Sum of Random Integers

up vote
2
down vote

favorite

I am trying to get a list of numbers with at least 1000 sums as possible, using random numbers from 1 to 10. But I don't want to get blank list output.

How can I only get a list of numbers greater than 1000 in total?
I think my loop is insufficient.

Here is my code:

m = ;
n = RandomInteger[1, 10, 170];
liste = Total[n];
While[liste >= 1000, AppendTo[m, n]; Break;]
m
liste

asked 3 hours ago

ithilquessirr

132

New contributor

2

Yes, I think your loop is insufficient, too :) It's not perfectly clear to me what you want, but maybe it's this?: While[n = RandomInteger[1, 10, 170]; Total[n] < 1000,]; n
â€“Â Michael E2
3 hours ago

RandomInteger[1, 10, 1000] is guaranteed to work.
â€“Â AccidentalFourierTransform
2 hours ago

Or: Reap[sum = 0; While[sum < 1000, sum += (n = RandomInteger[1, 10]); Sow[n]]][[2, 1]]
â€“Â Daniel Lichtblau
2 hours ago

add a commentÂ |Â

up vote
2
down vote

favorite

I am trying to get a list of numbers with at least 1000 sums as possible, using random numbers from 1 to 10. But I don't want to get blank list output.

How can I only get a list of numbers greater than 1000 in total?
I think my loop is insufficient.

Here is my code:

m = ;
n = RandomInteger[1, 10, 170];
liste = Total[n];
While[liste >= 1000, AppendTo[m, n]; Break;]
m
liste

asked 3 hours ago

ithilquessirr

132

New contributor

2

Yes, I think your loop is insufficient, too :) It's not perfectly clear to me what you want, but maybe it's this?: While[n = RandomInteger[1, 10, 170]; Total[n] < 1000,]; n
â€“Â Michael E2
3 hours ago

RandomInteger[1, 10, 1000] is guaranteed to work.
â€“Â AccidentalFourierTransform
2 hours ago

Or: Reap[sum = 0; While[sum < 1000, sum += (n = RandomInteger[1, 10]); Sow[n]]][[2, 1]]
â€“Â Daniel Lichtblau
2 hours ago

add a commentÂ |Â

up vote
2
down vote

favorite

I am trying to get a list of numbers with at least 1000 sums as possible, using random numbers from 1 to 10. But I don't want to get blank list output.

How can I only get a list of numbers greater than 1000 in total?
I think my loop is insufficient.

Here is my code:

m = ;
n = RandomInteger[1, 10, 170];
liste = Total[n];
While[liste >= 1000, AppendTo[m, n]; Break;]
m
liste

asked 3 hours ago

ithilquessirr

132

New contributor

I am trying to get a list of numbers with at least 1000 sums as possible, using random numbers from 1 to 10. But I don't want to get blank list output.

How can I only get a list of numbers greater than 1000 in total?
I think my loop is insufficient.

Here is my code:

m = ;
n = RandomInteger[1, 10, 170];
liste = Total[n];
While[liste >= 1000, AppendTo[m, n]; Break;]
m
liste

random education infinite-loop

asked 3 hours ago

ithilquessirr

132

New contributor

asked 3 hours ago

ithilquessirr

132

New contributor

asked 3 hours ago

ithilquessirr

132

New contributor

asked 3 hours ago

ithilquessirr

132

asked 3 hours ago

ithilquessirr

132

New contributor

ithilquessirr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

2

Yes, I think your loop is insufficient, too :) It's not perfectly clear to me what you want, but maybe it's this?: While[n = RandomInteger[1, 10, 170]; Total[n] < 1000,]; n
â€“Â Michael E2
3 hours ago

RandomInteger[1, 10, 1000] is guaranteed to work.
â€“Â AccidentalFourierTransform
2 hours ago

Or: Reap[sum = 0; While[sum < 1000, sum += (n = RandomInteger[1, 10]); Sow[n]]][[2, 1]]
â€“Â Daniel Lichtblau
2 hours ago

add a commentÂ |Â

2

Yes, I think your loop is insufficient, too :) It's not perfectly clear to me what you want, but maybe it's this?: While[n = RandomInteger[1, 10, 170]; Total[n] < 1000,]; n
â€“Â Michael E2
3 hours ago

RandomInteger[1, 10, 1000] is guaranteed to work.
â€“Â AccidentalFourierTransform
2 hours ago

