I give up
-
There are problems with floating point rounding errors in your solution which is why you did not get a match. Steve
-
a for apple, o for orange 29a + 39o = 1999 29(a + o) + 10o = 1999 lets call a + o = t (total) 29t + 10o = 1999 we know o is a whole number, o = (1999 - 29t)/10, for this to be a whole number 29t would be ***9 (has 9 in units place) so t would be *1 (has 1 in units place) so our probable answer set (for total fruits) are 1, 11, 21, 31, 41, 51, 61, 71 ... for lower limit a=0, so t = o (all fruits are oranges, since oranges are costlier) 39t = 1999, t = 1999/39 = 51.256 (just more than 51) for upper limit o = 0, so t =a (all fruits are apples, since apples are cheaper) 29t = 1999, t = 1999/29 = 68.931 (just less than 70). the only number that ends with *1 in this range is 61, so t = 61. so, 29*61 + 10o = 1999, 10o = 1999 - 1769, o = 230/10 = 23. a = 61- 23 = 38.
That's a beautifully worked argument! :-D
-
Today I realised that math is definitively not for me. For fun I was trying to solve some puzzles, and I've been stuck with this one for about an hour. I know some of you really get it, so could I ask for some help? Thanks! A fruitstand in my neighborhood sells apples for 29 cents apiece, and oranges for 39 cents apiece. One day I went there and filled a basket with a mixture of apples and oranges. When I paid for them I handed the clerk a $20 bill, and she handed me back my change ... one penny. How many apples did I buy? (Assume there was no sales tax charged.) Matias
29a + 39o = 1999 a + o = 61 <= because if we assume that only apples you bought, then we get 1999/29=68. and if we assume that only oranges: 1999/39=51.25 meaning that you got >=52 and <=68 overall. However, the price you paid ends with 9 (1999) -- as each cost ends with 9 (29,39) --> the multiplier must finish with 1. The only value between 52 and 68 with 1 at the end is 61. So, resolve the system 29a + 39o =1999 a+o=61 =========== 29a + 39(61-a)=1999 29a + 39*61-39a=1999 -10a=1999-39*61 -10a=1999-2379 -10a=-380 a=38 o=61-38=23 :laugh: "...Ability to type is not enough to become a Programmer. Unless you type in VB. But then again you have to type really fast..." Me
-
Today I realised that math is definitively not for me. For fun I was trying to solve some puzzles, and I've been stuck with this one for about an hour. I know some of you really get it, so could I ask for some help? Thanks! A fruitstand in my neighborhood sells apples for 29 cents apiece, and oranges for 39 cents apiece. One day I went there and filled a basket with a mixture of apples and oranges. When I paid for them I handed the clerk a $20 bill, and she handed me back my change ... one penny. How many apples did I buy? (Assume there was no sales tax charged.) Matias
I'll let the solution speak for itself: (Using Haskell set logic)
solution :: [(Int, Int)] solution = [ (apples, oranges) | apples <- [0..100], oranges <- [0..100], apples * 29 + oranges * 39 == 1999 ]Result:[(38,23)]
Hence, the answer must be 38 apples. Haskell really shines sometimes. I don't think solutions really get more elegant than this. :-D
-
a for apple, o for orange 29a + 39o = 1999 29(a + o) + 10o = 1999 lets call a + o = t (total) 29t + 10o = 1999 we know o is a whole number, o = (1999 - 29t)/10, for this to be a whole number 29t would be ***9 (has 9 in units place) so t would be *1 (has 1 in units place) so our probable answer set (for total fruits) are 1, 11, 21, 31, 41, 51, 61, 71 ... for lower limit a=0, so t = o (all fruits are oranges, since oranges are costlier) 39t = 1999, t = 1999/39 = 51.256 (just more than 51) for upper limit o = 0, so t =a (all fruits are apples, since apples are cheaper) 29t = 1999, t = 1999/29 = 68.931 (just less than 70). the only number that ends with *1 in this range is 61, so t = 61. so, 29*61 + 10o = 1999, 10o = 1999 - 1769, o = 230/10 = 23. a = 61- 23 = 38.
