Can You Cross Like A Boss?

Six three-digit numbers fill the rows of a CrossProduct table, one digit per cell. The product of each row’s three digits is given on the right ($210, 144, 54, 135, 4, 49$), and the products down the hundreds, tens and ones columns are $6{,}615$, $15{,}552$ and $420$. Find all six numbers.

Solution

Each row product factors into three single digits in only a handful of ways, and the three column products decide which factorization each row must take. Searching across the rows and pruning the moment a running column product overshoots its target leaves exactly one consistent table: $$\boxed{765,\ 982,\ 392,\ 593,\ 141,\ 717.}$$ Reading down the columns confirms it: the hundreds digits multiply to $7\cdot9\cdot3\cdot5\cdot1\cdot7 = 6{,}615$, the tens to $6\cdot8\cdot9\cdot9\cdot4\cdot1 = 15{,}552$, and the ones to $5\cdot2\cdot2\cdot3\cdot1\cdot7 = 420$.

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 hit the three column totals.

rows = [210, 144, 54, 135, 4, 49]
cols = (6615, 15552, 420)
opts = [[(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] 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, ['765','982','392','593','141','717']