Problem Sum

liukawa

New member
Joined
Feb 5, 2012
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3
Hi Guys,

Not sure whether is this the right place to post my questions.

Nevertheless, would appreciate if you guys could help me with my question !

The question goes like this...

Angei had a bag of beads in 12 different colours, 4 different sizes and 3 different shapes.She was able to group them according to colour, size of shape equally.
If the total number of beads was below 80,what was the greatest possible number of beads in the bag?


Appreciate if you guys can help !

Thanks in advance !
 
Hello, liukawa!

I can think of no formula for this problem . . .


Angei had a bag of beads in 12 different colours, 4 different sizes and 3 different shapes.
She was able to group them according to colour, size of shape equally. .If the total number
of beads was below 80, what was the greatest possible number of beads in the bag?

Suppose the three shapes are: circles, squares, triangles.
And they come in four sizes: 1, 2, 3, 4.

Make one of each of the twelve combinations:. .\(\displaystyle \begin{Bmatrix} C_1 & C_2 & C_3 & C_4 \\ S_1 & S_2 & S_3 & S_4 \\ T_1 & T_2 & T_3 & T_4 \end{Bmatrix}\)


Suppose the twelve colors are: .\(\displaystyle a, b, c, \hdots l\)

Now distribute the 12 colors over these 12 objects
. . (a different color for each object).

. . Then we have: .\(\displaystyle \begin{Bmatrix}C_{1a} & C_{2b} & C_{3c} & C_{4d} \\ S_{1e} & S_{2f} & S_{3g} & S_{4h} \\ T_{1i} & T_{2j} & T_{3k} & T_{4l} \end{Bmatrix}\)


Now imagine that we have six of these sets.

These 72 objects can be partitioned into:
. . 12 equal sets by color,
. . 4 equal sets by size, and
. . 3 equal sets by shape.
 
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