Solution to Problem 17: Do or Die

Solution to 17.1

There are no two dice that can satisfy column P, so there must be three dice in that row.  One is centered in rows B or C, one in E or F, and one in H or I.  

There are no three dice that could simultaneously satisfy columns P and Q, so there is at least one die centered in column O and another in column R.  Specifically, we need to have the 5 and 6 (horizonally) centered in P and Q, the 4 on one side, and either the 3 or the 2 on the other.  Very specifically, to keep O and R from being too filled, we need to have the 5 centered in Q, the 6 centered in P,the 4 centered in O, and either the 2 or 3 centered in R.

If the middle die that passed through column P were centered at F, then it must be a 1 centered at PF.  We have concluded that the 1 cannot be anywhere near here, so the die is centered in row E.  This can either be the 6 centered at EP or the 4 centered at EO, which would have a 2 next to it centered at ER.

 
If it were this latter case, then the 6 would have to be centered at BP (or else either the I or J would have too many pips in it).  Two of the pips in column R will come from the 5 centered in Q, and the third must come from a 3 centered at BS.  The third pip in O must come from the 1, but there would be no way to put that and the 5 down without them overlapping.  Therefore, it must be the case that the 6 is centered at EP instead.
Recall that the 5 is centered at column Q and the 4 at O.  The only way to satisfy Q and R is to have the 3 centered at R, and also the 2 must be centered at N to satisfy O.  If the 5 were centered at B, then the 4 and 3 would be below the 6 and the 2 above, but there would be no place for the 1.  Therefore, the 4 is centered at BO, and the 3 at BR.  The only way to satisfy the J is to have the 5 centered at IQ.  Then the 2 is centered at HN, and the only space available for the 1 is centered at GS.
The dice are centered at FS, HN, BR, BO, IQ, EP.

Solution to 17.2

The 3 or 5 die is not part of column S, since R has 0 pips.  Column S must be either a 6 and 4 or a 6, 1, and 2.  It could not be the former, because then there would be no way to get 5 pips into column L.  Therefore, 6 is centered in column R, 2 in R, and 1 is S, and also we must have 5 centered in column M, 4 in M, and 3 in L to make columns K and L work.  Two of these dice must also be centered in rows B or C, two in E or F, and two in H or I.

The die on the left centered at B or C cannot be 4, because the 1 and 2 on the right cannot fill the 2 pips needed in the "middle" of the 4, and the 6 on the right would overload at least one of the other rows.  Similarly, the die in the left top cannot be 3.  We see that the 5 die must be centered at BM. 

 
The remaining pip in row B is at BS.  It could either be part of the 1 die or the 2 die centered at CR.  If it were the latter, then 3 would have to be centered at EL to make the row D sum work, but then there would be no way to make row E work.  Therefore, the 1 is centered at BS.  

If the 3 die were centered at EL or FL, then there would be no way to make rows D or E work by placing a die on the right.  So the 4 die must be in the center left position.  The only way to make the row D and E counts work is if 4 is centered at EL and 6 at FR.

The only way to make the row J count work is to have the 3 at IL and the 2 at IR.

The dice are centered at BS, IR, IL, EM, BM, FR.

Notes: I probably spent about five minutes working on the first of these problems before deciding that the sample problem was much easier than the two in the test.  The second one is probably easier than the first, and I might have gotten somewhere if I had spent that five minutes there.