
Originally Posted by
Dragon
The number on a standard six faced die are arranged such that numbers on opposite faces always add to 7.The product of the number appearing on the four lateral faces of a rolled die is calculated(ignoring the numbers on the top and bottom) What is the maximum possible value of this product?
here is the rolled out version of the cube:
Code:
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: 1 :
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: 2 : 3 : 5 : 4 :
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: 6 :
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another one:
Code:
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: 4 :
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: 1 : 2 : 6 : 5 :
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: 3 :
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and the last one:
Code:
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: 2 :
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: 1 : 3 : 6 : 4 :
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: 5 :
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Your job is to find the product of all those middle rows.
So for the first one: $\displaystyle 2\times3\times5\times4=120$
Second One: $\displaystyle 1\times2\times6\times5=60$
Last One: $\displaystyle 1\times3\times6\times4=74$
So the maximum possible value is: $\displaystyle \boxed{120}$