Or: Reap[sum = 0; While[sum < 1000, sum += (n = RandomInteger[1, 10]); Sow[n]]][[2, 1]]
â€“Â Daniel Lichtblau
2 hours ago

Yes, I think your loop is insufficient, too :) It's not perfectly clear to me what you want, but maybe it's this?: While[n = RandomInteger[1, 10, 170]; Total[n] < 1000,]; n
â€“Â Michael E2
3 hours ago

RandomInteger[1, 10, 1000] is guaranteed to work.
â€“Â AccidentalFourierTransform
2 hours ago

Or: Reap[sum = 0; While[sum < 1000, sum += (n = RandomInteger[1, 10]); Sow[n]]][[2, 1]]
â€“Â Daniel Lichtblau
2 hours ago

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
2
down vote

accepted

if you want to output just one number with your conditions try

liste = 0;
While[liste < 1000, n = RandomInteger[1, 10, 170];
liste = Total[n]]
liste
n

here is the result and the 170 numbers that add up to this result

If you don't want all these numbers to be displayed, just remove the last n from the code

1020

4,4,10,4,4,9,9,4,2,2,10,8,8,2,2,1,3,6,3,8,7,8,4,1,6,1,5,3,9,9,1,9,2,3,10,10,3,10,6,10,10,9,10,10,2,6,7,1,5,9,8,5,3,8,10,8,1,1,10,6,7,10,3,7,7,5,9,3,10,10,6,6,5,3,9,10,1,8,1,8,4,9,1,4,9,8,7,2,9,5,6,3,10,10,10,5,3,6,6,4,1,10,3,4,10,4,10,5,10,10,5,5,9,1,10,5,9,5,5,5,7,8,1,3,2,8,10,9,9,6,2,9,5,5,6,10,1,5,7,7,7,3,7,7,4,10,5,8,2,1,8,9,8,2,5,10,1,2,7,9,8,9,3,8,8,3,10,2,6,4

edited 2 hours ago

answered 2 hours ago

J42161217

2,417218

add a commentÂ |Â

up vote
1
down vote

var = Array[x, 170];

dud = DiscreteUniformDistribution[1, 10];

Since you want the sum (Total) of the 170 variables to be at least 1000 and the most the sum could be is 1700, then the distribution for the truncated sum is

dist = TruncatedDistribution[1000, 1700,
 TransformedDistribution[Total[var],
 Thread[Distributed[var, dud]]]];

To get a list of 1000 random draws from this distribution

SeedRandom[0]

list = RandomVariate[dist, 1000];

Mean[list] // N

(* 1015.98 *)

Median[list]

(* 1012 *)

Histogram[list]

enter image description here

answered 2 hours ago

Bob Hanlon

56.6k23591

add a commentÂ |Â

up vote
0
down vote

One way of looking at your problem could be in terms of making a series of experiment. Then understanding the likelihood that the sum of 170 integers is greater than 1000. The code would then be:

res = Table[i, RandomVariate[DiscreteUniformDistribution[1, 10], 170], i,
 1, 100]; 
Select[(Total@#[[2]]) >= 1000 &]@ res

answered 1 hour ago

FredrikD

8071822

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
2
down vote

accepted

if you want to output just one number with your conditions try

liste = 0;
While[liste < 1000, n = RandomInteger[1, 10, 170];
liste = Total[n]]
liste
n

here is the result and the 170 numbers that add up to this result

If you don't want all these numbers to be displayed, just remove the last n from the code

1020

4,4,10,4,4,9,9,4,2,2,10,8,8,2,2,1,3,6,3,8,7,8,4,1,6,1,5,3,9,9,1,9,2,3,10,10,3,10,6,10,10,9,10,10,2,6,7,1,5,9,8,5,3,8,10,8,1,1,10,6,7,10,3,7,7,5,9,3,10,10,6,6,5,3,9,10,1,8,1,8,4,9,1,4,9,8,7,2,9,5,6,3,10,10,10,5,3,6,6,4,1,10,3,4,10,4,10,5,10,10,5,5,9,1,10,5,9,5,5,5,7,8,1,3,2,8,10,9,9,6,2,9,5,5,6,10,1,5,7,7,7,3,7,7,4,10,5,8,2,1,8,9,8,2,5,10,1,2,7,9,8,9,3,8,8,3,10,2,6,4

edited 2 hours ago

answered 2 hours ago

J42161217

2,417218

add a commentÂ |Â

up vote
2
down vote

accepted

if you want to output just one number with your conditions try

liste = 0;
While[liste < 1000, n = RandomInteger[1, 10, 170];
liste = Total[n]]
liste
n

here is the result and the 170 numbers that add up to this result

If you don't want all these numbers to be displayed, just remove the last n from the code

