Henry Ernest Dudeney/Puzzles and Curious Problems/110 - An Absolute Skeleton/Solution 2

Puzzles and Curious Problems by Henry Ernest Dudeney: $110$

An Absolute Skeleton

Here is a good skeleton puzzle.

The only conditions are:

$(1)$ No digit appears twice in any row of figures except the dividend.

$(2)$ If $2$ be added to the last figure in the quotient it equals the last but one,

and if $2$ be added to the third figure from the end it gives the last figure but $3$ in the quotient.

That is to say, the quotient might end in, say, $9742$, or in $3186$.

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    ------------
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      ***
       ---
       ***
       ***
       ----
       ****
       ****
       -----
         ***
         ***
         ----
         ****
         ****
         -----
          ****
          ****
          -----
           ****
           ****
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            ****
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            ----

Solution

        32819764
    ------------
 310)10174126840
      930
       ---
       874
       620
       ----
       2541
       2480
       -----
         612
         310
         ----
         3026
         2790
         -----
          2368
          2170
          -----
           1984
           1860
           -----
            1240
            1240
            ----

Proof

From the initial deductions, we have determined that:

$(1): \quad$ It is possible that there is a solution in which the divisor is $310$

$(2): \quad$ If this is the case, then the quotient is one of:

$32418697$

$32419786$

$32816497$

$32819764$

Thus the dividend in each case will be:

$10049796070$

$10050133660$

$10173114070$

$10174126840$

It remains to investigate each one.

Hence we set up a long division and check the partial dividends in each case.

        32418697
    ------------
 310)10049796070
      930
       ---
       749
       620
       ----
       1297
       1240
       -----
         579
         310
         ----
         2696
         ****

and we need go no further.

        32419786
    ------------
 310)10050133660
      930
       ---
       750
       620
       ----
       1301
       1240
       -----
         613
         310
         ----
         3033
         ****
         -----

and we need go no further.

        32816497
    ------------
 310)10173114070
      930
       ---
       873
       620
       ----
       2531
       2480
       -----
         511
         ***
         ----

and we need go no further.

Finally:

        32819764
    ------------
 310)10174126840
      930
       ---
       874
       620
       ----
       2541
       2480
       -----
         612
         310
         ----
         3026
         2790
         -----
          2368
          2170
          -----
           1984
           1860
           -----
            1240
            1240
            ----

and we have found a solution.

Hence the only solution with $D = 310$ is:

$\dfrac {10174126840} {310} = 32819764$

$\blacksquare$