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#1
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1+3+5+7+9+11+13+15+17+19=100 make a sum of 5 numbers make 50
1+3+5+7+9+11+13+15+17+19=100
Choose 5 numbers from these 10 figures so there sum is = 50 |
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#2
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1+3+5+7+9+11+13+15+17+19=100 make a sum of 5 numbers make 50
Ali -
Tushar Mehta describes two methods with references to others: http://www.tushar-mehta.com/excel/te...ues/index.html - Mike Middleton http://www.DecisionToolworks.com Decision Analysis Add-ins for Excel "" wrote in message ... 1+3+5+7+9+11+13+15+17+19=100 Choose 5 numbers from these 10 figures so there sum is = 50 |
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#3
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1+3+5+7+9+11+13+15+17+19=100 make a sum of 5 numbers make 50
Ali wrote:
1+3+5+7+9+11+13+15+17+19=100 Choose 5 numbers from these 10 figures so there sum is = 50 The sum of 5 Odd numbers is Odd also, so I don't believe there is a solution as stated that equals an even number (ie 50) - - - HTH Dana DeLouis |
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#4
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1+3+5+7+9+11+13+15+17+19=100 make a sum of 5 numbers make 50
1+3+5+7+15+19
-- Gary''s Student - gsnu200815 |
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#5
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1+3+5+7+9+11+13+15+17+19=100 make a sum of 5 numbers make 50
On Nov 22, 4:24*am, Dana DeLouis wrote:
Ali wrote: 1+3+5+7+9+11+13+15+17+19=100 * Choose 5 numbers from these 10 figures so there sum is = 50 The sum of 5 Odd numbers is Odd also, so I don't believe there is a solution as stated that equals an even number (ie 50) Why use well-founded reasoning when brute force will do the trick? Just kidding. But the following might be a useful paradigm for solving such problems when the answer is not so obvious. Sub doit() x = Array(1, 3, 5, 7, 9, 11, 13, 15, 17, 19) For i1 = 0 To UBound(x) - 4 For i2 = i1 + 1 To UBound(x) - 3 For i3 = i2 + 1 To UBound(x) - 2 For i4 = i3 + 1 To UBound(x) - 1 For i5 = i4 + 1 To UBound(x) Sum = x(i1) + x(i2) + x(i3) + x(i4) + x(i5) If Sum = 50 Then Debug.Print Sum; x(i1); x(i2); x(i3); x(i4); x(i5) Next i5: Next i4: Next i3: Next i2: Next i1 End Sub PS: Lots of solutions when choosing 6. I wonder if the OP simply mistyped. |
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#6
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1+3+5+7+9+11+13+15+17+19=100 make a sum of 5 numbers make 50
That looks like 6 numbers to me.<g
See Dana's post as to why the problem as asked can't be solved. -- Rick (MVP - Excel) "Gary''s Student" wrote in message ... 1+3+5+7+15+19 -- Gary''s Student - gsnu200815 |
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#7
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1+3+5+7+9+11+13+15+17+19=100 make a sum of 5 numbers make 50
??? The OP's initial post said "Choose 5 numbers from these 10 figures so
there sum is = 50" (where he misspelled "their")... how do you read that as allowing more than 5 numbers? I read it, to paraphrase, as this... Choose 5 number such that their sum equals 50" -- Rick (MVP - Excel) "rebel" wrote in message ... On Sat, 22 Nov 2008 16:05:07 -0500, "Rick Rothstein" wrote: That looks like 6 numbers to me.<g but the "conditions" didn't say *only* five. It just required use of five, which the poster has complied with. See Dana's post as to why the problem as asked can't be solved. That was my initial response, but GS's solution appears to be compliant. |
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#8
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1+3+5+7+9+11+13+15+17+19=100 make a sum of 5 numbers make 50
On Sat, 22 Nov 2008 16:05:07 -0500, "Rick Rothstein"
wrote: That looks like 6 numbers to me.<g but the "conditions" didn't say *only* five. It just required use of five, which the poster has complied with. See Dana's post as to why the problem as asked can't be solved. That was my initial response, but GS's solution appears to be compliant. |
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#9
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1+3+5+7+9+11+13+15+17+19=100 make a sum of 5 numbers make 50
A clear demonstration that just because a person can add does not mean they
can count. -- Gary''s Student - gsnu200815 "Rick Rothstein" wrote: That looks like 6 numbers to me.<g See Dana's post as to why the problem as asked can't be solved. -- Rick (MVP - Excel) "Gary''s Student" wrote in message ... 1+3+5+7+15+19 -- Gary''s Student - gsnu200815 |
