The value of 24! is written out in full, but one of its digits has been smudged and cannot be read. The digits that are still visible add up to 73. What is the smudged digit?

24! contains plenty of factors of 3, so it is divisible by 9, which means its full digit sum must also be a multiple of 9. If the missing digit is $d$ then $73+d$ must be a multiple of 9. Since $73=9\times 8+1$, we need $d=8$, giving a digit sum of 81. No other single digit works, because the next multiple of 9 after 81 is 90 and 90 − 73 = 17 is not a digit. Testing divisibility by 3 alone is not enough — it would allow $d=2$, 5 or 8.