We say fenders(factors enders) of 156 are 1, 2, 3, 4, 6, 8, 9, and that 156 is a 7-fender. \n\na) Find a 1-fender which is compostite\nb) Show that a number which has 0 and 9 as fenders has at least four more fenders.\n\nc) the smallest 10-fender is less than 700. Find it and explain why it is the smallest.\n\nAnother 7-fender is 460, its set of fenders is 0, 1,2,3,4,5,6, which is different from the set of fenders of 156.\n\nd) Find three 9-fenders less than 1000 with different sets of fenders.\n\n\n\nPlease answer to the best of your capabilities. The best answer would receive 10 points and a personal thank you message to express my gratitudes to the person. This teaser is extremely crucial in preparing me for my exam and i would really urge all readers to contribute the best to their abilities. Thanks a million to the readers, a billion the contributors and finally a zillion to the person who gives me the most accurate and correct answers.
128237
Basically this question is one of trial and error. You have to think of the many possibilities and try eliminating some to get the desired results. It would be mind-boggling to finish these questions off in half an hour.
353598
f = 30 x 25 x 20 x 15 x 10 x 5 x the other 24 integers. \n\nSo f contains 7 factors of 5 (25 supplies 2 of them) and more than 7 factors of 2. \n\n(The 15 even numbers supply a total of more than 15 such factors as there are three powers of 2 and 3 of the other even numbers (12, 20 & 28) are also divisble by 4 and 24 is divisible by 8)\n\nBut 7 factors of 2 is all we need for our purposes and even if we had a multitude of surplus 2s, they cannot give us another 10\n\ni.e. f contains 7 factors of 10. And ONLY 7 such factors as each 10 can only be made up of a 5 and a 2 \n\nso d = 7 is the only possible answer, given that we are told that d >6\n\nGENERALISING:\n\n50, 75, 100, also supply two 5s to a factorial they are part of, \n\nso for f = 55! and d> 12, d must = 13\nso for f = 80! and d> 18, d must = 19\nso for f = 105! and d> 24, d must = 25\n\nhowever\n\nfor f = 130! and d> 30, d is not 31 but 32\nCAN YOU SEE WHY?
46211
As there is 22 elements in first call,\nOur first is 0\nmiddle is 11\nend is 21\nNow 11th element is 35 which is less than 37.\nSo now our first is changed to 12\nand middle is 17\nend is 21.\n17th element is 42 which is greater than 37\nSo now we will chage end to 16.\nfirst is 12 and middle is 14.\n14th element is 42>37\nSo we change end to 13, first is 12 and middle is 13.\nNow 13th element is 37. So we will now reurn 13.
477333
Hi, \n\nI'll help with the first one, the rest are similar so this should help:\n\n8(k+3)=12k-4\n\n8*k +8*3 = 12k - 4\n\n8k + 24 = 12k - 4\n\n8k + 28 = 12k \n\n28 = 4k\n\nk = 28/4 =7\n\nGood Luck!\nMatt\n\nP.S. I think it's ironic that someone who misspells 2/3 of the words in her post is calling someone else retarded. Seriously, is her time that valuable that she needs to abbreviate the word 'your'?
457875
Boy, this is tricky.  I think the probability is (8! * 41c2 * 10c2)/49!, where 41c2 = 41!/39!2! and 10c2 = 10!/8!2!  In this number, 8! represents the correct picks, 41c2 the incorrect picks, and 10c2 the number of ways to order the picks.\n\nFor exactly seven correct numbers in 10 picks , only a minor change is needed.  It becomes (7! * 42c3 * 10c3)/49!.  Follow this pattern for 6 of 10, 5 of 10, etc
264094
Let's visualise these by size, Water Melons cost 40 schillings, apples cost 10 schillings and cherries cost 1 schilling each.\n\nSuppose our Austrian fruit and veg merchant buys 100 cherries at 1 schilling each, he spends 100 schillings. For the bigger fruit he puts a cherry back on the stall and takes a water melon instead for 39 further schillings, or ditto an apple for 9 further schillings.\n\nSo we ask "is 39 a factor of 159 (259 - 100)?" and "is 9 a factor of 159?" and the answer is No as 159 is prime. So we have to rule out solutions that involve buying no water melons and solutions that involved buying no apples.\n\nWe notice that 4 water melons is the maximum as 4 x 39 = 156 and more than that would burst the budget.\n\n1 water melon would mean 120 schillings for apples but 120 (modulo 9) = 3\n\n3 water melons would mean 42 schillings for apples but 42 (modulo 9) = 6\n\n2 water melons would mean 81 schillings for apples and 81 (modulo 9) = 0 (it is 9 squared)\n\nSo clearly the answer is 2 water melons, 9 apples and 89 cherries remain having swopped 11 pf them for larger items of fruit.
