Can a known sum and average determine a set of natural numbers?
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Say we are given a predetermined sum and average of $n$ distinct natural numbers ranging from $0-50.$ Knowing the sum and average of any such set of natural numbers, is it be possible to determine what that set is? If so, would the solution be unique?
E.g., Let's say $n=5.$ It looks like the solution is not unique, but is there any way to functionally determine the set, or is it just guess and check?
discrete-mathematics
edited Dec 14 '18 at 18:30
amWhy
192k28225439
asked Dec 14 '18 at 18:15
MathematicalMemesterMathematicalMemester
103
$endgroup$
add a comment |
$begingroup$
Say we are given a predetermined sum and average of $n$ distinct natural numbers ranging from $0-50.$ Knowing the sum and average of any such set of natural numbers, is it be possible to determine what that set is? If so, would the solution be unique?
E.g., Let's say $n=5.$ It looks like the solution is not unique, but is there any way to functionally determine the set, or is it just guess and check?
discrete-mathematics
edited Dec 14 '18 at 18:30
amWhy
192k28225439
asked Dec 14 '18 at 18:15
MathematicalMemesterMathematicalMemester
103
$endgroup$
$begingroup$
Are they $n$ distinct natural numbers? E.g. (if you don't count $0$ as a natural number) the only set of natural numbers with sum $6$ and average $2$ is ${1,2,3}$.
$endgroup$
– Robert Israel
Dec 14 '18 at 18:19
$begingroup$
Let's say n is equal to 5. It looks like the solution is not unique, but is there any way to functionally determine the set, or is it just guess and check?
$endgroup$
– MathematicalMemester
Dec 14 '18 at 18:21
$begingroup$
@RobertIsrael And yes, they should be distinct natural numbers.
$endgroup$
– MathematicalMemester
Dec 14 '18 at 18:24
$begingroup$
@MathematicalMemester I improved your post (question) by including information you posted in the copies. Ideally, this is the kind of information you should include in a question post, so it does not read like you copied it from an exercise, and shows some thought on your part.
$endgroup$
– amWhy
Dec 14 '18 at 18:32
add a comment |
$begingroup$
Say we are given a predetermined sum and average of $n$ distinct natural numbers ranging from $0-50.$ Knowing the sum and average of any such set of natural numbers, is it be possible to determine what that set is? If so, would the solution be unique?
E.g., Let's say $n=5.$ It looks like the solution is not unique, but is there any way to functionally determine the set, or is it just guess and check?
discrete-mathematics
edited Dec 14 '18 at 18:30
amWhy
192k28225439
asked Dec 14 '18 at 18:15
MathematicalMemesterMathematicalMemester
103
$endgroup$
Say we are given a predetermined sum and average of $n$ distinct natural numbers ranging from $0-50.$ Knowing the sum and average of any such set of natural numbers, is it be possible to determine what that set is? If so, would the solution be unique?
E.g., Let's say $n=5.$ It looks like the solution is not unique, but is there any way to functionally determine the set, or is it just guess and check?
discrete-mathematics
discrete-mathematics
edited Dec 14 '18 at 18:30
amWhy
192k28225439
asked Dec 14 '18 at 18:15
MathematicalMemesterMathematicalMemester
103
edited Dec 14 '18 at 18:30
amWhy
192k28225439
asked Dec 14 '18 at 18:15
MathematicalMemesterMathematicalMemester
103
edited Dec 14 '18 at 18:30
amWhy
192k28225439
edited Dec 14 '18 at 18:30
amWhy
192k28225439
edited Dec 14 '18 at 18:30
amWhy
192k28225439
192k28225439
asked Dec 14 '18 at 18:15
MathematicalMemesterMathematicalMemester
103
asked Dec 14 '18 at 18:15
MathematicalMemesterMathematicalMemester
103
asked Dec 14 '18 at 18:15
MathematicalMemesterMathematicalMemester
103
103
$begingroup$
Are they $n$ distinct natural numbers? E.g. (if you don't count $0$ as a natural number) the only set of natural numbers with sum $6$ and average $2$ is ${1,2,3}$.
