Grades 11-12 Video Solutions 2012-2012_11-12

2012_11-12_20

Video Transcription

A kangaroo wants to build a row of standard dice. On a standard die, each pair of opposite  faces has a total of seven dots. He can glue two faces together if they have the same number of dots. He would like the total number of dots on the outer faces of the dice in the  row to be 2012. How many dice does he need?  So let's examine this die right here. The way that I'm drawing it is that this D corresponds to like this face. It's like this one right here. And this seven minus D is like the one  opposite to that. Then there's the top face, the bottom face, and you're looking at the  front face and the back face is like behind. So the top, bottom, front and back both contribute  a total of 14 because like these two add up to seven, and then the front and back also add up to seven. So that's 14. And then finally we have this extra D and seven minus D, which  contribute another seven to the total. And now if we extend this by one more die, we have to glue this seven minus D to the seven minus D because they have to be the same number  of dots. And so now this dice, this die right here has a top, bottom, front and back, which contributes 14. This one also contributes 14 in the same way. And finally we have to  add on these two D at the end. So the way we're going to count this is look at the contribution from each one of the dice, from the top, bottom, front and back, and then add up the  two side pieces. So if we have an odd number of dice like this, we have to start with D and end with a seven minus D. And if we have an even number of dice, we start with D and  end with the same D, which means that for an odd number of dice, let's say two N plus one dice, each die contributes 14 from the front, from the top, bottom, front and back.  And then this seven from this D and the seven minus D. And now if we have an even number of dice like this, let's say two N dice, each one contributes 14 as well as this two D from  here and here. Now we need to see if either one of these can be equal to 2012. So first, this guy has 14 times something and then plus seven. So in total, this is a multiple of  seven, but 2012 just isn't a multiple of seven. Okay, so this one is out the window. Now we just need to try it with this one. So let's set the two equal. Now I think this  reminds me of taking 2012, dividing it by 28, getting some quotient and some sort of remainder because this D right here ranges from one to six. So two D is less than 28.  We know that quotients and remainders are unique, so the only way that we're going to get this term to be less than 28 is if N is the quotient when you divide 2012 by 28 and  this is the remainder. So let's do that and we get the remainder of 24. That tells us that D is 12, which means that there's one side of the die which has a 12 on it, which  is impossible. So it's impossible.

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