330071
this is 100% correct\n\nhttp://www.mxtabs.net/tab_versions.php?path=Guitar,g,563,Green+Day,Good+Riddance+%28time+Of+Your+Life%29,234265
83501
ok so for A, the first set has 2 seats. so between the 2 seats, theres 2.5feet of steel. and then from each seat to the end, theres 2 more feet (on each side) so thatz 4 feet + 2.5feet = 6.5 feet for the first set.\nfor the 2nd set, 3 seats so 2*2.5 between the seats (2.5 between seats 1 and 2 and then 2.5 more between seats 2 and 3) + 4 =9feet of steel for the 2nd set.\nfor the 3rd set, theres 4 seats so that means (3*2.5) + 4 = 11.5feet of steel.\n\nso you should get the pattern hopefully. if theres 5 seats, then 4*2.5 + 4. 6 seats is 5*2.5 + 4.\nso then the expression would be [(n-1)*2.5] + 4. i dont really know how to explain why its correct. i guess you could just say  it works and it makez sense. for n seats, theres n-1 gaps between the seats and each gap is 2.5feet long. and then theres alwayz 2 feet on both ends added to it which is the +4. \n\nanywayz hope this helps. let me know if u need mroe help or whatever. laterz
439972
a) 4 is in the tens place, so it is worth 4 * 10 = 40\nb) 6 is in the ones place, so it is worth 6 * 1 = 6\nc) 3 is in the thousands place, so it is worth 3 * 1000 = 3000\nd) first 2 is in the hundreds place, so it is worth 2 * 100 = 200\ne) 7 is in the hundreds place, so it is worth 7 * 100 = 700\n\nOr you can just say the numbers and see what you say when you are on the underlined number...\n\n42 = "forty two" so 4 is "forty"\n176 = "one hundred seventy six" so 6 is "six"\n3049 = "three thousand forty nine" so 3 is "three thousand"\n1223 = "one thousand two hundred twenty three" so first 2 is "two hundred"\n10731 = "ten thousand seven hundred thirty one" so 7 is "seven hundred"
654475
The two numbers for Rusty are because he drove for two different owners in his career, thus two different cars and two different numbers. As far as the gift goes, it sounds good. My husband is a huge fan and he would love that. If he collects already you may want to find out what the ratio is. 1:64 is Hot Wheels size. 1:24 is the larger size. Most guys collect the bigger ones. (go figure...they need to have the bigger ones!) If it is autographed make sure it has a certificate of authenticity! Even cooler is a picture of the driver signing that thing....the person selling the item sometimes also has a picture. Good luck!