$endgroup$
– Robert Israel
Dec 14 '18 at 18:19
$begingroup$
Let's say n is equal to 5. It looks like the solution is not unique, but is there any way to functionally determine the set, or is it just guess and check?
$endgroup$
– MathematicalMemester
Dec 14 '18 at 18:21
$begingroup$
@RobertIsrael And yes, they should be distinct natural numbers.
$endgroup$
– MathematicalMemester
Dec 14 '18 at 18:24
$begingroup$
@MathematicalMemester I improved your post (question) by including information you posted in the copies. Ideally, this is the kind of information you should include in a question post, so it does not read like you copied it from an exercise, and shows some thought on your part.
$endgroup$
– amWhy
Dec 14 '18 at 18:32
add a comment |
$begingroup$
Are they $n$ distinct natural numbers? E.g. (if you don't count $0$ as a natural number) the only set of natural numbers with sum $6$ and average $2$ is ${1,2,3}$.
$endgroup$
– Robert Israel
Dec 14 '18 at 18:19
$begingroup$
Let's say n is equal to 5. It looks like the solution is not unique, but is there any way to functionally determine the set, or is it just guess and check?
$endgroup$
– MathematicalMemester
Dec 14 '18 at 18:21
$begingroup$
@RobertIsrael And yes, they should be distinct natural numbers.
$endgroup$
– MathematicalMemester
Dec 14 '18 at 18:24
$begingroup$
@MathematicalMemester I improved your post (question) by including information you posted in the copies. Ideally, this is the kind of information you should include in a question post, so it does not read like you copied it from an exercise, and shows some thought on your part.
$endgroup$
– amWhy
Dec 14 '18 at 18:32
$begingroup$
Are they $n$ distinct natural numbers? E.g. (if you don't count $0$ as a natural number) the only set of natural numbers with sum $6$ and average $2$ is ${1,2,3}$.
$endgroup$
– Robert Israel
Dec 14 '18 at 18:19
$begingroup$
Are they $n$ distinct natural numbers? E.g. (if you don't count $0$ as a natural number) the only set of natural numbers with sum $6$ and average $2$ is ${1,2,3}$.
$endgroup$
– Robert Israel
Dec 14 '18 at 18:19
$begingroup$
Let's say n is equal to 5. It looks like the solution is not unique, but is there any way to functionally determine the set, or is it just guess and check?
$endgroup$
– MathematicalMemester
Dec 14 '18 at 18:21
$begingroup$
Let's say n is equal to 5. It looks like the solution is not unique, but is there any way to functionally determine the set, or is it just guess and check?
$endgroup$
– MathematicalMemester
Dec 14 '18 at 18:21
$begingroup$
@RobertIsrael And yes, they should be distinct natural numbers.
$endgroup$
– MathematicalMemester
Dec 14 '18 at 18:24
$begingroup$
@RobertIsrael And yes, they should be distinct natural numbers.
$endgroup$
– MathematicalMemester
Dec 14 '18 at 18:24
$begingroup$
@MathematicalMemester I improved your post (question) by including information you posted in the copies. Ideally, this is the kind of information you should include in a question post, so it does not read like you copied it from an exercise, and shows some thought on your part.
$endgroup$
– amWhy
Dec 14 '18 at 18:32
$begingroup$
@MathematicalMemester I improved your post (question) by including information you posted in the copies. Ideally, this is the kind of information you should include in a question post, so it does not read like you copied it from an exercise, and shows some thought on your part.
$endgroup$
– amWhy
Dec 14 '18 at 18:32
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
For a set of $n$ numbers, you have $n$ "degrees of freedom".
Generally speaking, each condition you impose loses a degree of freedom.
Naively, you would assume $(n-2)$ degrees of freedom after imposing two conditions on $n$ variables.
However, the sum $S$ and mean $M$ are closely related: $S=nM$.
