Melody

Oh, If you want to know how I came up with this answer then just ask.  

What about 111?

111+3+3=117 = 0 (mod9)

I came up with the digtts must add to one of the following

3,6,9,12,15,18,21,24,27

The smallest is 111  

the biggest is 999

and there are lots in between.

111  

114

117

120

123

126

129

...

just keep adding 3

so there will be   (999-111)/3  +1 = 297 of them      

That is what I get anyway.

Maybe you can find one of mine that does not work.  I certainly have not checked them.

This is a weird question.

If a+b+c  = 6 or 15 or 24     then    x could be equal to 1 or 4 or 7

If a+b+c  = 3 or 12 or 21    then     x could be equal to  2 or 5 or 8

If a+b+c = 0 or 9 or 18 or 27 then  x could be equal to  0,3,6 or 9     There are 4 of them so that is no good.

So 3 possible values of x will happen when 

a+b+c = 3,6,  12 15,  21,24

Now you can finish it.

Hi Ginger,

It is very nice to see you.   

 So in the USA a kite would qualify as a Trapezium.  In Britain it would qualify as a Trapezoid.

And here I would only ever have called it a kite or, more generally, a quadrilateral.

Before I started using this forum I had never seen the work Trapezoid used.  

Oblong is another strange word that I see sometimes.  I think it just is a rectangle.

Cheap tablecloths from China (in our shops) are often referred to as oblong in shape.

Hi Cal,

I always balk when I see the word Trapezoid.

In Australia we call the Trapeziums.

Your answer is excellent.  :)

Hi Cal,

Yes, it refused to display as code. It displayed as not properly formatted Latex and I did not want that.

So I made the writing white so it would disappear.

I think you can highlight and copy it if anyone wanted to see it.

No one has ever asked me why I usually add the coding at the bottom like this. This has surprised me a little.

I do it for 2 reasons.

1) It may help some people to learn LaTex Coding.

2) Web 2 often will not allow me to get in and edit my code, especially not when it is a big block of code.

   If I have the code at the bottom I can copy it into a new Latex box and edit that instead (Then i would delete the original)

I do not think that it exists.

Let

A = \(\begin{pmatrix} a\;b\;c\\ d\;e\;f\\ h\;i\;j\\ \end{pmatrix}\)

\(\begin{pmatrix} a\;b\;c\\ d\;e\;f\\ h\;i\;j\\ \end{pmatrix} \cdot \begin{pmatrix} 1\\0\\0 \end{pmatrix}= \begin{pmatrix} a\\d\\h \end{pmatrix}= \begin{pmatrix} -3\\4\\0 \end{pmatrix}\\ so\\ a=-3\qquad d=4\qquad h=0\)

\(\begin{pmatrix} -3\;b\;c\\ 4\;e\;f\\ 0\;i\;j\\ \end{pmatrix} \cdot  \begin{pmatrix} 0\\1\\1 \end{pmatrix}= \begin{pmatrix} b+c\\e+f\\i+j \end{pmatrix}= \begin{pmatrix} 1\\2\\3 \end{pmatrix}\\ so\\ c=1-b \qquad f=2-e\qquad j=3-i\)

\(\begin{pmatrix} -3\;\;b\;\;1-b\\ 4\;\;e\;\;2-e\\ 0\;i\;\;\;3-i\\ \end{pmatrix} \cdot  \begin{pmatrix} 1\\1\\1 \end{pmatrix}= \begin{pmatrix} -3+1\\4+2\\0+3 \end{pmatrix}= \begin{pmatrix} -2\\6\\3 \end{pmatrix}\ne \begin{pmatrix} 3\\2\\1 \end{pmatrix}\\ \)

LaTex:

\text{\begin{pmatrix}
a\;b\;c\\
d\;e\;f\\
h\;i\;j\\
\end{pmatrix}

\begin{pmatrix}
a\;b\;c\\
d\;e\;f\\
h\;i\;j\\
\end{pmatrix}
\cdot 
\begin{pmatrix}
1\\0\\0
\end{pmatrix}=
\begin{pmatrix}
a\\d\\h
\end{pmatrix}=
\begin{pmatrix}
-3\\4\\0
\end{pmatrix}\\
so\\ a=-3\qquad b=4\qquad c=0

\begin{pmatrix}
-3\;b\;c\\
4\;e\;f\\
0\;i\;j\\
\end{pmatrix}
\cdot 
\begin{pmatrix}
0\\1\\1
\end{pmatrix}=
\begin{pmatrix}
b+c\\e+f\\i+j
\end{pmatrix}=
\begin{pmatrix}
1\\2\\3
\end{pmatrix}\\
so\\ c=1-b \qquad f=2-e\qquad j=3-i

\begin{pmatrix}
-3\;\;b\;\;1-b\\
4\;\;e\;\;2-e\\
0\;i\;\;\;3-i\\
\end{pmatrix}
\cdot 
\begin{pmatrix}
1\\1\\1
\end{pmatrix}=
\begin{pmatrix}
-3+1\\4+2\\0+3
\end{pmatrix}=
\begin{pmatrix}
-2\\6\\3
\end{pmatrix}\ne
\begin{pmatrix}
3\\2\\1
\end{pmatrix}\\}
 

I doubt this is correct.

Scenarios

1 Purple in 1 and green in 3

2 Purple in 1 and green in 6

3 Purple in 1 and green in 9

4 Green in 1 and purple in 6

5 Purple in 2 and green in 7

6 Purple in 2 and green in 8

Maybe that is it

7!*6 = 30240

Yes you are right, thanks.

I'd work out each selection separately and then add them together

2 on committee      10*9=90ways

3 on committee       90*8=720 ways

4 on the committee  90*8C2 ways

etc