Enter your email address:
Any number which is not a multiple of 6,9,20,54,120,180,1080.
I caculated that 49 is the greatest number of chicken pieces which can be ordered for which there there are no combinations of mutiples of 6,9,20. I calculated all the combinations which fulfill orders from 50 to 100 and conjecture that all chicken orders greater than 49 can be fulfilled! - GAK
49 is not correct - two 20's and a9 equal 49. Also, the first comment can't be correct, either. 15 is not a multiple of the listed numbers, but 6+9 does the trick easily. I think the answer is 37. This is not an elegant proof, but I think it works. The problem is of the form 6A+9B+20C = D. We must find Max D' where D' <> D and D' and D are in the set of whole numbers. It can be shown for any order over 6 that every number divisible by 3 is covered just by considering the 6 and 9 piece orders...if 6A+9B = 3C where C is a whole number greater than 1...then 2A+3B = C and it can easily be seen that for every C over 2 there is a combination of A and B that reaches it (set B = 0 and you can cover every even number with some number A and every odd number is covered by A-1 and B=1). So our existing gap is orders below 6 pieces and any number not divisible by 3.I think we can agree that if by magic we were given or could reach the numbers -1 and -2, then we would cover every existing whole number 4 and over (all multiples of 3 and by using -1 and -2 all the numbers in between). By that logic, consider whole number mulitple orders (C) of the larger 20-piece product. So 20C = D. For C=1, 20 mod 3 = 2 (20 divided by 3 has a remainder of 2). So by adding 20 to sets of 6 and 9 we can cover all multiples of 3 over 20 plus 2. (for instance we had already covered multiples of 3 - 6,9,12,15,18,21,24,27, etc...now we also cover 20+3=23 which is 21+2, 20+6=26 which is 24+2, 20+9=29 which is 27+2..etc...) So now we have covered all whole numbers N over 6 divisible by 3 and whole numbers M over 20 where M mod 3 = 2 (M divided by 3 is remainder 2). Not covered is all whole numbers P over 20 with P mod 3 = 1 ( for instance 43 is 42 divided by 3 = 14 with a remainder of 1). Now consider the sets of 20 again where C=2. You can reach 40 with 2sets. 40 mod 3 = 1 (39/3 = 13 remainder 1). So now I can reach all remainder 1 numbers over 40. I have now covered all numbers divisible by 3, all numbers divisble by 3 with a remainder 2 over 20 and all numbers divisble by 3 with a remainder 1 over 40. So all numbers over 40 are accounted for. Under 40, what is't covered is numbers under 6, numbers under 20 mod 3 not equal to zero (remainder 1 or 2) and numbers under 40 mod 3 = 1. The largest number D under 40 where D mod 3 = 1 is 37.
infinite its a vegetarian restaurant
Let us read the problem again. It is a NON vegaterian resturant. It also states how many pieces CAN NOT be ordered, not how many Can. The answer is Any number that is NOT 6, 9, or 20 or a multiple of 6, 9, or 20 can NOT be ordered.
answer is 20
I think the answer is 37. I really appreciated your effort. Online study now very useful for the students. Students like these types of resources. I already found a website of math practice for my kids, Both my kids start learning from this website free of cost. Learn Math Fast
I was thinking about my proof and realized I made an error...so I am surprised no one called me out on this, but the answer turns out not to be 37. It is 43. The logic above is right, but I missed something important. This gives me a chance to adjust and summarize my rambling proof to make it a bit clearer. Basically, all multiples of 3 EXCEPT 3 itself are covered by 6 and 9. If you do a little bit of work, you can prove this for yourself. Add the pack of 20. Since 20 is a "remainder 2" number when dividing by 3, using 20 covers all numbers when dividing by 3 that give a remainder of 2 (26, 29,32,35 etc) EXCEPT the first one, 23 (just like the first 3), because we can use 20 to start at a remainder 2 number (20) and just use 6 and 9 to cover all the numbers of the form "20+multiple of 3" (remainder 2 numbers) greater than 26. So we have covered all multiples of 3 over 6, and all "remainder 2" numbers when dividing by 3 over 26. So far, we haven't covered any "remainder 1" numbers when dividing by 3. Now add another 20-pack and start at 40. 40 is a "Remainder 1" number when dividing by 3. So we can now cover all "Remainder 1" numbers when dividing by 3 over 40 (46,49,52,55,etc..)EXCEPT the first one - 43. So 43 is the highest number. My mistake was not incorporating the fact that the first 3 was not covered when starting at 20 and 40, just like when starting from 0.
It's better to close that restaurant.. :-P
Get Hourly profit for 200 hours on every hour without any risk and without any work, best business plans everAllTimeProfit.com
0 it is a vegitarian restaurant and therefore will not sell chicken
7*10*21 = 1470?
1073?(6*9*20) -7Can we get confirmation of the correct answer please? ... then I'll wrap my head around the proof.
