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Each line in the following list of equalities may be made true by moving, according to a common rule, the numbers at the left of the = equals sign in that line.

      What is the value of X in X 11 10 9 = 11? 

                        0 1 2 3 4 5 6 7 8 9  =  0
                          1 2 3 4 5 6 7 8 9  =  1
                            2 3 4 5 6 7 8 9  =  2
                              3 4 5 6 7 8 9  =  3
                                4 5 6 7 8 9  =  4
                                  5 6 7 8 9  =  5
                                    6 7 8 9  =  6
                                      7 8 9  =  7
                                        8 9  =  8
                                       10 9  =  9
                                    11 10 9  =  10
                                  X 11 10 9  =  11
                                 12 11 10 9  =  12
                           10 11 12 11 10 9  =  13
                                             .
                                             .
                                             .
              ... including, in case it’s not obvious, ...
 10 11 12 13 14 15 16 17 18 19 20 19 18 17 16 15 14 13 12 11 10 9  =  153
                                             .
                                             .
                                             .

No symbols or markings need to be added and the numbers may move left, right, up and down as long as they remain whole, separate, and to the left of the = equals sign.

The rule for moving numbers is so simple that it can be stated in an English sentence of 9 words or fewer. Understanding the result, though, requires remembering a convention familiar to a vast majority of mathematicians and taught to even more students who do not go on to pursue that field.

This list of lines begins with = 0 but has no end and, incidentally, the entry for each line has infinitely many alternatives. The present equations were chosen to emphasize some patterns into which the value for X fits in a surprising way.

(Note: This puzzle’s presentation was overhauled in April, 2020, and many comments from 2017 no longer apply exactly.)

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15
  • $\begingroup$ Does "to a unique integer result" mean (1) that each line's result is well-defined or (2) that all the lines' results are different? $\endgroup$
    – Gareth McCaughan ♦
    Jul 5, 2017 at 11:21
  • $\begingroup$ Thanks once again, @Gareth, edited. I was trying to avert unintended utterly simple solutions, like every line = C. $\endgroup$
    – humn
    Jul 5, 2017 at 11:30
  • 1
    $\begingroup$ I was looking for an 'x' in the first line (line 1). It's actually lower-case 'L'. $\endgroup$
    – Lawrence
    Jul 5, 2017 at 12:02
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    $\begingroup$ @humn No problem :) . Upper-case line references is a good idea in this question, actually. $\endgroup$
    – Lawrence
    Jul 5, 2017 at 12:10
  • 1
    $\begingroup$ Thank you for a great puzzle idea or two, @Paul Karam! Something to seem so simple that it's almost impossible. $\endgroup$
    – humn
    Jul 7, 2017 at 17:08

3 Answers 3

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3
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The composition is

A string of right-associative $x_y$ operations, where $y$ is the base to which $x$ is expressed. The RHS is expressed in base 10 (the familiar '10', as opposed to the one where 'every base is base 10' :) ).

With an ascending string of single-digit numbers, the whole string reduces to the left-most operand because we start on the right and each pair reduces to the left operand. For example,

$3_4 = 3$.

The given examples all check out, including the monster that evaluates to 153.

We can start with [12 11 10 9] = 12, as given, then continue with:

$13_{12}$ = 12 + 3 = 15
$14_{15}$ = 15 + 4 = 19
...
$10_{153}$ = 153

Now, we have the following reduction since [11 10 9] = 10:

$X_{[11,10,9]} = X_{10}$

which we are told evaluates to 11.

Therefore $X=11$.

$$ \require{begingroup}\begingroup \def\b#1{_{\large\,#1}} \begin{array}{rcl} 0\b{1\b{2\b{3\b{4\b{5\b{6\b{7\b{8\b9}}}}}}}} & = & 0 \\ 1\b{2\b{3\b{4\b{5\b{6\b{7\b{8\b9}}}}}}} & = & 1 \\ 2\b{3\b{4\b{5\b{6\b{7\b{8\b9}}}}}} & = & 2 \\ 3\b{4\b{5\b{6\b{7\b{8\b9}}}}} & = & 3 \\ 4\b{5\b{6\b{7\b{8\b9}}}} & = & 4 \\ 5\b{6\b{7\b{8\b9}}} & = & 5 \\ 6\b{7\b{8\b9}} & = & 6 \\ 7\b{8\b9} & = & 7 \\ 8\b9 & = & 8 \\ 10\b9 & = & 9 \\ 11\b{10\b9} & = & 10 \\ {\large\bf 11}\b{11\b{10\b9}} & = & 11 \\ 12\b{11\b{10\b9}} & = & 12 \\ 10\b{11\b{12\b{11\b{10\b9}}}} & = & 13 \\ & \vdots \\ 10\b{11\b{12\b{13\b{14\b{15\b{16\b{17\b{18\rlap{\b{19\b{20\b{19\b{18\b{17\b{16\b{15\b{14\b{13\b{12\b{11\b{10\b9}}}}}}}}}}}}}}}}}}}}} & = & 153 \\ & \vdots \\ \end{array} \endgroup $$

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2
  • 1
    $\begingroup$ One for the road. :) $\endgroup$
    – Lawrence
    Apr 12, 2020 at 16:34
  • 1
    $\begingroup$ And a picture as a chaser $\endgroup$
    – humn
    Apr 16, 2020 at 17:48
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+100
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Note: The puzzle’s presentation has been revised since the time of this posting. The version from that time is appended to this answer.


