I give up
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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
38 apples and 23 oranges 38 * 29 + 23 * 39 = 1999 okay, okay - I cheated. I wrote a quick VB program to figure it out for me.
Private Sub Command1_Click() Dim i As Integer Dim j As Integer Dim intApples As Integer Dim intOranges As Integer intApples = 29 intOranges = 39 For i = 1 To 40 For j = 1 To 40 If intApples * i + intOranges * j = 1999 Then MsgBox CStr(i) & " apples and " & CStr(j) & " oranges" Exit Sub End If Next Next End SubMy articles www.stillwaterexpress.com BlackDice
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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
Read what Michael wrote... Two
fors should solve it quickly :) pseudocode:// 100 is naive upper estimation for sulution, 100*29 > 1999
for i= 0 to 100 do for j=0 to 100 do if(29*i+39*j == 1999) then print(i, j);There (probably) is logical way how to find solution... but brute force is OK too :) Don't give up math after hour or so... better take a rest, solution will come to your mind suddenly - on toilet, on the way to work or while playing some 3D-action ;) Ahh yeah and I forgot about mirror solutions... (j,i) is same as (i,j) :-O David Never forget: "Stay kul and happy" (I.A.)
David's thoughts / dnhsoftware.org / MyHTMLTidy -
Read what Michael wrote... Two
fors should solve it quickly :) pseudocode:// 100 is naive upper estimation for sulution, 100*29 > 1999
for i= 0 to 100 do for j=0 to 100 do if(29*i+39*j == 1999) then print(i, j);There (probably) is logical way how to find solution... but brute force is OK too :) Don't give up math after hour or so... better take a rest, solution will come to your mind suddenly - on toilet, on the way to work or while playing some 3D-action ;) Ahh yeah and I forgot about mirror solutions... (j,i) is same as (i,j) :-O David Never forget: "Stay kul and happy" (I.A.)
David's thoughts / dnhsoftware.org / MyHTMLTidyMirror solution will not work because the cost is different for apples and oranges. 38*29+23*39=1999 23*29+38*39=2149 Steve
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Mirror solution will not work because the cost is different for apples and oranges. 38*29+23*39=1999 23*29+38*39=2149 Steve
:doh: that's true David Never forget: "Stay kul and happy" (I.A.)
David's thoughts / dnhsoftware.org / MyHTMLTidy -
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
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.
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There isn't a single equation - but you do have constraints The maximum number of apples is 68 (if no oranges are bought) The maximum number of oranges is 51 (if no apples are bought) so,
for(a = 0; a <= 68; a++)
{
if((1999 - a * 29) % 39 == 0)
printf("apples = %d, oranges = %d\n", a, (1999 - a * 29) / 39)
}There is only one match a = 38 and o = 23 Steve
Does your solution still work if you run the application on a MacTM? (ducks) :laugh:
Software Zen:
delete this; -
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.
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There are problems with floating point rounding errors in your solution which is why you did not get a match. Steve
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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
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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
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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
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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
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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[^]
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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
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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
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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:
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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!
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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
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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
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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