Nine Red Bricks is a puzzle in Professor Layton and the Miracle Mask.


Lionel is contemplating a row of nine bricks for his mother's garden wall. Two adjacent bricks in the row are much heavier than the rest, but he can't remember which.

He's decided the only way to identify them is to lift the bricks one by one, but to save time, he wants to do it in the most efficient way possible.

Can you work out the maximum number of bricks he might need to lift in order to find the two heavy ones?


Click a Tab to reveal the Hint.

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?

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.

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?

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.



Too bad.

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


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.


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