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Sweet Sums is a puzzle in Professor Layton and the Last Specter.

Puzzle

"My children have been very kind to each other lately, so I decided to give them some candy as a reward.

I have four jars of candy, A, B, C, and D. The combined number of candies in jars A and B is equal to twice the number in jar C. The combined number of candies in jars B and D equals twice the number in jar A. If you take three candies from jar D and put them in jar A, jar A will contain twice the number in jar B.

Which jar contains six pieces of candy?"

Hints




Click a Tab to reveal the Hint.

All you need to do is figure out which jar contains six pieces of candy.


Even if you can't work it out using complicated mathematical calculations, you should still be able to solve it by figuring out an alternate method for determining the number of candies in each jar.

Here's the puzzle in equations:


A + B = 2C

B + D = 2A

A + 3 = 2B

D > 3


This might help you spot a detail that you may have missed.

A + 3 = 2B


2B must be an even number 2 is one of its factors. That means A + 3 is also an even number.


Subtracting 3 from an even number will always result in an odd number, which means A must be odd.

Odd + Odd = Even

Odd + Even = Odd


From this simple rule, you can figure out that number of candies in jar B must be odd that the number in jar D must also be odd.


This means that all of the jars except C contain an odd numbers of candies. Therefore, the number of candies in jar C must be...


Solution

Correct

Correct!

You know that A + 3 = 2B. Now, as 2B must be even and 3 is odd, A must also be odd.

Similarly,
A (odd) + B = 2C (even), so B = odd
A (odd) + D = 2C (even), so D = odd

The jar you are looking for contains an even number of sweets. As C is the only remaining possibility, it must be the answer by default.

LS079S

A big thanks to http://professorlayton4walkthrough.blogspot.com