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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 $3$.

How many positive integers N are there such that the least common multiple of N and 1000 is 1000?What is the probability of covering all $6$ faces of a die after rolling it $14$ times or less?What is the chance of rolling a die and getting the number six three times at exactly 10 rolls?$n$-sided die thrown $n$ times. Probability that sum of results equals sum of prime factors of $n$.Maximum number of integers $(m,n)$ in $Atimes B$ such that $|m-n|leq 1000$Rolling a die 100 times and probability of product of resultsHow many order triple (a,b,c) are there such that $a.b.c le 1000$Probability that the minimum is $2$ and the maximum is $5$ after rolling a die $4$ timesRolling a fair $k$-side die $n$ times, what is the expected number of distinct outcomes?Rolling $4$ dice and multiplying the results. What is the probability that the product is divisible by $5$ or has $5$ as the least significant digit?

2

3

$begingroup$

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 :(

combinatorics

share|cite|improve this question

edited Apr 25 at 9:55

Asaf Karagila♦

309k33442776

asked Apr 25 at 4:01

Randin DRandin D

1026

$endgroup$

add a comment | 

2

3

$begingroup$

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 :(

combinatorics

share|cite|improve this question

edited Apr 25 at 9:55

Asaf Karagila♦

309k33442776

asked Apr 25 at 4:01

Randin DRandin D

1026

$endgroup$

add a comment | 

$begingroup$

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 :(

combinatorics

share|cite|improve this question

edited Apr 25 at 9:55

Asaf Karagila♦

309k33442776

asked Apr 25 at 4:01

Randin DRandin D

1026

$endgroup$

share|cite|improve this question

edited Apr 25 at 9:55

Asaf Karagila♦

309k33442776

asked Apr 25 at 4:01

Randin DRandin D

1026

share|cite|improve this question

edited Apr 25 at 9:55

Asaf Karagila♦

309k33442776

asked Apr 25 at 4:01

Randin DRandin D

1026

edited Apr 25 at 9:55

Asaf Karagila♦

309k33442776

edited Apr 25 at 9:55

Asaf Karagila♦

309k33442776

asked Apr 25 at 4:01

Randin DRandin D

1026

asked Apr 25 at 4:01

Randin DRandin D

1026

2 Answers
2

active

oldest

votes

10

$begingroup$

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 $i<j$ 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.

share|cite|improve this answer

answered Apr 25 at 4:22

Mike EarnestMike Earnest

28.9k22256

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i dont get it, how again can u use pigeonhole? and how does this work for cubes not 3k
  $endgroup$
  – Randin D
  Apr 25 at 4:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (a) There are $1000$ numbers $iin1,2,dots,1000$, 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 $x^3iy^3jcdots t^3k$, which is the cube of $x^iy^jcdots t^k$.
  $endgroup$
  – Mike Earnest
  Apr 25 at 4:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i get pigeonhole ..1000 into 729 means 3 will be in one pigeonhole right? and thats the cube?
  $endgroup$
  – Randin D
  Apr 25 at 4:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how do u get the 729 again?
  $endgroup$
  – Randin D
  Apr 25 at 4:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $3$.
  $endgroup$
  – Mike Earnest
  Apr 25 at 16:02
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

Strengthening the result another way:

For each $i=1,dots,1000$ let $2^a_icdot 3^b_icdot 5^c_i$ be the product of the first $i$ rolls. Then there are only $3^3=27$ possible values for the 3-tuple $(a_ibmod 3, b_ibmod 3, c_ibmod 3)$, 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.

share|cite|improve this answer

answered Apr 25 at 8:31

Rosie FRosie F

1,468416

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Ilmari Karonen
  Apr 25 at 9:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @IlmariKaronen OK, good point, now that I see that in the OP.
  $endgroup$
  – Rosie F
  Apr 25 at 9:05
  

  
  
  
  

add a comment | 

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10

$begingroup$

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 $i<j$ 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.

