The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. # Practice Paper 3 Question 8

A whiteboard has $$p$$ $$A"$$ symbols and $$m$$ $$B"$$ symbols written on it. You choose any two symbols to erase and replace them with another one according to the following rules:

• $$A A \Rightarrow B$$
• $$A B\text{ or } BA \Rightarrow A$$
• $$B B \Rightarrow B$$

If you continue to apply replacements for as long as possible, which values of $$p$$ and $$m$$ result in a single $$A$$ remaining at the end?

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## Hints

• Hint 1
Can the number of $$A$$s on the whiteboard increase?
• Hint 2
What restrictions exist on how we may reduce the number of $$A$$s?
• Hint 3
... and what does that tell you about $$p?$$
• Hint 4
The question essentially asks to remove all $$B$$s. Couild there be a situation when you cannot remove a B? Justify.

## Solution

The number of $$A$$s can never be increased, so $$p\geq1.$$ Also, as the number of $$A$$s may decrease only by $$2$$ at a time via the first rule, $$p$$ must be odd (since we need one $$A$$ at the end). In contrast, there is no constraint on the initial number of $$B$$s. Since the number of $$A$$s is always greater than $$0,$$ we will always be able to apply a replacement if there is at least one $$B.$$ Every replacement that involves a $$B$$ reduces the number of $$B$$s by one. Consequently, the number of $$B$$s will be reduced to $$0$$ no matter how many we started with.

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