7
$\begingroup$

Given these:

Puzzle

Find the rule, and provide the answer to "?" (the last row).

EDIT:Even though "+" denotes addition, it should not be taken in a tradition sense. The positioning and pattern could affect the answer. There should be one (or a few) numerical answers at the end.

$\endgroup$
7
  • $\begingroup$ Is the + symbol addition, or can it be interpreted differently? $\endgroup$
    – Phylyp
    Jan 30, 2018 at 2:17
  • 1
    $\begingroup$ @Phylyp It can be anything. For example, in the first row, a combination of one, three, five, six produces the output 15. $\endgroup$
    – John Doe
    Jan 30, 2018 at 2:24
  • 2
    $\begingroup$ Is this a puzzle you created yourself? (And if not, could you provide attribution?) $\endgroup$
    – puzzledPig
    Jan 30, 2018 at 4:20
  • $\begingroup$ Does the alignment of the dice matter or is that just an accident of whatever graphics program you used to make the image? $\endgroup$ Jan 30, 2018 at 20:03
  • $\begingroup$ @Bachrach44 The fact that the dices are slightly off center is just human error. The ordering does, however, matter. $\endgroup$
    – John Doe
    Jan 30, 2018 at 22:39

3 Answers 3

5
$\begingroup$

This puzzle actually has infinitely many solutions. Using the notation by @JamalSenjaya, we have the matrix

$$\left[\begin{array}{cccccc|c}D_1&D_2&D_3&D_4&D_5&D_6\\\hline1&0&1&0&1&1&15\\ 1&1&1&1&0&0&8\\0&1&1&1&0&1&5\\1&0&0&1&2&0&17\\0&2&1&0&0&1&5\\3&0&0&1&0&0&x\end{array}\right]$$

where $?$ is denoted by $x$. Reducing this into its triangular form gives

$$\left[\begin{array}{cccccc|c}D_1&D_2&D_3&D_4&D_5&D_6\\\hline1&0&0&-\frac13&0&0&\frac x3\\0&1&\frac12&0&0&\frac12&\frac52\\0&0&1&-\frac13&1&1&\frac{45-x}3\\0&0&0&1&-\frac37&0&\frac{x-30}7\\0&0&0&0&1&0&\frac{139-3x}{16}\\0&0&0&0&0&1&\frac{5x-45}{16}\end{array}\right]$$

Hence the set of solutions is

$$\begin{align}D_6&=\frac5{16}(x-9)\\D_5&=7-\frac3{16}(x-9)\\D_4&=\frac1{16}(x-9)\\D_3&=5-\frac7{16}(x-9)\\D_2&=\frac1{16}(x-9)\\D_1&=3+\frac5{16}(x-9)\end{align}$$

In other words, we want $x$ such that all of $D_1,D_2,D_3,D_4,D_5,D_6$ to be integers, and there are infinitely many such $x$ of the form $16k + 9$, $k\in\mathbb{N}$.

$$x = \cdots, -39, -23, -7, 9, 25, 41, \cdots$$

$\endgroup$
1
  • 1
    $\begingroup$ This is assuming that each die stands for a unique value. I feel like there's probably some odd trick going on that still uses the actual values on the die to get to those values. $\endgroup$
    – DqwertyC
    Jan 31, 2018 at 0:43
4
$\begingroup$

This puzzle can have many solutions

lets denote the dice 1 = D1, dice 2 as D2, ... , to dice 6 as D6.

so the solutions can be :

D1 = 3, D2 = 0, D3 = 5, D4 = 0, D5 = 7, D6 = 0, so the answer is 9.
I think this is the most suitable answer, because it do not use negative numbers, it also have pattern. The pattern is odd dices = (number of dots) + 2, even dices = 0.

or if we can use negative numbers

D1 = -12, D2 = -3, D3 = 26, D4 = -3, D5 = 16, D6 = -15, so the answer is -39

or

D1 = -7, D2 = -2, D3 = 19, D4 = -2, D5 = 13, D6 = -10, so the answer is -23

and there are still some more solutions we can search for.

$\endgroup$
3
$\begingroup$

Given the details in the puzzle there are an infinite number of answers (can't say yet if countable many or an uncountable infinite).

For example, a solution which uses the fact that you gave 5 examples with 4 dice:

-369/141 * first_dice + 669/141 * second_dice - 355 / 141 * third_dice - 13/141 * fourth_dice + 2330/141

So, for the ?, we get

2223/141

Of course, this is most likely not the answer you're looking for, so the puzzle needs to be made better.

$\endgroup$
2
  • 2
    $\begingroup$ Welcome to Puzzling. Your answer seems to extrapolate a lot, try to find a simple answer. Occam's Razor :) $\endgroup$
    – ABcDexter
    Jan 30, 2018 at 6:07
  • 1
    $\begingroup$ I'm sorry, but I don't see how this is more complex than the other answers here. $\endgroup$ Jan 30, 2018 at 14:22

Not the answer you're looking for? Browse other questions tagged or ask your own question.