# Rolling a die $1000$ times, there is a time interval such that the product of all the results in that interval is a cube of $3$.

Each of the six faces of a die is marked with an integer, not necessarily positive. The die is rolled $$1000$$ times. Show that there is a time interval such that the product of all rolls in this interval is a cube of an integer. (For example, it could happen that the product of all outcomes between 5th and 20th throws is a cube; obviously, the interval has to include at least one throw!)

I looked at abcdef as a possible product , somehow $$1000$$ of these 'terms' no matter how we shuffle them , a product like ababab or adadad or acdacdacd will be in the string but i cant prove how :(

We can show something stronger; there exists an interval of rolls where each side has appeared a number of times which is a multiple of $$3$$.

For each $$i=1,2,\dots,1000$$, let $$a_i$$ be number of times that the first face has appeared in rolls numbered $$1,2,\dots,i$$. Same for $$b_i,c_i,d_i,e_i,f_i$$. Consider the $$1000$$ ordered six-tuples of values $$T_i:=(a_i,b_i,\dots,f_i)$$, where each coordinate is only recorded modulo $$3$$. There are $$3^6=729$$ possible six-tuples, and there are $$1000$$ rolls, so by the pigeonhole principle, there are two indices $$i for which $$T_i=T_j$$. This implies that among rolls numbered $$i+1,i+2,\dots,j$$, each face appears a number of times which is a multiple of three.

• i dont get it, how again can u use pigeonhole? and how does this work for cubes not 3k – Randin D Apr 25 at 4:32
• (a) There are $1000$ numbers $i\in\{1,2,\dots,1000\}$, and $729$ possible tuples. $1000$ pigeons in $729$ holes. (b) If each face appears a multiple of three times, and the numbers of the faces are $x,y,z,w,s,t$, then the product of the numbers in that interval $[i+1,j]$ is $x^{3i}y^{3j}\cdots t^{3k}$, which is the cube of $x^iy^j\cdots t^k$. – Mike Earnest Apr 25 at 4:36
• i get pigeonhole ..1000 into 729 means 3 will be in one pigeonhole right? and thats the cube? – Randin D Apr 25 at 4:39
• how do u get the 729 again? – Randin D Apr 25 at 4:41
• Do you understand my sentence that starts with “Consider...”? We keep track of the remainder of the number of times each face appears in the first $i$ rolls modulo 3. There are three possible remainders, and six faces, so $3^6$. The reason this is helpful is because if the number of times the $a$ face appears in the first $i$ rolls is $3m+r$, and in the first $j$ rolls is $3n+r$, then subtracting, the number of times $a$ appears after $i$ and up to $j$ is $3(n-m)$, which is a multiple of $3$. – Mike Earnest Apr 25 at 16:02

Strengthening the result another way:

For each $$i=1,\dots,1000$$ let $$2^{a_i}\cdot 3^{b_i}\cdot 5^{c_i}$$ be the product of the first $$i$$ rolls. Then there are only $$3^3=27$$ possible values for the 3-tuple $$(a_i\bmod 3, b_i\bmod 3, c_i\bmod 3)$$, so, by a pigeonhole principle argument similar to Mike Earnest's in his answer, a cube must appear if there are more than 27 rolls.

• The question specifies that the sides of the die might be numbered with any integers, not necessarily just 1, 2, 3, 4, 5 and 6. I believe the worst case would be one where each side is numbered with a distinct prime. – Ilmari Karonen Apr 25 at 9:04
• @IlmariKaronen OK, good point, now that I see that in the OP. – Rosie F Apr 25 at 9:05