Problem 20 · 2012 Math Kangaroo
Stretch
Logic & Word Problems
caseworksum-constraint
Renate wants to glue together a number of ordinary dice (whose number of points on opposite sides always adds up to 7) to form a “dicebar” as shown. Doing this she only wants to glue sides together with an equal number of points. She wants to make sure that the sum of all points on the non-glued sides equals 2012. How many dice does she have to glue together?
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Answer: E — It is impossible to obtain exactly 2012 points on the non-glued together sides.
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Hint 1 of 2
Glued faces show equal pips, and the two glued faces of an inner die are opposite, so they sum to 7.
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Hint 2 of 2
That forces the exposed total into a fixed form — work out its possible values and check whether 2012 can ever be hit.
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Approach: track the exposed pip total under the opposite-faces rule
- Each die has \(21\) pips; a bar of \(n\) dice has \(n-1\) glued joints, each hiding \(2v\) pips for the equal glued value \(v\).
- Each inner die's two glued faces are opposite, summing to 7, so consecutive joint values alternate \(v\) and \(7-v\) and adjacent joints sum to 7.
- The exposed total is \(21n - 2(\text{joint sum})\): for even \(n-1\) this is \(7(2n+1)\), an odd multiple of 7 (and \(2012\) is not a multiple of 7); for odd \(n-1\) it is even and forces \(v = 7n - 999\), which has no valid \(n\) with \(1\le v\le 6\).
- So no \(n\) gives exactly 2012, making it impossible, choice E.
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