I give up

[Michael Dunn]

Let a = number of apples bought, and o = number of oranges bought. Then you have: 29_a_ + 39_o_ = 1999 You can't solve this equation in isolation, because to solve for 2 unknowns you need 2 equations. You have the restriction that both a and o must be integers, but I can't think of an equation for that in the 30 seconds I've been considering this problem ;) --Mike-- Visual C++ MVP :cool: LINKS~! Ericahist | 1ClickPicGrabber | CP SearchBar v2.0.2 | C++ Forum FAQ Strange things are afoot at the U+004B U+20DD

[lucy 0]

don't think the clerk gave the correct change. I did this and it returned no answer: bool bAnswer = false; for ( nApple = 0; !bAnswer && nApple < 19.99/0.29; nApple ++ ) { nOrange = (19.99 - nApple * 0.29) / 0.39; if ( (nOrange * 0.39 + nApple * 0.29) == 19.99 ) bAnswer = true; } return bAnswer;

[Steve Mayfield]

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

[Matias Szulman]

Thanks!

[Steve Mayfield]

There are problems with floating point rounding errors in your solution which is why you did not get a match. Steve

[BlackDice]

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 Sub

My articles www.stillwaterexpress.com BlackDice

[DavidNohejl]

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

[Steve Mayfield]

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

[DavidNohejl]

:doh: that's true David Never forget: "Stay kul and happy" (I.A.)
David's thoughts / dnhsoftware.org / MyHTMLTidy

[Srikanth Remani]

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.

[Gary Wheeler]

Does your solution still work if you run the application on a MacTM? (ducks) :laugh:

---

Software Zen: delete this;

[lucy 0]

brilliant!

[lucy 0]

ouch, :-O

[Graham Bradshaw]

That's a beautifully worked argument! :-D

[igor1960]

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

[generic_user_id]

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

[Rui A Rebelo]

Very nice! :) Rui A. Rebelo Computers are useless, they can only provide answers. Pablo Picasso

[Brit]

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[^]

[Bassam Abdul Baki]

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

[Steve Mayfield]

Only if the case color is orange :laugh: Steve