# Maths Problem

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#1
Over easterwe've been given a maths problems sheet from college (the fun!) and I just don't get how to even start solving one of the problems.

There are 3 types of amoebae A, B and C in a test tube. At the start of the experiment there are 20 amoebae of type A, 21 of type B and 22 of type C. These amoebae can combine together with any two amoebabe of different types merging into one amoeba of the third type. After a lot of such merges, only one amoeba remains in the test tube. What is its' type?
I know I should probably solve it myself but can anyone give me any ideas on how to get started???
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#2
Has anyone done any questions like this before? I'm sure there must be similar ones... its still puzzling me tho
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17 years ago
#3
B
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17 years ago
#4
(Original post by MyNameIsNeo)
B
why?
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17 years ago
#5
(Original post by Pixelfairy #1)
Has anyone done any questions like this before? I'm sure there must be similar ones... its still puzzling me tho
Umm, are you sure it's possible?

For example, suppose the last amoebas are all C's. Then just prior to their last mating ritual there must have been an even number of A's and B's.
But there are 63 amoebas in total, which is an odd number. So if an even number "reproduced" with each other, there was still an odd one left over??

I imagine I'm missing something quite elementary
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17 years ago
#6
Start with a simplified version. Say you have 1 of type a, 2 of B and 3 of C.

So there are 6 Aomeba. It would go something like this:

A+B = C

Now you have 4 Cs, and 1 B.

B+C = A

1 A, 3 C

A+C = B

1 B, 2 C

B+C = A

1 A, 1 C

A+C = B

Try looking at higher ones, eg. 2 As, 3 Bs and 4Cs. But I think you will find the general, the middle one is the winner.
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17 years ago
#7
Thanks. I gotta better idea of what's happening now.
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17 years ago
#8
Each combination increments (ie, increases by 1) one of A, B, C and decrements (ie, decreases by 1) the other two. So each even amoebae-count becomes odd, and each odd amoebae-count becomes even.

Initially (A, B, C) is (even, odd, even). [(20, 21, 22) is (even, odd, even).]

After one combination, (A, B, C) is (odd, even, odd).

After two combinations, (A, B, C) is (even, odd, even).

After three combinations, (A, B, C) is (odd, even, odd).

And so on.

So, after any number of combinations, (A, B, C) is either (odd, even, odd) or (even, odd, even).

When there's only one amoeba left (A, B, C) must be (1, 0, 0), (0, 1, 0) or (0, 0, 1). But only one of these has the pattern (odd, even, odd) or (even, odd, even).
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#9
(Original post by Jonny W)
Each combination increments (ie, increases by 1) one of A, B, C and decrements (ie, decreases by 1) the other two. So each even amoebae-count becomes odd, and each odd amoebae-count becomes even.

Initially (A, B, C) is (even, odd, even). [(20, 21, 22) is (even, odd, even).]

After one combination, (A, B, C) is (odd, even, odd).

After two combinations, (A, B, C) is (even, odd, even).

After three combinations, (A, B, C) is (odd, even, odd).

And so on.

So, after any number of combinations, (A, B, C) is either (odd, even, odd) or (even, odd, even).

When there's only one amoeba left (A, B, C) must be (1, 0, 0), (0, 1, 0) or (0, 0, 1). But only one of these has the pattern (odd, even, odd) or (even, odd, even).
Thanks I can see what to do now! Id've never have guessed to do that on my own
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