Very nice! :) Rui A. Rebelo Computers are useless, they can only provide answers. Pablo Picasso
-
Today I realised that math is definitively not for me. For fun I was trying to solve some puzzles, and I've been stuck with this one for about an hour. I know some of you really get it, so could I ask for some help? Thanks! A fruitstand in my neighborhood sells apples for 29 cents apiece, and oranges for 39 cents apiece. One day I went there and filled a basket with a mixture of apples and oranges. When I paid for them I handed the clerk a $20 bill, and she handed me back my change ... one penny. How many apples did I buy? (Assume there was no sales tax charged.) Matias
Srikanth had the same idea I did, so I thought I'd offer a different solution. How many apples can I get for 1999 or less? 68 (with 27 pennies left over). Knowing that, (a) 1 apple + 10 pennies = 1 orange (b) 4 apples + 1 penny = 3 oranges In order to use up my 27 pennies, I do two "a" exchanges (costing a total of 20 pennies) to get me down to 7 pennies and then do 7 "b" exchanges (costing 7 pennies). 68 apples + 27 pennies --> 66 apples + 2 oranges + 7 pennies --> (66-28) apples + (2+21) oranges + 0 pennies = 38 apples + 23 oranges + 0 pennies left ----------------------------------------------------- Empires Of Steel[^]
-
Today I realised that math is definitively not for me. For fun I was trying to solve some puzzles, and I've been stuck with this one for about an hour. I know some of you really get it, so could I ask for some help? Thanks! A fruitstand in my neighborhood sells apples for 29 cents apiece, and oranges for 39 cents apiece. One day I went there and filled a basket with a mixture of apples and oranges. When I paid for them I handed the clerk a $20 bill, and she handed me back my change ... one penny. How many apples did I buy? (Assume there was no sales tax charged.) Matias
29a + 39o = 1999 29a ≤ 1999 ⇒ 0 ≤ a ≤ 68 39o ≤ 1999 ⇒ 0 ≤ o ≤ 51 29a + 39o ≡ 9a + 9o = 9(a + o) ≡ 9 (mod 10) ∴ a + o ≡ 1 (mod 10) Since 29 and 39 do not divide 1999, 29(a + o) < 29a + 39o < 39(a + 0) ⇒ 51 < a + o < 68. ∴ a + o = 61. a + o = 61 and 29a + 39o = 1999 ⇒ a = 38 and o = 23. "For that one fraction of a second, you were open to options you would never have considered. That is the exploration that awaits you. Not mapping stars and studying nebula, but charting the unknown possibilities of existence." - Q (Star Trek: The Next Generation) Web - Blog
-
Does your solution still work if you run the application on a MacTM? (ducks) :laugh:
Software Zen:
delete this;Only if the case color is orange :laugh: Steve
-
Srikanth had the same idea I did, so I thought I'd offer a different solution. How many apples can I get for 1999 or less? 68 (with 27 pennies left over). Knowing that, (a) 1 apple + 10 pennies = 1 orange (b) 4 apples + 1 penny = 3 oranges In order to use up my 27 pennies, I do two "a" exchanges (costing a total of 20 pennies) to get me down to 7 pennies and then do 7 "b" exchanges (costing 7 pennies). 68 apples + 27 pennies --> 66 apples + 2 oranges + 7 pennies --> (66-28) apples + (2+21) oranges + 0 pennies = 38 apples + 23 oranges + 0 pennies left ----------------------------------------------------- Empires Of Steel[^]
The combination of one apple and one orange costs 68 cents. 10 apples and 10 oranges will cost you $6.80. If you picked 30 apples and 30 oranges, the price comes to $20.40. You remove one orange and the price becomes $20.01, tantalizingly close to $19.99! Now, you need to figure out how many oranges you have to take away and how many apples to add till you get a price difference of 2 cents. 8 apples will cost you $2.32 and 6 oranges will cost you $2.34. Better still, when you round the prices to 30 cents an apple and 40 cents an orange, you see immediately that 8x30 = 6x40 and the difference in count (between 8 and 6) is precisely the 2 cents that you are looking for. Hence 38 apples and 23 (29-6) oranges. No programs. No pencil and paper. No simultaneous equations. Just simple mental arithmetic. Next question please! :cool:
-
I'll let the solution speak for itself: (Using Haskell set logic)
solution :: [(Int, Int)] solution = [ (apples, oranges) | apples <- [0..100], oranges <- [0..100], apples * 29 + oranges * 39 == 1999 ]Result:[(38,23)]
Hence, the answer must be 38 apples. Haskell really shines sometimes. I don't think solutions really get more elegant than this. :-D
That's cheating. ;) -- Schni Schna Schnappi! Schnappi Schnappi Schnapp!
