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The Testament of Sherlock Holmes Hints
How can I solve the Roman numeral puzzle box?
1 of 13: The first thing to do is decide where the one X (10) that you're allowed should go.
2 of 13: None of the sums or differences can be larger than 10, so it's obvious that you can't use the X in any of the parts with a + sign.
3 of 13: And you can't subtract 10 from a larger number, since you're not allowed by make any numbers that are larger than 10.
4 of 13: So the only place you could put the X is on the left side of the third part's minus sign.
5 of 13: To figure out what number to put on the right side of the minus sign, you'll need to use the facts that no sum or difference can be greater than 10, and each part (except the first one) has to have a sum or difference that's greater than the previous part's.
6 of 13: To start with, notice that the two previous parts both have a +3 in them. That means the first part can't add up to less than 4 (1 + 3), and the second part can't add up to less than 5 (2 + 3). That means the third part can't end up being less than 6 (10 - 4).
7 of 13: So start by putting a one (I) into the first part's empty space, a two (II) into the second part's empty space, a four (IV) into the third part's empty space, and another four into the fourth part's empty space. That'll make the four-part number sequence be 4, 5, 6, 7.
8 of 13: That would work except for one thing: you've got two shafts left over. That's no good, since you're required to use all of them.
9 of 13: So add one to each of the four parts' sums/differences to see if that'll work. You'll end up with II + III (2 + 3) in the first part, III + III (3 + 3) in the second part, X - III (10 - 3) in the third part, and V + III (5 + 3) in the third part.
10 of 13: That'll give you another acceptable sequence, namely 5, 6, 7, 8. However, there'll be one shaft left unused.
11 of 13: The most obvious thing to do is change the V in the fourth part into a VI. That'll use up the last shaft, and your number sequence (5, 6, 7, 9) won't violate any of the stated rules. Unfortunately, there's a hidden rule that says all of the numbers in the sequence (except the first one) have to be *exactly* one more than the previous one.
12 of 13: So take the I back from the VI in the fourth part, then do the "+1 across the board" thing again.
13 of 13: That'll give you III + III (3 + 3) in the first part, IV + III (4 + 3) in the second part, X - II (10 - 2) in the third part, and VI + III (6 + 3) in the fourth part. That forms the acceptable sequence of 6, 7, 8, 9 while using up all the shafts.
