Problem 3 · Multiplying digits down to zero
To end at zero you must hit a multiple of 10 or carry a 0.
Integer answer, at most 4 digitsWe choose a two-digit number and multiply its digits. If the result has two digits, we multiply its digits again, repeating the process until we get a single-digit number. With how many different numbers can we start if we want to end at zero?
Reasoned solution
Key idea: the final result is $0$ exactly when a digit $0$ shows up at some step, i.e. when we reach a multiple of $10$.
The $9$ multiples of $10$ (from $10$ to $90$) get there immediately (product $0$). Then, tracing backwards which products land on a multiple of $10$:
$25, 52, 55 \to 10$; $45, 54, 59, 69, 96 \to 20$ (via $45 \to 20$ and $54 \to 20$); $56, 65, 78, 87 \to 30$; $58, 85 \to 40$; $95 \to 45 \to 20$.
In total: $9$ multiples of $10$ plus $15$ more numbers ($25, 45, 52, 54, 55, 56, 58, 59, 65, 69, 78, 85, 87, 95, 96$).