Can You Hunt For The Mysterious Numbers?

Riddler Express

On “You Bet Your Fife,” a real number is chosen uniformly between $0$ and $100$. You guess a value; if your guess is less than the number you win a fife worth exactly your guess, otherwise nothing. What should you guess to maximize your expected winnings?

Solution

If you guess $g$, you win $g$ when the hidden number $X$ exceeds $g$, which happens with probability $\tfrac{100-g}{100}$, and nothing otherwise. So your expected winnings are $$\mathbb{E}[\text{winnings}] = g\cdot\frac{100-g}{100} = \frac{g(100-g)}{100},$$ a downward parabola in $g$ maximized where $g = 100 - g$, that is at $$g = \boxed{50}, \qquad \mathbb{E}[\text{winnings}] = \frac{50\cdot 50}{100} = \$25.$$ The balance is between asking for more and the rising chance of winning nothing, and it tips exactly at the halfway mark.

The computation

Maximize the expected-winnings parabola.

best = max(range(101), key=lambda g: g * (100 - g) / 100)
print(best, best * (100 - best) / 100)        # 50, 25.0

Riddler Classic

Eight three-digit numbers fill the rows of a table. The product of each number’s three digits is given at the right ($294, 216, 135, 98, 112, 84, 245, 40$), and the products of the hundreds, tens and ones digits down the columns are $8{,}890{,}560$, $156{,}800$ and $55{,}566$. Find all eight numbers.

Solution

Each row’s product factors into three digits in only a few ways, and the three column products pin down which factorization each row must use. A short backtracking search, trying each row’s digit triples and pruning whenever a running column product overshoots its target, finds a unique solution: $$\boxed{776,\ 983,\ 953,\ 727,\ 827,\ 743,\ 577,\ 851.}$$ A quick check: the hundreds digits multiply to $7\cdot9\cdot9\cdot7\cdot8\cdot7\cdot5\cdot8 = 8{,}890{,}560$, and the row and column products all match.

The computation

For each row list the digit triples whose product matches; then search across rows for the one combination whose hundreds, tens and ones digits multiply to the three column totals.

rows = [294, 216, 135, 98, 112, 84, 245, 40]
cols = (8890560, 156800, 55566)
def triples(p):
    return [(h, t, o) for h in range(1, 10) for t in range(1, 10)
            for o in range(1, 10) if h*t*o == p]
opts = [triples(p) for p in rows]
sols = []
def search(i, prod, acc):
    if any(prod[c] > cols[c] for c in range(3)):
        return
    if i == len(rows):
        if prod == list(cols): sols.append(acc)
        return
    for h, t, o in opts[i]:
        search(i + 1, [prod[0]*h, prod[1]*t, prod[2]*o], acc + [f"{h}{t}{o}"])
search(0, [1, 1, 1], [])
print(len(sols), sols[0])     # 1, ['776','983','953','727','827','743','577','851']