11
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An alphametic puzzle containing 42. Don't panic although this is so big. Note that it contains the decimal point and thus several * must be 0.

enter image description here

Here's an ascii version for easier editing:

          * * * * * * 4 2.* * * * *
    _______________________________
* * | * * * * * * * * * *.
        * *              .
        ---              .
          * *            .
          * *            .
          ---            .
          * * *          .
            * *          .
          -----          .
            * * *        .
              * *        .
              ---        .
                * *      .
                * *      .
                ---      .
                  * * *  .
                  * * *  .
                  -----  .
    D                 * *
     O                * *
      N               ---
       '              * * *
        T             * * *
                      -----
          P               * * *
           A              * * *
            N             -----
             I              * * *
              C             * * *
                            -----
                              * * *
                              * * *
                              -----
                                  *
$\endgroup$
  • $\begingroup$ Each asterisk can be a digit between 0 and 9 right? Independantly? (that is a dumb comment but we need to be really sure) $\endgroup$ – Saeïdryl Jun 29 '18 at 13:23
  • 2
    $\begingroup$ I'm not sure this quite qualifies as an alphametic. $\endgroup$ – Ian MacDonald Jun 29 '18 at 13:24
  • $\begingroup$ @Saeïdryl Sure. Besides, the first asterisk of each line is nonzero. $\endgroup$ – P.-S. Park Jun 29 '18 at 13:29
  • 2
    $\begingroup$ Do you think you could provide a bit more context for why I would want to bother solving this? Long division of a 10-digit number just to produce a small segment of the answer as having "42" is rather arbitrary and not very interesting. Does the rest of the number have some link to Douglas Adams' work? $\endgroup$ – Ian MacDonald Jun 29 '18 at 13:38
  • 4
    $\begingroup$ Three minutes into this puzzle I already have definite solutions for the first digits of every number, and the first digit of the dividend (which I believe could be also called the numerator) in particular should be blindingly obvious from the first subtraction only. VTC crew, please do spend at least a token effort to verify the reason before voting to close. Voting to reopen, of course. $\endgroup$ – Bass Jul 5 '18 at 8:42
11
$\begingroup$

The original calculation is

$\frac{1002324327}{31} = 32333042.80645 \text{ (remainder } 0.00005 \text).$

The text suggestion (Don't Panic) is certainly helpful, since the road to the solution has quite a few twists in it.

I'm not going to be spoiler-tagging everything (it would be a solid spoiler block all the way), so please avert your eyes, if you want to solve this yourself. (And you should want to, it's an excellent puzzle,)

The basic idea is that all information is encoded in the various patterns of the long division. From these, it's possible to extract constraints on the values, and other information nuggets, and from these it's possible to reconstruct the calculation completely.

First, start by filling in the more obvious bits:

If you subtract a two-digit number from a three-digit number, and get a one-digit answer, the numbers must be 10x and 9y. The same logic applies, if the subtraction result is 10, so you can use it a second time.

Also, fill in the zeroes in the decimal part, and the two zeroes in the quotient.

          * * * * * 0 4 2.* 0 * * *
    _______________________________
* * | 1 0 * * * * * * * *.
        9 *              .
        ---              .
          * *            .
          * *            .
          ---            .
          1 0 *          .
            9 *          .
          -----          .
            1 0 *        .
              9 *        .
              ---        .
                * *      .
                * *      .
                ---      .
                  * * *  .
                  * * *  .
                  -----  .
    D                 * *  
     O                * *
      N               ---  
       '              * * 0
        T             * * *
                      -----
          P               * 0 0
           A              * * *
            N             -----
             I              * * 0
              C             * * *
                            -----
                              * * 0
                              * * *
                              -----
                                  *

Then, we must reason about the divisor.

We know that the "divisor times 2" has 2 digits, and that "divisor times 4" has 3 digits. This limits the possible divisor values quite nicely.

Since the number in the first subtraction has two digits, the first digit of the quotient cannot therefore be greater than 3. Check the other possible cases against the numbers in the first subtraction:
* if the quotient starts with 1, the divisor must be 90-99. This times two has too many digits (see under the 42), so 1 is right out.
* if the quotient starts with 2, then the divisor must be 45-49. This times two is at least 90. Directly under the 42, there is a subtraction, where two times the divisor is taken from a 2-digit number, and the difference has two digits, so this is also impossible.

Therefore, the quotient must start with 3, and the divisor is either 31,32 or 33.