share|cite|improve this answer

answered Apr 25 at 4:22

Mike EarnestMike Earnest

28.9k22256

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i dont get it, how again can u use pigeonhole? and how does this work for cubes not 3k
  $endgroup$
  – Randin D
  Apr 25 at 4:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (a) There are $1000$ numbers $iin1,2,dots,1000$, 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 $x^3iy^3jcdots t^3k$, which is the cube of $x^iy^jcdots t^k$.
  $endgroup$
  – Mike Earnest
  Apr 25 at 4:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i get pigeonhole ..1000 into 729 means 3 will be in one pigeonhole right? and thats the cube?
  $endgroup$
  – Randin D
  Apr 25 at 4:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how do u get the 729 again?
  $endgroup$
  – Randin D
  Apr 25 at 4:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $3$.
  $endgroup$
  – Mike Earnest
  Apr 25 at 16:02
  

  
  
  
  

 | 
show 1 more comment

10

$begingroup$

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 $i<j$ 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.

share|cite|improve this answer

answered Apr 25 at 4:22

Mike EarnestMike Earnest

28.9k22256

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i dont get it, how again can u use pigeonhole? and how does this work for cubes not 3k
  $endgroup$
  – Randin D
  Apr 25 at 4:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (a) There are $1000$ numbers $iin1,2,dots,1000$, 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 $x^3iy^3jcdots t^3k$, which is the cube of $x^iy^jcdots t^k$.
  $endgroup$
  – Mike Earnest
  Apr 25 at 4:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i get pigeonhole ..1000 into 729 means 3 will be in one pigeonhole right? and thats the cube?
  $endgroup$
  – Randin D
  Apr 25 at 4:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how do u get the 729 again?
  $endgroup$
  – Randin D
  Apr 25 at 4:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $3$.
  $endgroup$
  – Mike Earnest
  Apr 25 at 16:02
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

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 $i<j$ 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.

share|cite|improve this answer

answered Apr 25 at 4:22

Mike EarnestMike Earnest

28.9k22256

$endgroup$

share|cite|improve this answer

answered Apr 25 at 4:22

Mike EarnestMike Earnest

28.9k22256

answered Apr 25 at 4:22

Mike EarnestMike Earnest

28.9k22256

answered Apr 25 at 4:22

Mike EarnestMike Earnest

28.9k22256

1

$begingroup$

Strengthening the result another way:

For each $i=1,dots,1000$ let $2^a_icdot 3^b_icdot 5^c_i$ be the product of the first $i$ rolls. Then there are only $3^3=27$ possible values for the 3-tuple $(a_ibmod 3, b_ibmod 3, c_ibmod 3)$, 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.

share|cite|improve this answer

answered Apr 25 at 8:31

Rosie FRosie F

1,468416

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Ilmari Karonen
  Apr 25 at 9:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @IlmariKaronen OK, good point, now that I see that in the OP.
  $endgroup$
  – Rosie F
  Apr 25 at 9:05
  

  
  
  
  

add a comment | 

1

$begingroup$

Strengthening the result another way:

For each $i=1,dots,1000$ let $2^a_icdot 3^b_icdot 5^c_i$ be the product of the first $i$ rolls. Then there are only $3^3=27$ possible values for the 3-tuple $(a_ibmod 3, b_ibmod 3, c_ibmod 3)$, 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.

share|cite|improve this answer

answered Apr 25 at 8:31

Rosie FRosie F

1,468416

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Ilmari Karonen
  Apr 25 at 9:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @IlmariKaronen OK, good point, now that I see that in the OP.
  $endgroup$
  – Rosie F
  Apr 25 at 9:05
  

  
  
  
  

add a comment | 

$begingroup$

Strengthening the result another way:

For each $i=1,dots,1000$ let $2^a_icdot 3^b_icdot 5^c_i$ be the product of the first $i$ rolls. Then there are only $3^3=27$ possible values for the 3-tuple $(a_ibmod 3, b_ibmod 3, c_ibmod 3)$, 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.

share|cite|improve this answer

answered Apr 25 at 8:31

Rosie FRosie F

1,468416

$endgroup$

share|cite|improve this answer

answered Apr 25 at 8:31

Rosie FRosie F

1,468416

answered Apr 25 at 8:31

Rosie FRosie F

1,468416

answered Apr 25 at 8:31

Rosie FRosie F

1,468416