# Find the value of the addition of dice [closed]

Given these:

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.

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

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 $16k + 9$, $k\in\mathbb{N}$.

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

• 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. Commented Jan 31, 2018 at 0:43

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.

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.

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