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From: Rick Rothstein on 22 Nov 2008 16:05
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" <GarysStudent(a)discussions.microsoft.com> wrote in message
news:6E603516-B130-484F-A1AB-0B75BB0084B2(a)microsoft.com...
> 1+3+5+7+15+19
> --
> Gary''s Student - gsnu200815

From: rebel on 22 Nov 2008 19:14
On Sat, 22 Nov 2008 16:05:07 -0500, "Rick Rothstein"
<rick.newsNO.SPAM(a)NO.SPAMverizon.net> 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.
From: Rick Rothstein on 22 Nov 2008 18:38
??? 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" <me(a)privacy.net> wrote in message
news:k08hi4tibckdhu8di4qd6qv3fcj4nk3j2f(a)4ax.com...
> On Sat, 22 Nov 2008 16:05:07 -0500, "Rick Rothstein"
> <rick.newsNO.SPAM(a)NO.SPAMverizon.net> 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.

From: Gary''s Student on 22 Nov 2008 20:15
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" <GarysStudent(a)discussions.microsoft.com> wrote in message
> news:6E603516-B130-484F-A1AB-0B75BB0084B2(a)microsoft.com...
> > 1+3+5+7+15+19
> > --
> > Gary''s Student - gsnu200815
>
>
From: smartin on 22 Nov 2008 21:13
joeu2004 wrote:
> On Nov 22, 4:24 am, Dana DeLouis <delo...(a)bellsouth.net> 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
'[/code]
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