For example, say you have three numbers with sum 30 and mean 10. Then you need $x_1+x_2+x_3=30$ and $frac{1}{3}(x_1+x_2+x_3)=10$. But this last condition is equivalent to $x_1+x_2+x_3=30$.
Say you have three numbers with the sum 30 and the mean 5. Then you need $x_1+x_2+x_3=30$ and $frac{1}{3}(x_1+x_2+x_3)=5$. This last condition is equivalent to $x_1+x_2+x_3=15$, and so you need $15=30$ which is impossible; there are no such numbers.
In short, if the sum and mean are compatible, i.e. $S=nM$, then you have one, and only one, condition:
$$x_1+cdots+x_n=S$$
You need to know the number of partitions of $S$.
answered Dec 14 '18 at 18:48
Fly by NightFly by Night
25.8k32978
$endgroup$
add a comment |
$begingroup$
No. First notice that the two numbers you are given are not independent from one another. If $x_1 + ... + x_n = alpha$ then the average is $ frac{x_1+ ... +x_n}{n} = frac{alpha}{n}$ so you only have the information
$$x_1 + ... + x_n = alpha qquad 0<alpha<50, ; alphageq x_i geq 0.$$
An equation which as $binom{alpha + (n-1)}{n-1}$ different solutions.
answered Dec 14 '18 at 18:28
DigitalisDigitalis
528216
$endgroup$
$begingroup$
Are you suggesting that this is the number of required sets? You might be overcounting: ${1,2,3}$ and ${1,3,2}$ are different ordered triples but the same set really
$endgroup$
– Shubham Johri
Dec 14 '18 at 20:23
$begingroup$
You are right I am counting the number of $(x_1,x_2,...,x_n) in mathbb{N}^n$ that verify $x_1 + ... +x_n = alpha$.
$endgroup$
– Digitalis
Dec 15 '18 at 10:16
$begingroup$
Say $x_1+x_2=2$. The possible tuples are $(0,2),(2,0),(1,1); 3$ in total. But ${2,0}, {0,2}$ is the same set. So we only have $2$ distinct sets
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:19
add a comment |
$begingroup$
If we have $n$ distinct numbers in ${a,a+1,...,b-1,b}$ then $ngeq a+(a+1)+...+(a+n-1)=an+n(n-1)/2$ and $nleq b+(b-1)+...+(b-n+1)=bn-n(n-1)/2$. We find that a set can be found if, and only if, $an+n(n-1)/2leq nleq bn+n(n-1)/2$. The set will be unique if one of the inequalities is an equality, or off-by-one, or if $n=1$.
answered Dec 14 '18 at 18:30
SmileyCraftSmileyCraft
3,376516
$endgroup$
add a comment |
$begingroup$
No. To start, if you know the sum and the amount of digits then the average doesn’t give you any more information since the average is the sum over the amount of digits (given that they are all distinct). So you could imagine different sets of $ n $ numbers that add up to a given number $ S $. Ex.:
10=1+2+7=1+4+5.
answered Dec 14 '18 at 18:38
KirtpoleKirtpole
1012
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For a set of $n$ numbers, you have $n$ "degrees of freedom".
Generally speaking, each condition you impose loses a degree of freedom.
Naively, you would assume $(n-2)$ degrees of freedom after imposing two conditions on $n$ variables.
However, the sum $S$ and mean $M$ are closely related: $S=nM$.
For example, say you have three numbers with sum 30 and mean 10. Then you need $x_1+x_2+x_3=30$ and $frac{1}{3}(x_1+x_2+x_3)=10$. But this last condition is equivalent to $x_1+x_2+x_3=30$.
Say you have three numbers with the sum 30 and the mean 5. Then you need $x_1+x_2+x_3=30$ and $frac{1}{3}(x_1+x_2+x_3)=5$. This last condition is equivalent to $x_1+x_2+x_3=15$, and so you need $15=30$ which is impossible; there are no such numbers.
