MM020 - Nine Red Bricks

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Nine Red Bricks
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Nine Red Bricks
Puzzle Number 020
Puzzle Name Nine Red Bricks
Picarats Given 40 Picarats
Type Write Answer
Location Market Street
Previous Puzzle MM019 - Spinning Maize
Next Puzzle MM021 - A Lick of Paint

This is the twentieth puzzle you'll encounter in Professor Layton and the Miracle Mask. To access this puzzle, you must talk to Lionel. In order to solve this puzzle, you must figure out the maximum number of bricks that would need to be lifted to find the two heavy bricks.


[edit] Hints

Hint One
    The key here is the fact that the two heavy bricks are next to each other.

    For example, say Lionel lifted bricks 2 and 4. If they both weighed the same, then which bricks could he rule out?

Hint Two
    Let's say Lionel lifted 2 and 4, and 4 was heavier. The adjacent heavy brick would be 3 or 5, and he could find out which by lifting just one of them.

    If 2 and 4 both weighed the same, then he could rule out bricks 1 to 4.

Hint Three
    Let's carry on from Hint 2. If bricks 2 and 4 weighed the same, then Lionel could try lifting 6. If that one was heavier, then the other heavy brick would have to be 5 or 7. He could determine which by lifting either brick, for a total of four lifts.

    So what would he do if 6 wasn't a heavy brick?

Super Hint
    Carrying on from Hint 3, if 6 was a normal brick as well, then Lionel would lift 9. If 9 was heavier, then he'd know that 8 and 9 were the heavy bricks.

    If 9 was not heavier, that would leave only 7 and 8 as the possible heavy bricks--he wouldn't even need to lift them to know!

    That's all the possibilities covered. At most, he'd only have to lift 2, 4, 6, and 9.

[edit] Messages

[edit] When Failed

Too bad.

Remember that Lionel want to lift the bricks in the most efficient way possible.

[edit] When Completed

Correct! It's four!

For example, he might start by lifting bricks 2, 4 and 6. If one of them is heavier, he can then lift a neighboring brick. If that brick isn't also heavy, then he knows the one on the other side is.

If none of the bricks out of 2, 4 and 6 are heavier than the rest, he can simply lift 9. If 9 is heavier, then 8 must be the other heavy brick. If 9 isn't heavier, then it has to be 7 and 8.

[edit] Solution

Four bricks is the maximum number of bricks that can be lifted.

[edit] Progress

580 Picarats and 90 Hint Coins.

Last edited by Squiggle on 19 August 2015 at 05:35
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