=Feynman Puzzle =

This is the solution to the long division puzzle on page 5 of "Perfectly Reasonable Deviations from the Beaten Path", letters of Richard Feynman collected by his daughter Michelle and published in 2005 .

Here is the long division puzzle:

          ##A#
     ---------
 #A# ) ####A##
       ##AA
       ----
        ###A
         ##A
        -----
         ####
         #A##
         -----
          ####
          ####
          ----
             0

A can be any digit; each # can be any digit from 0 through 9 except A. There is only one solution.

How to solve it:

Use a computer as a big dumb hammer. Boring, though more difficult to program than you might imagine.

How to solve it the smart way:

Let's assign some algebraic values, b through f, to the digits in the two factors:

          deAf
     ---------
 bAc ) ####A## 
       ##AA
       ----
        ###A
         ##A
        -----
         ####
         #A##
         -----
          ####
          ####
          ----
             0

We know quite a few things here. We know that ( d * Ac ) mod100 = AA , and that ( e * Ac ) mod 10 = A . Since d * bAc is 4 digits, and e * bAc is three digits, we know that d > e . So let's make a table of the possible A's, c's, d's, and e's (keeping in mind that c,d, and e cannot equal A.

We start by looking at all the possible A's. A is mod10 of two pairs of multiplied digits ( c*d and c*e ), d>e, not containing A.

 A  digit pairs
 0  2*5                         no
 1  3*7 9*9                     no
 2  3*4 4*8 6*7 8*9             maybe (4*8 and 4*3) or (8*9 and 8*4)
 3  7*9                         no
 4  2*2 2*7 3*8 8*8 6*9         maybe (2*7 and 2*2) or (8*8 and 8*3)
 5  ( nothing, all include 5 )  no
 6  2*3 2*8 4*4 4*9 7*8         maybe (2*8 and 2*3) or (4*9 and 4*4) or (8*7 and 8*2)
 7  3*9                         no
 8  2*4 2*9 3*6 4*7             maybe (2*9 and 2*4) or (4*7 and 4*2)
 9  3*3 7*7                     no

We need two pairs of multipled digits, with one digit in common, to make c, d, and e. Here are the possibilities in tabular form:

 A  c  d  e  (Ac * d)       mod100  =AA?
 -------------------------------------
 2  4  8  3   24 * 8   192      92  no
 2  8  9  4   28 * 9   252      52  no
 4  2  7  2   42 * 7   294      94  no
 4  8  8  3   48 * 8   384      84  no
 6  2  8  3   62 * 8   496      96  no
 6  4  9  4   64 * 9   596      96  no
 6  8  7  2   68 * 7   476      76  no
 8  2  9  4   82 * 9   738      38  no
 8  4  7  2   84 * 7   588      88  yes!!!

There is only one solution for A, c, d, and e. We have solved most of the problem! So, here is the partially solved puzzle:

          728f
     ---------
 b84 ) ####8## 
       ##88
       ----
        ###8
         ##8
        -----
         ##0#
         #8##
         -----
          ####
          ####
          ----
             0

Now we can figure out b. We know that 8 * b84 = #8## , and since 8*84 is 672, we know that ( 8 * b ) mod 10 is (8-6)=2. Thus, b must be either 4 or 9. It cannot be 9, or d*bAc = 2*984 would be a 4 digit number, not 3 digits. So b is 4. One variable to go!

We know that 728f * 484 is ####8## . 7280 * 484 is 3523520, so ( f * 484 + 520 ) = #8## . Here is a table of ( f * 484 + 520 ):

  f                    #8##?
  1 * 484 + 520 = 1004  no
  2 * 484 + 520 = 1488  no
  3 * 484 + 520 = 1972  no
  4 * 484 + 520 = 1456  no
  5 * 484 + 520 = 2940  no
  6 * 484 + 520 = 3424  no
  7 * 484 + 520 = 3908  no
  8 * 484 + 520 = 4392  no
  9 * 484 + 520 = 4876  yes!!

so f = 9. Our two factors are *484* and *7289* . Done! Here is the long division puzzle, solved:

          7289
     ---------
 484 ) 3527876 
       3388
       ----
        1398
         968
        -----
         4307
         3872
         -----
          4356
          4356
          ----
             0

Discovering this puzzle in an age before calculators is impressive. Apparently, Richard Feynman and his father traded puzzles like this all the time, so they must have spent a lot of time working on them, and they were both rapid and accurate paper and mental calculators!

FeynmanPuzzle (last edited 2011-01-23 02:02:12 by KeithLofstrom)