-
29a + 39o = 1999 29a ≤ 1999 ⇒ 0 ≤ a ≤ 68 39o ≤ 1999 ⇒ 0 ≤ o ≤ 51 29a + 39o ≡ 9a + 9o = 9(a + o) ≡ 9 (mod 10) ∴ a + o ≡ 1 (mod 10) Since 29 and 39 do not divide 1999, 29(a + o) < 29a + 39o < 39(a + 0) ⇒ 51 < a + o < 68. ∴ a + o = 61. a + o = 61 and 29a + 39o = 1999 ⇒ a = 38 and o = 23. "For that one fraction of a second, you were open to options you would never have considered. That is the exploration that awaits you. Not mapping stars and studying nebula, but charting the unknown possibilities of existence." - Q (Star Trek: The Next Generation) Web - Blog
This is exactly the method I used to solve the problem. (congruency and equivalence relations). Ahh, the good old days of discrete mathematics! -Sean ---- Shag a Lizard
-
This is exactly the method I used to solve the problem. (congruency and equivalence relations). Ahh, the good old days of discrete mathematics! -Sean ---- Shag a Lizard
Gotta love Mathematics. Nothing you can't prove or counter-prove with it. :-D "For that one fraction of a second, you were open to options you would never have considered. That is the exploration that awaits you. Not mapping stars and studying nebula, but charting the unknown possibilities of existence." - Q (Star Trek: The Next Generation) Web - Blog
-
Gotta love Mathematics. Nothing you can't prove or counter-prove with it. :-D "For that one fraction of a second, you were open to options you would never have considered. That is the exploration that awaits you. Not mapping stars and studying nebula, but charting the unknown possibilities of existence." - Q (Star Trek: The Next Generation) Web - Blog
Especially Goedel's theorem.;P
I can imagine the sinking feeling one would have after ordering my book, only to find a laughably ridiculous theory with demented logic once the book arrives - Mark McCutcheon
-
Especially Goedel's theorem.;P
I can imagine the sinking feeling one would have after ordering my book, only to find a laughably ridiculous theory with demented logic once the book arrives - Mark McCutcheon
Reminds me of a cousin of mine who used to say, if God is all-powerful, why can't he create a rock that is so big, that even he can't lift it, and if he can't lift it, how is he so powerful? I think that Goedel's theorem would fall under the prove that we can't prove it, in which case we have a proof (of sorts). :) And yes, you're a smart-ass. :-D "For that one fraction of a second, you were open to options you would never have considered. That is the exploration that awaits you. Not mapping stars and studying nebula, but charting the unknown possibilities of existence." - Q (Star Trek: The Next Generation) Web - Blog
-
a for apple, o for orange 29a + 39o = 1999 29(a + o) + 10o = 1999 lets call a + o = t (total) 29t + 10o = 1999 we know o is a whole number, o = (1999 - 29t)/10, for this to be a whole number 29t would be ***9 (has 9 in units place) so t would be *1 (has 1 in units place) so our probable answer set (for total fruits) are 1, 11, 21, 31, 41, 51, 61, 71 ... for lower limit a=0, so t = o (all fruits are oranges, since oranges are costlier) 39t = 1999, t = 1999/39 = 51.256 (just more than 51) for upper limit o = 0, so t =a (all fruits are apples, since apples are cheaper) 29t = 1999, t = 1999/29 = 68.931 (just less than 70). the only number that ends with *1 in this range is 61, so t = 61. so, 29*61 + 10o = 1999, 10o = 1999 - 1769, o = 230/10 = 23. a = 61- 23 = 38.
