**Hard Maths Logic Question - 25 February**

There is a jar in which there are two types of candies – 20 blueberry and 16 strawberry. You perform the following steps:

1) You take out two candies.

2) If the two candies are of same flavor, you add a blueberry one otherwise, you add the strawberry one.

You repeat these two steps till there is just one candy remaining in the jar. Which flavored candy will be left?

Also, if you began with 100 blueberry candies and 93 strawberry candies, which flavor would have been the last one to remain ?

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Thankyou for the above information.

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Let these symbols designate the fruits:

ReplyDeleteB=blueberry

S=strawberry

The rules are:

1) Remove B&B: add B, net -B.

2) Remove B&S: add S, net -B.

3) Remove S&S: add B, net -2*S+B.

That is, remove a B, or remove two S and add a B.

Another way of looking at this is strawberries can only be removed two at a time, whereas blueberries can increase or decrease by one.

Starting with 20*B + 16*S, you'll see that after removing two strawberries 8 times, you'll end up with only blueberries, regardless of how many times a blueberry and blueberry/strawberry were removed. Once you end up with only blueberries, there is only one kind of move left, and that is removing blueberries one at a time until only a single blueberry is left.

Starting with 100*B + 93*S, you'll see that some point, you'll end up with only one strawberry and some number of blueberries; at that point, each successive draw will eliminate only one blueberry. Eventually, you'll end up with only one strawberry.

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