1020

4,4,10,4,4,9,9,4,2,2,10,8,8,2,2,1,3,6,3,8,7,8,4,1,6,1,5,3,9,9,1,9,2,3,10,10,3,10,6,10,10,9,10,10,2,6,7,1,5,9,8,5,3,8,10,8,1,1,10,6,7,10,3,7,7,5,9,3,10,10,6,6,5,3,9,10,1,8,1,8,4,9,1,4,9,8,7,2,9,5,6,3,10,10,10,5,3,6,6,4,1,10,3,4,10,4,10,5,10,10,5,5,9,1,10,5,9,5,5,5,7,8,1,3,2,8,10,9,9,6,2,9,5,5,6,10,1,5,7,7,7,3,7,7,4,10,5,8,2,1,8,9,8,2,5,10,1,2,7,9,8,9,3,8,8,3,10,2,6,4

edited 2 hours ago

answered 2 hours ago

J42161217

2,417218

add a commentÂ |Â

up vote
2
down vote

accepted

if you want to output just one number with your conditions try

liste = 0;
While[liste < 1000, n = RandomInteger[1, 10, 170];
liste = Total[n]]
liste
n

here is the result and the 170 numbers that add up to this result

If you don't want all these numbers to be displayed, just remove the last n from the code

1020

4,4,10,4,4,9,9,4,2,2,10,8,8,2,2,1,3,6,3,8,7,8,4,1,6,1,5,3,9,9,1,9,2,3,10,10,3,10,6,10,10,9,10,10,2,6,7,1,5,9,8,5,3,8,10,8,1,1,10,6,7,10,3,7,7,5,9,3,10,10,6,6,5,3,9,10,1,8,1,8,4,9,1,4,9,8,7,2,9,5,6,3,10,10,10,5,3,6,6,4,1,10,3,4,10,4,10,5,10,10,5,5,9,1,10,5,9,5,5,5,7,8,1,3,2,8,10,9,9,6,2,9,5,5,6,10,1,5,7,7,7,3,7,7,4,10,5,8,2,1,8,9,8,2,5,10,1,2,7,9,8,9,3,8,8,3,10,2,6,4

edited 2 hours ago

answered 2 hours ago

J42161217

2,417218

if you want to output just one number with your conditions try

liste = 0;
While[liste < 1000, n = RandomInteger[1, 10, 170];
liste = Total[n]]
liste
n

here is the result and the 170 numbers that add up to this result

If you don't want all these numbers to be displayed, just remove the last n from the code

1020

4,4,10,4,4,9,9,4,2,2,10,8,8,2,2,1,3,6,3,8,7,8,4,1,6,1,5,3,9,9,1,9,2,3,10,10,3,10,6,10,10,9,10,10,2,6,7,1,5,9,8,5,3,8,10,8,1,1,10,6,7,10,3,7,7,5,9,3,10,10,6,6,5,3,9,10,1,8,1,8,4,9,1,4,9,8,7,2,9,5,6,3,10,10,10,5,3,6,6,4,1,10,3,4,10,4,10,5,10,10,5,5,9,1,10,5,9,5,5,5,7,8,1,3,2,8,10,9,9,6,2,9,5,5,6,10,1,5,7,7,7,3,7,7,4,10,5,8,2,1,8,9,8,2,5,10,1,2,7,9,8,9,3,8,8,3,10,2,6,4

edited 2 hours ago

answered 2 hours ago

J42161217

2,417218

edited 2 hours ago

answered 2 hours ago

J42161217

2,417218

answered 2 hours ago

J42161217

2,417218

answered 2 hours ago

J42161217

2,417218

add a commentÂ |Â

up vote
1
down vote

var = Array[x, 170];

dud = DiscreteUniformDistribution[1, 10];

Since you want the sum (Total) of the 170 variables to be at least 1000 and the most the sum could be is 1700, then the distribution for the truncated sum is

dist = TruncatedDistribution[1000, 1700,
 TransformedDistribution[Total[var],
 Thread[Distributed[var, dud]]]];

To get a list of 1000 random draws from this distribution

SeedRandom[0]

list = RandomVariate[dist, 1000];