43Good explanation herehttp://en.wikipedia.org/wiki/Coin_problem
oakley sunglasses, oakley vault, jordan shoes, polo ralph lauren outlet, burberry outlet online, michael kors outlet online, true religion, christian louboutin shoes, tiffany and co jewelry, louis vuitton outlet, gucci handbags, christian louboutin outlet, nike shoes, michael kors outlet online, longchamp outlet, tory burch outlet, coach outlet, red bottom shoes, louboutin shoes, ray ban sunglasses, coach purses, michael kors outlet store, air max, louis vuitton outlet online, cheap oakley sunglasses, coach outlet store online, coach factory outlet, prada handbags, chanel handbags, ray ban outlet, longchamp outlet online, kate spade outlet online, burberry outlet online, louis vuitton, polo ralph lauren, nike free, prada outlet, louis vuitton handbags, michael kors outlet online, michael kors outlet, michael kors outlet online, louis vuitton outlet, kate spade handbags, longchamp handbags, nike air max
barbour, oakley pas cher, abercrombie and fitch, nike air force, mac cosmetics, ray ban uk, air max, guess pas cher, vans outlet, hermes pas cher, michael kors uk, nike roshe, ralph lauren pas cher, timberland, hollister, longchamp pas cher, sac michael kors, lululemon, sac vanessa bruno, michael kors canada, sac louis vuitton, mulberry, north face pas cher, ray ban pas cher, longchamp, converse pas cher, chaussure louboutin, nike blazer pas cher, longchamp, nike trainers, nike roshe run, lacoste pas cher, louis vuitton uk, scarpe hogan, nike huarache, tn pas cher, louis vuitton, hollister, louis vuitton pas cher, north face, ralph lauren, nike air max, new balance pas cher, vans pas cher, roshe run, abercrombie and fitch, nike free, nike free pas cher, hollister, burberry pas cher
north face outlet, bottega veneta, rolex watches, canada goose, reebok outlet, north face jackets, moncler outlet, beats headphones, ugg boots, nfl jerseys, marc jacobs outlet, chi flat iron, canada goose uk, babyliss pro, insanity workout, timberland shoes, birkin bag, canada goose pas cher, giuseppe zanotti, lululemon outlet, jimmy choo shoes, soccer jerseys, canada goose outlet, new balance outlet, asics shoes, canada goose outlet, celine handbags, moncler, ugg outlet, moncler, mont blanc pens, instyler ionic styler, iphone 6 case, hollister clothing, ghd, p90x workout, ugg soldes, soccer shoes, uggs on sale, nike air max, herve leger, wedding dresses, canada goose outlet, baseball bats, mcm handbags, moncler, ferragamo shoes, valentino shoes, uggs outlet, ugg
The answer is 43.1) Let N >= 44 be an integer number. We shall show that there exists non-negative integers A, B, C such that 6*A + 9*B + 20*C = N. We consider 6 cases depending on the remainder of N when divided by 6.1.0) N = 0 mod 6.Let A = N / 6, B = 0 and C = 0. Since N >= 44 we have A >= 7. In addition 6*A + 9*B + 20*C = 6*N/6 + 9*0 + 20*0 = N.1.1) N = 1 mod 6.Let A = (N - 1)/6 - 8, B = 1 and C = 2. Since N = 1 mod 6 and N >= 44 we have N >= 49. Hence A >= (49 - 1)/6 - 8 = 0. In addition 6*A + 9*B + 20*C = (N - 1) - 48 + 9*1 + 20*2 = N.1.2) N = 2 mod 6.Let A = (N - 2)/6 - 3, B = 0 and C = 1. With arguments similar as before we obtain N >= 44, A >= 4 and 6*A + 9*B + 20*C = N.1.3) N = 3 mod 6.A = (N - 3)/6 - 1, B = 1 and C = 0. We have N >= 45, A >= 6 and 6*A + 9*B + 20*C = N.1.4) N = 4 mod 6.A = (N - 4)/6 - 6, B = 0, C = 2. We have N >= 46, A >= 1 and 6*A + 9*B + 20*C = N.1.5) N = 5 mod 6.A = (N - 5)/6 - 4, B = 1, C = 1. We have N >= 47, A >= 3 and 6*A + 9*B + 20*C = N.Now, we shall show that it's impossible to have 6*A + 9*B + 20*C = 43 with A, B and C non-negative integer numbers. Suppose by contradiction that this is possible.Since A, B >=0 we have 20*C <= 43 which implies C <= 2.If C = 0, then 6*A + 9*B = 43 - 20*C = 43. This cannot be possible since 6*A + 9*B is divisible by 3 and 43 is not.If C = 1, then 6*A + 9*B = 43 - 20*C = 23. This cannot be possible since 6*A + 9*B is divisible by 3 and 23 is not.If C = 2, then 6*A + 9*B = 43 - 20*C = 3, hence 2*A + 3*B = 1. Therefore, 2*A <= 1, i.e., A = 0 and 3*B <= 1, i.e., B = 0. But this leads to another contradiction since 1 = 2*A + 3*B = 2*0 + 3*0 = 0.
Online Easy Jobs from home with data entry, copy pasting, facebook jobswww.jobzcorner.com
Post a Comment