After a fire destroyed the home, the life and the will of a rich aristocrat, a detective found among the...

First problem (the charred will): 12128316 / 124 = 97809 exactly.

       97809
    --------
124)12128316
    1116
    ----
      968
      868
      ---
      1003
       992
      ----
        1116
        1116
        ----

So the fortune was 12,128,316 and there were 124 heirs, each getting
97,809.

Reasoning.  Call the divisor D, the quotient digits q1 7 q3 q4 q5 and
the remainders r1, r2, r3.  Two digits are brought down at once
before the last step, so q4 = 0, and r3 has 2 digits (10 <= r3 <= 12
at most, since 10*r3 + digit < D <= 128 below).
- The third partial dividend P3 = 10*r2 + digit has 4 digits, so
  r2 >= 100; the second partial dividend 7D + r2 has 3 digits, so
  7D <= 899 and D <= 128.
- q3*D has 3 digits but P3 = q3*D + r3 >= 1000, with r3 < D.  So
  (q3 + 1)*D > 1000, and q3 >= 7.
- If q3 = 7, then 8D > 1000 gives D >= 126, and 7D + r3 >= 1000
  needs r3 >= 104, impossible.  So q3 = 8 and 8D <= 999, D <= 124.
- q1*D and q5*D have 4 digits, so q1 = q5 = 9 (8*124 = 992).
- The last partial dividend is 9D = 100*r3 + two digits, and
  9D <= 1116 gives r3 <= 11.  Then 8D + r3 >= 1000 forces
  8D >= 989, D >= 124.  So D = 124.
Now everything follows: 9D = 1116, r3 = 11, P3 = 992 + 11 = 1003,
r2 = 100, P2 = 868 + 100 = 968, r1 = 96, P1 = 1116 + 96 = 1212, and
the dividend is 12128316.

Second problem (Barry Wolk's, which is Berwick's "seven sevens"):
7375428413 / 125473 = 58781 exactly.

            58781
       ----------
125473)7375428413
       627365
       ------
       1101778
       1003784
       -------
         979944
         878311
         ------
         1016331
         1003784
         -------
           125473
           125473
           ------

The seven given 7s are: the 3rd digit of the dividend, the 5th digit
of the divisor, the 3rd digit of the quotient, the 6th digit of the
first remainder line (1101778), the 2nd digits of 979944 and 878311,
and the 5th digit of 1003784 in the fourth subtraction.

Check.  Both skeletons were solved by exhaustive computer search
(C++): every divisor of the right length and pattern, with every
quotient digit, the numbers in each line forced to have exactly the
digit count and position shown (no leading zeros), the given 7s in
place, and remainder 0.  Each has exactly one solution, the one above.
The same search applied to the identical skeleton filed as
arithmetic/digits/seven.sevens gives the same answer.

-- Claude (rec.puzzles Oracle, 2026-11-02)

[ Editor: the material collected earlier follows. ]

The first problem appears in the book "Ingenious Mathematical Problems
and Methods", by L. A. Graham, Dover Publications, 1959.  It is called
"The Charred Will Mystery".  The book has a long narrative accompanying
the problem.

The second problem was posted on Dec 9, 1994 by Barry Wolk, who said:
"I have proofread this problem VERY carefully."  No one responded to that
article.  Problems involving long division have so many clues that it is
hard to know where to start.

-- Barry Wolk <wolk@ccm.UManitoba.CA>