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#10
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1+3+5+7+9+11+13+15+17+19=100 make a sum of 5 numbers make 50
joeu2004 wrote:
On Nov 22, 4:24 am, Dana DeLouis wrote: Ali wrote: 1+3+5+7+9+11+13+15+17+19=100 Choose 5 numbers from these 10 figures so there sum is = 50 The sum of 5 Odd numbers is Odd also, so I don't believe there is a solution as stated that equals an even number (ie 50) Why use well-founded reasoning when brute force will do the trick? Just kidding. But the following might be a useful paradigm for solving such problems when the answer is not so obvious. Sub doit() x = Array(1, 3, 5, 7, 9, 11, 13, 15, 17, 19) For i1 = 0 To UBound(x) - 4 For i2 = i1 + 1 To UBound(x) - 3 For i3 = i2 + 1 To UBound(x) - 2 For i4 = i3 + 1 To UBound(x) - 1 For i5 = i4 + 1 To UBound(x) Sum = x(i1) + x(i2) + x(i3) + x(i4) + x(i5) If Sum = 50 Then Debug.Print Sum; x(i1); x(i2); x(i3); x(i4); x(i5) Next i5: Next i4: Next i3: Next i2: Next i1 End Sub PS: Lots of solutions when choosing 6. I wonder if the OP simply mistyped. How cool! Has anyone noticed that plotting x = Sum v. y = count of combinations making Sum resembles a normal distribution? (Is it?) Ok, perhaps that is not so fascinating, but try this: Make x() an array of Fibonacci numbers and try the plot again. A fractal-like pattern emerges. Evaluate the following for Test in (0..4000) to appreciate. ' Code:
Public Function doit(Test As Long) As Long
On Error GoTo ErrBreak
Dim x(41) As Long
Dim i1 As Long
Dim i2 As Long
Dim i3 As Long
Dim i4 As Long
Dim i5 As Long
Dim Sum As Long
Dim R As Long
' set up Fibonacci sequence
x(0) = 0
x(1) = 1
For i1 = 2 To 41
x(i1) = x(i1 - 2) + x(i1 - 1)
Next i1
For i1 = 0 To UBound(x) - 4
For i2 = i1 + 1 To UBound(x) - 3
For i3 = i2 + 1 To UBound(x) - 2
For i4 = i3 + 1 To UBound(x) - 1
For i5 = i4 + 1 To UBound(x)
Sum = x(i1) + x(i2) + x(i3) + x(i4) + x(i5)
If Sum = Test Then
R = R + 1
End If
Next i5: Next i4: Next i3: Next i2:
Debug.Print Test & "." & i1
Next i1
doit = R
Exit Function
ErrBreak: Stop
Resume Next
End Function
'
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#11
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1+3+5+7+9+11+13+15+17+19=100 make a sum of 5 numbers make 50
I got more chuckles off this post, than any in a long time!
Next we should square the circle, and trisect the angle with compus and ruler. 1+3+5+7+9+11+13+15+17+19=100 =SUM(100/(9-5-3+1)) = 50 A Mensa quiz! Some laughs, Shane Devenshire "Gary''s Student" wrote: A clear demonstration that just because a person can add does not mean they can count. -- Gary''s Student - gsnu200815 "Rick Rothstein" wrote: That looks like 6 numbers to me.<g See Dana's post as to why the problem as asked can't be solved. -- Rick (MVP - Excel) "Gary''s Student" wrote in message ... 1+3+5+7+15+19 -- Gary''s Student - gsnu200815 |
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#12
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1+3+5+7+9+11+13+15+17+19=100 make a sum of 5 numbers make 50
Hi Joeu,
Thanks for Your post, it was very helpful for me. But I would like to know if there's any way to make this code work for a set of x numbers in a sheet range allowing me to specify how many numbers should y add to get a result. for example: myfunction(LookUpRange as Range, NumberOfElements as integer, LookUpValue as long) The expected return of this function would be the address of cells that met the requirements. Thanks in Advance!! "joeu2004" wrote: On Nov 22, 4:24 am, Dana DeLouis wrote: Ali wrote: 1+3+5+7+9+11+13+15+17+19=100 Choose 5 numbers from these 10 figures so there sum is = 50 The sum of 5 Odd numbers is Odd also, so I don't believe there is a solution as stated that equals an even number (ie 50) Why use well-founded reasoning when brute force will do the trick? Just kidding. But the following might be a useful paradigm for solving such problems when the answer is not so obvious. Sub doit() x = Array(1, 3, 5, 7, 9, 11, 13, 15, 17, 19) For i1 = 0 To UBound(x) - 4 For i2 = i1 + 1 To UBound(x) - 3 For i3 = i2 + 1 To UBound(x) - 2 For i4 = i3 + 1 To UBound(x) - 1 For i5 = i4 + 1 To UBound(x) Sum = x(i1) + x(i2) + x(i3) + x(i4) + x(i5) If Sum = 50 Then Debug.Print Sum; x(i1); x(i2); x(i3); x(i4); x(i5) Next i5: Next i4: Next i3: Next i2: Next i1 End Sub PS: Lots of solutions when choosing 6. I wonder if the OP simply mistyped. |
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#13