Mean[list] // N

(* 1015.98 *)

Median[list]

(* 1012 *)

Histogram[list]

enter image description here

answered 2 hours ago

Bob Hanlon

56.6k23591

add a commentÂ |Â

up vote
1
down vote

var = Array[x, 170];

dud = DiscreteUniformDistribution[1, 10];

Since you want the sum (Total) of the 170 variables to be at least 1000 and the most the sum could be is 1700, then the distribution for the truncated sum is

dist = TruncatedDistribution[1000, 1700,
 TransformedDistribution[Total[var],
 Thread[Distributed[var, dud]]]];

To get a list of 1000 random draws from this distribution

SeedRandom[0]

list = RandomVariate[dist, 1000];

Mean[list] // N

(* 1015.98 *)

Median[list]

(* 1012 *)

Histogram[list]

enter image description here

answered 2 hours ago

Bob Hanlon

56.6k23591

add a commentÂ |Â

up vote
1
down vote

var = Array[x, 170];

dud = DiscreteUniformDistribution[1, 10];

Since you want the sum (Total) of the 170 variables to be at least 1000 and the most the sum could be is 1700, then the distribution for the truncated sum is

dist = TruncatedDistribution[1000, 1700,
 TransformedDistribution[Total[var],
 Thread[Distributed[var, dud]]]];

To get a list of 1000 random draws from this distribution

SeedRandom[0]

list = RandomVariate[dist, 1000];

Mean[list] // N

(* 1015.98 *)

Median[list]

(* 1012 *)

Histogram[list]

enter image description here

answered 2 hours ago

Bob Hanlon

56.6k23591

var = Array[x, 170];

dud = DiscreteUniformDistribution[1, 10];

Since you want the sum (Total) of the 170 variables to be at least 1000 and the most the sum could be is 1700, then the distribution for the truncated sum is

dist = TruncatedDistribution[1000, 1700,
 TransformedDistribution[Total[var],
 Thread[Distributed[var, dud]]]];

To get a list of 1000 random draws from this distribution

SeedRandom[0]

list = RandomVariate[dist, 1000];

Mean[list] // N

(* 1015.98 *)

Median[list]

(* 1012 *)

Histogram[list]

enter image description here

answered 2 hours ago

Bob Hanlon

56.6k23591

answered 2 hours ago

Bob Hanlon

56.6k23591

answered 2 hours ago

Bob Hanlon

56.6k23591

answered 2 hours ago

Bob Hanlon

56.6k23591

add a commentÂ |Â

up vote
0
down vote

One way of looking at your problem could be in terms of making a series of experiment. Then understanding the likelihood that the sum of 170 integers is greater than 1000. The code would then be:

res = Table[i, RandomVariate[DiscreteUniformDistribution[1, 10], 170], i,
 1, 100]; 
Select[(Total@#[[2]]) >= 1000 &]@ res

answered 1 hour ago

FredrikD

8071822

add a commentÂ |Â

up vote
0
down vote

One way of looking at your problem could be in terms of making a series of experiment. Then understanding the likelihood that the sum of 170 integers is greater than 1000. The code would then be:

res = Table[i, RandomVariate[DiscreteUniformDistribution[1, 10], 170], i,
 1, 100]; 
Select[(Total@#[[2]]) >= 1000 &]@ res

answered 1 hour ago

FredrikD

8071822

add a commentÂ |Â

up vote
0
down vote

One way of looking at your problem could be in terms of making a series of experiment. Then understanding the likelihood that the sum of 170 integers is greater than 1000. The code would then be:

res = Table[i, RandomVariate[DiscreteUniformDistribution[1, 10], 170], i,
 1, 100]; 
Select[(Total@#[[2]]) >= 1000 &]@ res

answered 1 hour ago

FredrikD

8071822

One way of looking at your problem could be in terms of making a series of experiment. Then understanding the likelihood that the sum of 170 integers is greater than 1000. The code would then be:

res = Table[i, RandomVariate[DiscreteUniformDistribution[1, 10], 170], i,
 1, 100]; 
Select[(Total@#[[2]]) >= 1000 &]@ res

answered 1 hour ago

FredrikD

8071822

answered 1 hour ago

FredrikD

8071822

answered 1 hour ago

FredrikD

8071822

answered 1 hour ago

FredrikD

8071822

add a commentÂ |Â

ithilquessirr is a new contributor. Be nice, and check out our Code of Conduct.

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