1+3+5+7+9+11+13+15+17+19=100 make a sum of 5 numbers make 50

[Ali]

1+3+5+7+9+11+13+15+17+19=100
Choose 5 numbers from these 10 figures so there sum is = 50

 

[Mike Middleton]

Ali -

Tushar Mehta describes two methods with references to others:

http://www.tushar-mehta.com/excel/templates/match_values/index.html

- Mike Middleton
http://www.DecisionToolworks.com
Decision Analysis Add-ins for Excel

 

[Dana DeLouis]

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

 

[Gary''s Student]

1+3+5+7+15+19

 

[joeu2004]

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.

 

[Rick Rothstein]

That looks like 6 numbers to me.<g>

See Dana's post as to why the problem as asked can't be solved.

 

[Rick Rothstein]

??? 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"

 

[rebel]

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.

 

[Gary''s Student]

A clear demonstration that just because a person can add does not mean they
can count.

 

[smartin]

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.

'

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
'

 

[Shane Devenshire]

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

 

[AshMorK]

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!!

 

[shahab qamar ansari]

great work thanks for your help

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/templates/match_values/index.html

- Mike Middleton
http://www.DecisionToolworks.com
Decision Analysis Add-ins for Excel



news:[email protected]...

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.

'

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
'