Amazing! :omg: What's even more amazing is that your very first post has garnered 11 5 votes and still counting. :cool: Vikram.
http://www.geocities.com/vpunathambekar "It's like hitting water with your fist. There's all sorts of motion and noise at impact, and no impression left whatsoever shortly thereafter." — gantww.
-
Reminds me of a cousin of mine who used to say, if God is all-powerful, why can't he create a rock that is so big, that even he can't lift it, and if he can't lift it, how is he so powerful? I think that Goedel's theorem would fall under the prove that we can't prove it, in which case we have a proof (of sorts). :) And yes, you're a smart-ass. :-D "For that one fraction of a second, you were open to options you would never have considered. That is the exploration that awaits you. Not mapping stars and studying nebula, but charting the unknown possibilities of existence." - Q (Star Trek: The Next Generation) Web - Blog
Bassam Abdul-Baki wrote: And yes, you're a smart-ass. Sorry, I just couldn't resist.:-D
I can imagine the sinking feeling one would have after ordering my book, only to find a laughably ridiculous theory with demented logic once the book arrives - Mark McCutcheon
-
Especially Goedel's theorem.;P
I can imagine the sinking feeling one would have after ordering my book, only to find a laughably ridiculous theory with demented logic once the book arrives - Mark McCutcheon
You beat me to it. :-D Vikram.
http://www.geocities.com/vpunathambekar "It's like hitting water with your fist. There's all sorts of motion and noise at impact, and no impression left whatsoever shortly thereafter." — gantww.
-
a for apple, o for orange 29a + 39o = 1999 29(a + o) + 10o = 1999 lets call a + o = t (total) 29t + 10o = 1999 we know o is a whole number, o = (1999 - 29t)/10, for this to be a whole number 29t would be ***9 (has 9 in units place) so t would be *1 (has 1 in units place) so our probable answer set (for total fruits) are 1, 11, 21, 31, 41, 51, 61, 71 ... for lower limit a=0, so t = o (all fruits are oranges, since oranges are costlier) 39t = 1999, t = 1999/39 = 51.256 (just more than 51) for upper limit o = 0, so t =a (all fruits are apples, since apples are cheaper) 29t = 1999, t = 1999/29 = 68.931 (just less than 70). the only number that ends with *1 in this range is 61, so t = 61. so, 29*61 + 10o = 1999, 10o = 1999 - 1769, o = 230/10 = 23. a = 61- 23 = 38.
-
29a + 39o = 1999 29a ≤ 1999 ⇒ 0 ≤ a ≤ 68 39o ≤ 1999 ⇒ 0 ≤ o ≤ 51 29a + 39o ≡ 9a + 9o = 9(a + o) ≡ 9 (mod 10) ∴ a + o ≡ 1 (mod 10) Since 29 and 39 do not divide 1999, 29(a + o) < 29a + 39o < 39(a + 0) ⇒ 51 < a + o < 68. ∴ a + o = 61. a + o = 61 and 29a + 39o = 1999 ⇒ a = 38 and o = 23. "For that one fraction of a second, you were open to options you would never have considered. That is the exploration that awaits you. Not mapping stars and studying nebula, but charting the unknown possibilities of existence." - Q (Star Trek: The Next Generation) Web - Blog
The solution is quite elegant. seems to me that I should go back and refine my discrete mathematics and modern algebra. Blood, Sweat and Code
-
Today I realised that math is definitively not for me. For fun I was trying to solve some puzzles, and I've been stuck with this one for about an hour. I know some of you really get it, so could I ask for some help? Thanks! A fruitstand in my neighborhood sells apples for 29 cents apiece, and oranges for 39 cents apiece. One day I went there and filled a basket with a mixture of apples and oranges. When I paid for them I handed the clerk a $20 bill, and she handed me back my change ... one penny. How many apples did I buy? (Assume there was no sales tax charged.) Matias
In .net and C#.... int apple,orange = 0; //surely apple wont be more than 69 for(apple=0;apple<69;apple++) { long temp1,temp2 = 0; temp1 = Math.DivRem((1999-39*apple),29,out temp2); if(temp2 == 0) { orange = (int)temp1; break; } } //Get the ornage and apple here! :-D