Fill that in, and do the same for the couple of other spots where the divisor times something is ninety-something.

While we're at it, add the first digit of 4 * divisor, and the 1 above it (deduced from the subtraction leaving only one digit)

          3 * 3 3 * 0 4 2.* 0 * * *
    _______________________________
3 * | 1 0 * * * * * * * *.
        9 *              .
        ---              .
          * *            .
          * *            .
          ---            .
          1 0 *          .
            9 *          .
          -----          .
            1 0 *        .
              9 *        .
              ---        .
                * *      .
                * *      .
                ---      .
                  1 * *  .
                  1 * *  .
                  -----  .
    D                 * *  
     O                6 *
      N               ---  
       '              * * 0
        T             * * *
                      -----
          P               * 0 0
           A              * * *
            N             -----
             I              * * 0
              C             * * *
                            -----
                              * * 0
                              * * *
                              -----
                                  *

Then, name the second digit of the divisor "X" (remembering from above that X must be either 1, 2 or 3), and attack the decimal portion of the calculation:

          3 * 3 3 * 0 4 2.* 0 * * *
    _______________________________
3 X | 1 0 * * * * * * * *.
        9 *              .
        ---              .
          * *            .
          * *            .
          ---            .
          1 0 *          .
            9 *          .
          -----          .
            1 0 *        .
              9 *        .
              ---        .
                * *      .
                * *      .
                ---      .
                  1 * *  .
                  1 * *  .
                  -----  .
    D                 * *  
     O                6 *
      N               ---                 Each of the following comments pertains the line it's on.
       '              * * 0               The logic proceeds in order from the top down.
        T             * * *               
                      -----
          P               * 0 0           200 or 300. (Not 100, because next line has 3 digits. Not 400+, because the divisor would fit more than 10 times.)
           A              * * *           If previous line is 200, quotient has 6, and this is 180+6X. If 300, quotient has 9, and this line is 270+9X. 
            N             -----           - X cannot be 3, because in both cases this subtraction would leave one digit only.
             I              * * 0         - X cannot be 2, because this line would have to be 120, and there's no three-digit multiple of 32 below 120. So X is 1, and this is 140 or 210                    
              C             * * *         If previous line is 140, this is 124 (4*31), leaving 16. If 210, this is 186(6*31), leaving 24
                            -----
                              * * 0       160 or 240.
                              * * *       155 (7*31 = 217 would leave too many digits after the subtraction)
                              -----
                                  *       this is therefore 5.

So we got the divisor, and a couple of the last numbers. Working upwards from there yields (mostly by simple arithmetic):

          3 * 3 3 * 0 4 2.8 0 6 4 5
    _______________________________
3 1 | 1 0 * * 3 * * 3 2 7.
        9 3              .
        ---              .
          * *            .
          * *            .
          ---            .
          1 0 3          .
            9 3          .
          -----          .
            1 0 *        .
              9 3        .
              ---        .
                * *      .
                * *      .
                ---      .
                  1 3 2  .
                  1 2 4  .
                  -----  .
    D                 8 7  
     O                6 2
      N               ---  
       '              2 5 0
        T             2 4 8        (if the last digit is 8, the quotient will also be that)
                      -----
          P               2 0 0   
           A              1 8 6   
            N             -----   
             I              1 4 0 
              C             1 2 4 
                            -----
                              1 6 0
                              1 5 5
                              -----
                                  5

and finally, with a couple more deductions, we are done:

          3 2 3 3 3 0 4 2.8 0 6 4 5
    _______________________________
3 1 | 1 0 0 2 3 2 4 3 2 7.
        9 3              .
        ---              .
          7 2            .      at least 70, so quotient is 2 or 3
          6 2            .      cannot be 93, because difference is 10, would need third digit above.
          ---            .
          1 0 3          .
            9 3          .
          -----          .
            1 0 2        .
              9 3        .
              ---        .
                9 4      .      at least 70, so quotient cannot be 1 
                9 3      .      if 62, the above would be 63, which is less than 70.
                ---      .
                  1 3 2  .
                  1 2 4  .
                  -----  .
    D                 8 7  
     O                6 2
      N               ---  
       '              2 5 0
        T             2 4 8    
                      -----
          P               2 0 0   
           A              1 8 6   
            N             -----   
             I              1 4 0 
              C             1 2 4 
                            -----
                              1 6 0
                              1 5 5
                              -----
                                  5
$\endgroup$

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