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A counter with buttons

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A counter shows $1$ . Each press of button A multiplies the shown number by $3$ , and each press of button B adds $1$ to it. What is the fewest number of presses that makes the counter show exactly $100$ ?

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Hint

Work backwards from $100$ : when can the last press have been a multiplication?

Solution

Backwards, a press of A divides by $3$ , which is possible only for a multiple of $3$ , and a press of B subtracts $1$ .
From $100$ , subtract to $99$ , divide to $33$ and $11$ , subtract to $10$ and $9$ , and divide to $3$ and $1$ .
The subtractions are forced, since $100$ , $11$ and $10$ are not multiples of $3$ , and dividing whenever possible keeps the route shortest.
So the fewest number of presses is $7$ .

Answer: $7$

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