Dice sum 3

Probability Level 4

I have a 2-sided dice with sides labelled $1,2$ , a 3-sided dice with sides labelled $1,2,3$ , and so on, up to a 49-sided dice with sides labelled $1,2,3,...,49$ . I simultaneously roll all 48 of them. Let $d_{i}$ denote the number that comes up on the dice with $i$ sides. The number of ways that I can roll the dice so that $3 \mid \displaystyle\sum_{i=2}^{49} d_{i}$ can be expressed in the form $\dfrac{n!}{3}$ . What is $n$ ?

This is the final (and hardest) of the dice sum problems. For the previous problem, see Dice Sums 2 .

The answer is 49.

3 solutions

Jon Haussmann
Apr 18, 2014

Roll all the dice, except the die with 3 sides. There are a total of $2 \times 4 \times 5 \times 6 \times \dots \times 49 = 49!/3$ possible outcomes. Then no matter what the previous rolls were, there is exactly one way to roll the die with 3 sides, so that the sum of the rolls is divisible by 3. So the $n$ we seek is 49.

Eddie The Head
Apr 20, 2014

First we roll all the dice except the third.We can notice that whatever be the remainder of the sum of these outcomes when divided by 3,we are only left with one possible choice from the 3-sided dice in order tho make the whole thing divisible by 3.

So the total number of possibilities = $2*4*5*......*49$ = $\boxed{\frac{49!}{3}}$

Kevin Bourrillion
Apr 26, 2014

Out of 49! possible rolls, we would expect very close to 1/3 of them to produce totals divisible by 3. The problem gives the answer away by telling us that it is indeed exactly n!/3 for some n. Either our hunch is right or it is WAY off, so it must be right.

Dice sum 3

The answer is 49.

3 solutions

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