666 dice

The only dice that really matters is the very last dice rolled because the sum of the first five must be either 0 mod 6, 1 mod 6, 2 mod 6, 3 mod 6, 4 mod 6, or 5 mod 6. (any integer must be one of these 6 due to the nature of mods) Thus, for each case to get the sum to evenly divide by 6, there is only one number to get the final sum to 0 mod 6. So, our answer is 1 in 6 or 1/6 which gives us a value for a+b of 7

Let us consider the case where five die are rolled. No matter what that sum is, there can only be one number the six dice can be to make the total sum a multiple of six. There is a $\frac{1}{6}$ chance that the one number will be rolled so the answer is $1+6=7$

Suppose we are rolling $n$ dice.

$\textbf{Claim:}$ The probability of rolling a sum that is a multiple of six is $\frac16$ for all integers $n\ge1$ .

$\textbf{Proof:}$

Let the result of the roll of the $n^{th}$ die be $r_n$ .

Case 1: $n=1$ .

The probability of rolling a multiple of six is creary $\frac16$ .

Case 2: $n>1$ .

Let $0\le a\le 5$ be the integer such that $\sum_{k=1}^{n-1}r_k\equiv a\pmod6$ Note that for $\displaystyle\sum_{k=1}^{n}r_k$ to be a multiple of six, we must have $r_n\equiv 6-a\pmod6\quad\implies\quad r_n=6-a,$ since $1\le r_n\le6$ . Hence for every $a$ there exists exactly one $r_n$ satisfying the conditions, so the probability is $\frac16$ .

Therefore, the final answer is $\boxed{\frac16}$ .

No matter what the first two dice are, if the third die is a certain number, and only one number, then the sum will be a multiple of 6. Therefore the probability is 1/6, and 1+6=7.

Each die can roll any number from 1 to 6. With six dice, the sum of the numbers rolled can vary from 6 (6x1) to 36 (6x6). Over this range of numbers, there are a = 6 possible multiples of 6: 6, 12, 18, 24, 30, and 36. There are b = 6x6 = 36 possible rolls. a/b = 6/36 which reduces to 1/6 as the probability of rolling a multiple of six. For a/b = 1/6, a+b = 7.

First dice can take 6 possible values ranging 1 to 6. So probability of getting a number divisible by 6 is 1/6. First 2 dice when rolled together they can take 36 possible values of which 6 are multiple of 6. so probability is 6/26=1/6. same argument goes on for 3-dice,4-dice,5-dice and 6-dice case also. in case of 6-dice rolled together there r 46656 possible values and 7776 r divisible by 6. again probability is 1/6. hence answer is 7

there six multiples of 6, (6,12,18,24,30,36), than the probability is sum of number of ways of multiplies divided by $6^6$ ,.

for calculating the number of ways of multiples like $d_1+d_2+d_3+d_4+d_5+d_6=24$ i used inclusion–exclusion principle.

(this is algorithm, i know isn't a programming problem it's just from simplifying calculation)

where $n = 24, k = 6, R = 6, 1 \leq d_i \leq R$

def numOfWays(n,k,R):
    n -= k
    p = 2
    i = 0
    s = 0
    while n - i * R >= 0:
            s += ((-1)**p) * C(k,k - i)* C(n + k - 1 - i * R, k - 1)
            p += 1
            i += 1
    return s

The only dice that really matters is the very last dice rolled because the sum of the first five must be either 0 mod 6, 1 mod 6, 2 mod 6, 3 mod 6, 4 mod 6, or 5 mod 6. (any integer must be one of these 6 due to the nature of mods) Thus, for each case to get the sum to evenly divide by 6, there is only one number to get the final sum to 0 mod 6. So, our answer is 1 in 6 or 1/6 which gives us a value for a+b of 7