510664
One way to attack this is to see how many ways 20 books can be put  in a row and then see how may ways that row can be divided into 5 shelves. \n\nFor the row of 20, there are 20 choices for the first book, for each of the 20 choices of the first book there are always 19 left for the second giving 20*19 ways of picking the first two books.  Extending this on there are 18 choices for the third, etc.  So thee are 20*19*18*17.......*4*3*2*1or 20! ways to arrange the 20 books in order. That's 2.43 x 10^18.\n\nNow for each ordering, there are different ways to divide the order, without shuffling, among the shelves.  Since we have the books in order, we will keep them that way with the first set going in order to shelf one, next group to two, etc.  That way, we only need to figure how many ways the 20 books can be divided without regard to which book is which.  Each ordering above can go into any of the schemes to divide the books among the shelves so you just multiply the two numbers together to get the total number of possibilities.\n\nSince each shelf must have two books minimum, start by putting two books on each shelf and leaving them there.  The problem then comes down to how many ways can the remaining 10 books be divided among 5 shelves.  We really want to put 4 divisions along the row of 10 books to tell where one shelf ends and the other begins.  This gives 14 locations to be occupied by either a book or divider.  The number of possible arrangements of the 14 locations is 14!.  But since all dividers look the same and all books (for now) look the same, there are only 14!/(10! * 4!) = 1001 possible ways to divide it onto the shelves.\n\nSo there are 1001 ways to divide the 20 books among the shelves for each of the 20! Arrangements of books.  Multiplying these together gives 1001*20! = 2.43 x 10^21 ways to arrange the books.  If you want to see all the arrangements, I suggest you start arranging books right away.  If you put together one arrangement of books per second it would take about 5000 times the age of the universe to go through all of them.
533434
Ok.. Don't laugh at me ... much.  I came up with 6 different possibiites of ages\n1x2x18  ----21 mangoes in the basket.  ( somewhere between 2-5 lbs each means the basket weighed around 60 lbs)  \n\n1x3x12-----16 mangoes in the basket (  48 lbs)\n1x6x6--- eliminate this answer because there is an "oldest" \n1x4x9--- 14 mangoes (42 lbs)\n2x2x9-- 13 mangoes ( 39 lbs)\n3x3x4---10 mangoes (30 lbs)  could you imagine the insanity of that household???\n\nThe reason for the third clue?  Probably not a native where the mangoes grow (latin america and south america) So maybe the basket is a shopping basket?\n\nWow. to narrow this down one would have to make many assumptions.\nSo any of the previous answers should work depeding on the type of basket.
118832
Ubergeek is right again, almost.  Half the cards are black, so prob that the first one is is 1/2.  After that, only 25 of the remaining 51 cards are, so prob that the second is black is 25/51, and prob both are is the product, 1/2 * 25/51 = 0.245098, roughly.  I used to teach this stuff.
102464
14 bolt is PLENTY strong!\n\nBut why not sell it to a 4 X 4 enthusiast, use that money to go towards a narrowed 9" rear?\n\nI question the availability of gear ratios for that particular style of rear...That's one of the reasons why more people  aren't doing what you are considering right now!  Also think about housing ends, brakes, axles with the proper bolt patterns, etc...
510483
I'm assuming these pipes are going to work like organ pipes and the resonant frequency of each will be inversely proportional to its length:\n\nTo cover 1 octave, the first (lowest note) pipe will be twice as long as the last (highest note) pipe.  There will be 13 pipes total including the 1st and last.  To get the length of the 12th pipe, multiply the length of the 13th pipe (the shortest) by the 12th root of 2.0\nDo the same for each of the pipes in order, making sure any given pipe is 2^(1/12) as long as the shorter one next to it.  This can all be done on paper beforehand and result in a list of exactly how long to cut each pipe.\nEach pipe will sound a note which is 1/2 step different from its neighbor.
35206
Hi again, just graph the three boundary lines and shade the correct sides.  This will give you a shaded triangle with vertices (2,3), (4,5), and (4,0)  [I think; I don't have graph paper handy].  The maximum and minimum values must occur at one of these 3 sets of coordinates.\n\nfor (2,3), 3x + y = 9\nfor (4,5), 3x + y = 17\nfor (4,0), 3x + y = 12\n\nSo 9 is the minimum value and 17 is the maximum.\n\nCheck it out yourself to be sure the graphs really intersected where I thought they did!
646154
Storage_ I corrected a miscount.\n\nThe answer is 12+24 = 36\n\nPlanes that include 4 vertices = 12\n\nobviously the six faces = 6\nPlanes that pass through the two parallel diagonals of opposing faces. There are two of these for each face but since they join two opposing faces there are 2*3 = 6\n\n24 planes that include only 3 vertices\nFor each face of the cube there are two diagonals and each digonal belongs to two planes that pass through the two vertices which are kitty corner to the end-points of the diagonals. this gives 2*2*6 =24