anonymous
  • anonymous
There are several letters e, a and b on a blackboard. We may replace two e's by one e, two a's by one b, two b's by one a, an a and a b by one e, an a and an e by one a,ab, and an e by one b. Prove that the last letter does not depend on the order of erasure.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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blockcolder
  • blockcolder
So to summarize: ee becomes e, aa becomes b, bb becomes a. ab becomes e, ae becomes a, be becomes b. Of course, the letters don't have to be adjacent.
anonymous
  • anonymous
Thank's for the simplification, but could someone please tell me how I am supposed to prove the last statement?
blockcolder
  • blockcolder
It took me a long time, but here's a solution. Let \(w_i\) be the number of a's, \(x_i\) the number of b's, and \(y_i\) the number of e's at step i. Step 0 is the initial configuration. I assert that if the quantity \(w_0+2x_0\equiv k \pmod{3}\), then \(w_i+2x_i\equiv k \pmod{3}\) for all \(i\leq w_0+x_0+y_0-1\), where k can be 0, 1, or 2. If k=0, then the final letter is e; if k=1, it's a; if k=2, it's b.

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blockcolder
  • blockcolder
You can prove this assertion by mathematical induction and considering all the 6 scenarios that can happen.
anonymous
  • anonymous
Thanks a lot!

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