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



"" <(E-Mail Removed)> wrote in message
news:92D81C03-8E58-491E-9C75-(E-Mail Removed)...
> 1+3+5+7+9+11+13+15+17+19=100
> Choose 5 numbers from these 10 figures so there sum is = 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

1+3+5+7+15+19
--
Gary''s Student - gsnu200815

On Nov 22, 4:24*am, Dana DeLouis <delo...@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.

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" <(E-Mail Removed)> wrote in message
news:6E603516-B130-484F-A1AB-(E-Mail Removed)...
> 1+3+5+7+15+19
> --
> Gary''s Student - gsnu200815

??? 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" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> On Sat, 22 Nov 2008 16:05:07 -0500, "Rick Rothstein"
> <(E-Mail Removed)> 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.

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" <(E-Mail Removed)> wrote in message
> news:6E603516-B130-484F-A1AB-(E-Mail Removed)...
> > 1+3+5+7+15+19
> > --
> > Gary''s Student - gsnu200815
>
>

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" <(E-Mail Removed)> wrote in message
> > news:6E603516-B130-484F-A1AB-(E-Mail Removed)...
> > > 1+3+5+7+15+19
> > > --
> > > Gary''s Student - gsnu200815
> >
> >

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 <delo...@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.
>
>