Lines $i$ and $j$ are the only other pair presented where consecutive rows have the same number of elements. There, the left-most digit is incremented by 2. Otherwise, the number in each column stays the same as the preceding row, except that new numbers can be introduced to the left. The rules for the new numbers isn't sought by this puzzle, and $x$ can be determined by the simple rule observed for lines $i$ and $j$.

Applying that simple rule to lines $l$ and $m$,

the increment-by-two rule gives us $12 = m + 2$, so $m = 10$.


Version of the puzzle when this answer was posted

Each of the following line a through line n evaluates, by one utterly simple secret rule, to a well-defined unique-from-each-other integer result. The resulting column of integers has an equally simple pattern.

line a:                         0 1 2 3 4 5 6 7 8 9
line b:                           1 2 3 4 5 6 7 8 9
line c:                             2 3 4 5 6 7 8 9
line d:                               3 4 5 6 7 8 9
line e:                                 4 5 6 7 8 9
line f:                                   5 6 7 8 9
line g:                                     6 7 8 9
line h:                                       7 8 9
line i:                                         8 9
line j:                                        10 9
line k:                                     11 10 9
line l:                                   X 11 10 9
line m:                                  12 11 10 9
line n:                            10 11 12 11 10 9

        What is the value of X in line l?

    (And what utterly simple rule applies to all lines?)

Notes from comments:
The l in line l is lowercase L, not numbercase one.
This sequence continues forward infinitely but not backward.

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16
  • $\begingroup$ Thank you for following one of my false leads. I'm trying to learn how to balance genuine clues and red herrings. Another puzzle was nothing but herrings. $\endgroup$
    – humn
    Jul 8, 2017 at 23:44
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    $\begingroup$ Apologies, Lawrence, I neglected to acknowledge that your rule does work, yet an even simpler rule produced these lines $\endgroup$
    – humn
    Jul 22, 2017 at 23:46
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    $\begingroup$ @humn I’m glad I did. I had about a 17h day today if I’ve counted my hours correctly, and this is such a nice surprise to find as I’m unwinding from the long day. Thanks again, and have a happy Easter. :) $\endgroup$
    – Lawrence
    Apr 12, 2020 at 15:59
  • 1
    $\begingroup$ Perhaps solving this puzzle was a nice palate cleanser for you, @Lawrence, and when I say that an answer gets a bounty, it gets its own bounty. It's the principle, not the points. Again, I am grateful for your helping me realize that the original presentation was too coyly ambiguous. What if I append a snapshot to the solution, as I did to your previous answers, but this time of the rearranged numbers to go along with your more-complete-than-I'd-hoped analysis? $\endgroup$
    – humn
    Apr 16, 2020 at 15:32
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    $\begingroup$ @humn Solving the years-long puzzle was certainly satisfying. Feel free to edit as you wish. :) $\endgroup$
    – Lawrence
    Apr 16, 2020 at 15:48
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+50
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Note: The puzzle’s presentation has been revised since the time of this posting. The version from that time is appended to this answer.


Rule: adjacent elements on each line differ by 1.

The only element adjacent to $x$ is 11. So if $x \ne 12$ by line $m$, we have $x=10$.


Version of the puzzle when this answer was posted

Each of the following line a through line n evaluates, by one utterly simple secret rule, to a well-defined unique-from-each-other integer result. The resulting column of integers has an equally simple pattern.

line a:                         0 1 2 3 4 5 6 7 8 9
line b:                           1 2 3 4 5 6 7 8 9
line c:                             2 3 4 5 6 7 8 9
line d:                               3 4 5 6 7 8 9
line e:                                 4 5 6 7 8 9
line f:                                   5 6 7 8 9
line g:                                     6 7 8 9
line h:                                       7 8 9
line i:                                         8 9
line j:                                        10 9
line k:                                     11 10 9
line l:                                   X 11 10 9
line m:                                  12 11 10 9
line n:                            10 11 12 11 10 9

        What is the value of X in line l?

    (And what utterly simple rule applies to all lines?)

Notes from comments:
The l in line l is lowercase L, not numbercase one.
This sequence continues forward infinitely but not backward.

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3
  • $\begingroup$ But that was "backward" I think, which OP said that it doesn't work like that... I suppose. $\endgroup$
    – Paul Karam
    Jul 23, 2017 at 0:37
  • $\begingroup$ I like it! That adjacent elements differ by 1 was an aesthetic choice, though. $\endgroup$
    – humn
    Jul 23, 2017 at 1:15
  • $\begingroup$ The puzzle statement has been clarified and hinted, but I realized that I've never seen the simple-looking mathematics in play be used like this, entirely valid though it is. $\endgroup$
    – humn
    Jul 23, 2017 at 8:44

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