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1+3+5+7+15+19-- Gary''s Student - gsnu200815
great work thanks for your help
On Saturday, November 22, 2008 3:31 AM Al wrote: 1+3+5+7+9+11+13+15+17+19=100 Choose 5 numbers from these 10 figures so there sum is = 50 On Saturday, November 22, 2008 3:51 AM Mike Middleton wrote: Ali - Tushar Mehta describes two methods with references to others: http://www.tushar-mehta.com/excel/te...ues/index.html - Mike Middleton http://www.DecisionToolworks.com Decision Analysis Add-ins for Excel "" wrote in message ... On Saturday, November 22, 2008 7:24 AM Dana DeLouis wrote: Ali wrote: The sum of 5 Odd numbers is Odd also, so I do not believe there is a solution as stated that equals an even number (ie 50) - - - HTH Dana DeLouis On Saturday, November 22, 2008 7:30 AM GarysStuden wrote: 1+3+5+7+15+19 -- Gary''s Student - gsnu200815 On Saturday, November 22, 2008 2:05 PM joeu2004 wrote: On Nov 22, 4:24=A0am, Dana DeLouis wrote: Why use well-founded reasoning when brute force will do the trick? Just kidding. But the following might be a useful paradigm for solving such problems when the answer is not so obvious. Sub doit() x =3D Array(1, 3, 5, 7, 9, 11, 13, 15, 17, 19) For i1 =3D 0 To UBound(x) - 4 For i2 =3D i1 + 1 To UBound(x) - 3 For i3 =3D i2 + 1 To UBound(x) - 2 For i4 =3D i3 + 1 To UBound(x) - 1 For i5 =3D i4 + 1 To UBound(x) Sum =3D x(i1) + x(i2) + x(i3) + x(i4) + x(i5) If Sum =3D 50 Then Debug.Print Sum; x(i1); x(i2); x(i3); x(i4); x(i5) Next i5: Next i4: Next i3: Next i2: Next i1 End Sub PS: Lots of solutions when choosing 6. I wonder if the OP simply mistyped. On Saturday, November 22, 2008 4:05 PM Rick Rothstein wrote: That looks like 6 numbers to me.<g See Dana's post as to why the problem as asked cannot be solved. -- Rick (MVP - Excel) On Saturday, November 22, 2008 6:38 PM Rick Rothstein wrote: ??? The OP's initial post said "Choose 5 numbers from these 10 figures so there sum is = 50" (where he misspelled "their")... how do you read that as allowing more than 5 numbers? I read it, to paraphrase, as this... Choose 5 number such that their sum equals 50" -- Rick (MVP - Excel) "rebel" wrote in message ... On Saturday, November 22, 2008 7:14 PM rebel wrote: but the "conditions" did not say *only* five. It just required use of five, which the poster has complied with. That was my initial response, but GS's solution appears to be compliant. On Saturday, November 22, 2008 8:15 PM GarysStuden wrote: A clear demonstration that just because a person can add does not mean they can count. -- Gary''s Student - gsnu200815 "Rick Rothstein" wrote: On Saturday, November 22, 2008 9:13 PM smartin wrote: joeu2004 wrote: How cool! Has anyone noticed that plotting x = Sum v. y = count of combinations making Sum resembles a normal distribution? (Is it?) Ok, perhaps that is not so fascinating, but try this: Make x() an array of Fibonacci numbers and try the plot again. A fractal-like pattern emerges. Evaluate the following for Test in (0..4000) to appreciate. ' Code:
Public Function doit(Test As Long) As Long
On Error GoTo ErrBreak
Dim x(41) As Long
Dim i1 As Long
Dim i2 As Long
Dim i3 As Long
Dim i4 As Long
Dim i5 As Long
Dim Sum As Long
Dim R As Long
' set up Fibonacci sequence
x(0) = 0
x(1) = 1
For i1 = 2 To 41
x(i1) = x(i1 - 2) + x(i1 - 1)
Next i1
For i1 = 0 To UBound(x) - 4
For i2 = i1 + 1 To UBound(x) - 3
For i3 = i2 + 1 To UBound(x) - 2
For i4 = i3 + 1 To UBound(x) - 1
For i5 = i4 + 1 To UBound(x)
Sum = x(i1) + x(i2) + x(i3) + x(i4) + x(i5)
If Sum = Test Then
R = R + 1
End If
Next i5: Next i4: Next i3: Next i2:
Debug.Print Test & "." & i1
Next i1
doit = R
Exit Function
ErrBreak: Stop
Resume Next
End Function
'
On Saturday, November 22, 2008 9:39 PM ShaneDevenshir wrote: I got more chuckles off this post, than any in a long time! Next we should square the circle, and trisect the angle with compus and ruler. 1+3+5+7+9+11+13+15+17+19=100 =SUM(100/(9-5-3+1)) = 50 A Mensa quiz! Some laughs, Shane Devenshire "Gary''s Student" wrote: On Thursday, December 04, 2008 7:19 AM AshMor wrote: Hi Joeu, Thanks for Your post, it was very helpful for me. But I would like to know if there's any way to make this code work for a set of x numbers in a sheet range allowing me to specify how many numbers should y add to get a result. for example: myfunction(LookUpRange as Range, NumberOfElements as integer, LookUpValue as long) The expected return of this function would be the address of cells that met the requirements. Thanks in Advance!! "joeu2004" wrote: |
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