get 11 + 97 = 108.  The pattern keeps repeating until we get 53 + 55 = 108.  How many 108's do we have?  Well, we have one for 10, 11, 13, 14, 16, 17, ..., 52, 53 (again, the later numbers share their 108's with these).  How many numbers are in this set?  In the set 10, 11, 12, 13, ..., 52, 53, there are 44 numbers.  The multiples of three range from 12 to 51--from 3*4 to 3*17.  Thus, there are 14 multiples of three in this range.  This means that there are 30 numbers in the set 10, 11, 13, 14, 16, 17, ..., 52, 53.  Thus, we have 30 108's to add together.\n\n30 * 108 = 3240.\n\n\n\n3) Here, I am assuming that the raise of 10% also applies to the previous year's raise and not just to the "principal" of $18000.  Thus, I am assuming a compound raise instead of 5 simple raises.\n\nDuring the first year, you make $18000.  During the second year, you make 18000 * 1.1 = $19800.  During the third year, you make 19800 * 1.1 = $21780.  During the fourth year, you make 21780 * 1.1 = $23958.  Finally, during the fifth year, you make 23958 * 1.1 = $26353.80.\n\nGood luck.
368117
That sounds familiar... Think I'm working on it now! But I don't remember q. It is n in my book. In which case, q (if I'm thinking right) is the number of samples. Then again what's up with the ^'s? If np was > 10, then it meant that the distribution was approx. normal. You might be ahead of me right now...
99191
In the experiment of testing for the number of students out of 37 who consider statistics exciting\n\nNo. of trials made \n==> No. of students tested = 37\n==>n = 37\n\nLet the event of a student tested being intersted in the subject of statistics be a success\n\nProbability of success in each trial\n= probability that a student tested is interested in the subject of statistics\n==> p = 0.75\n[Given that 75% of the population of all business statistics 1 students consider statistics exciting]\n\nTherefore, \nq = 1 - p ==> q = 1 - 0.75 ==> q = 0.25\n\nLet "x" indicate the number of successes in "n" trials\nsince the number of students who are interested in statistics can be either 0, 1, 2, .... 37\nThe values "x" can carry are 0, 1, 2, .... 37\n==> "X" is a discrete random variate with range = {0, 1, 2, .... 37}\n\nThe probability distribution of "x" is a binomial distribution..\n\n[Since there a number of trials all of which are identical and independent with only two possible outcomes in each trial either a success i.e. the student being interested or a failure i.e. the student not being interested, with the probability of success in each trial remaining the same all throught]\n\nThe probability function of a binomial distribution\n==> P(X=x) = nCx * p^x * q^(n-x)\n\nTherefore, for the above distribution\nP(X=x) = 37Cx * (0.75)^x * (0.25)^(37-x)\n==>P(X=x) = 37Cx * (0.75)^x * (0.25)^(37) * (0.25)^(-x)\n==>P(X=x) = 37Cx * (0.75)^x * (0.25)^(37) * 1/(0.25)^(x)\n==>P(X=x) = 37Cx * (0.75/0.25)^x * (0.25)^(37)\n==>P(X=x) = 37Cx * (3)^x * (0.25)^(37)\n\nTherefore, \nProbability of observiing 36 or more of the students who consider statistics to be an exciting subject\n==> P(X >= 36) = P(X=36) + P(X=37)\n\n\nP(X = 36) = C(37,36) * (3)^36 * (0.25)^(37)\nP(X = 37) = C(37,37) * (3)^37 * (0.25)^(37)\n\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/