In short, if the sum and mean are compatible, i.e. $S=nM$, then you have one, and only one, condition:
$$x_1+cdots+x_n=S$$
You need to know the number of partitions of $S$.
answered Dec 14 '18 at 18:48
Fly by NightFly by Night
25.8k32978
$endgroup$
add a comment |
$begingroup$
For a set of $n$ numbers, you have $n$ "degrees of freedom".
Generally speaking, each condition you impose loses a degree of freedom.
Naively, you would assume $(n-2)$ degrees of freedom after imposing two conditions on $n$ variables.
However, the sum $S$ and mean $M$ are closely related: $S=nM$.
For example, say you have three numbers with sum 30 and mean 10. Then you need $x_1+x_2+x_3=30$ and $frac{1}{3}(x_1+x_2+x_3)=10$. But this last condition is equivalent to $x_1+x_2+x_3=30$.
Say you have three numbers with the sum 30 and the mean 5. Then you need $x_1+x_2+x_3=30$ and $frac{1}{3}(x_1+x_2+x_3)=5$. This last condition is equivalent to $x_1+x_2+x_3=15$, and so you need $15=30$ which is impossible; there are no such numbers.
In short, if the sum and mean are compatible, i.e. $S=nM$, then you have one, and only one, condition:
$$x_1+cdots+x_n=S$$
You need to know the number of partitions of $S$.
answered Dec 14 '18 at 18:48
Fly by NightFly by Night
25.8k32978
$endgroup$
add a comment |
$begingroup$
For a set of $n$ numbers, you have $n$ "degrees of freedom".
Generally speaking, each condition you impose loses a degree of freedom.
Naively, you would assume $(n-2)$ degrees of freedom after imposing two conditions on $n$ variables.
However, the sum $S$ and mean $M$ are closely related: $S=nM$.
For example, say you have three numbers with sum 30 and mean 10. Then you need $x_1+x_2+x_3=30$ and $frac{1}{3}(x_1+x_2+x_3)=10$. But this last condition is equivalent to $x_1+x_2+x_3=30$.
Say you have three numbers with the sum 30 and the mean 5. Then you need $x_1+x_2+x_3=30$ and $frac{1}{3}(x_1+x_2+x_3)=5$. This last condition is equivalent to $x_1+x_2+x_3=15$, and so you need $15=30$ which is impossible; there are no such numbers.
In short, if the sum and mean are compatible, i.e. $S=nM$, then you have one, and only one, condition:
$$x_1+cdots+x_n=S$$
You need to know the number of partitions of $S$.
answered Dec 14 '18 at 18:48
Fly by NightFly by Night
25.8k32978
$endgroup$
For a set of $n$ numbers, you have $n$ "degrees of freedom".
Generally speaking, each condition you impose loses a degree of freedom.
Naively, you would assume $(n-2)$ degrees of freedom after imposing two conditions on $n$ variables.
However, the sum $S$ and mean $M$ are closely related: $S=nM$.
For example, say you have three numbers with sum 30 and mean 10. Then you need $x_1+x_2+x_3=30$ and $frac{1}{3}(x_1+x_2+x_3)=10$. But this last condition is equivalent to $x_1+x_2+x_3=30$.
Say you have three numbers with the sum 30 and the mean 5. Then you need $x_1+x_2+x_3=30$ and $frac{1}{3}(x_1+x_2+x_3)=5$. This last condition is equivalent to $x_1+x_2+x_3=15$, and so you need $15=30$ which is impossible; there are no such numbers.
In short, if the sum and mean are compatible, i.e. $S=nM$, then you have one, and only one, condition:
$$x_1+cdots+x_n=S$$
You need to know the number of partitions of $S$.
answered Dec 14 '18 at 18:48
Fly by NightFly by Night
25.8k32978
answered Dec 14 '18 at 18:48
Fly by NightFly by Night
25.8k32978
answered Dec 14 '18 at 18:48
Fly by NightFly by Night
25.8k32978
answered Dec 14 '18 at 18:48
Fly by NightFly by Night
25.8k32978
25.8k32978
add a comment |
add a comment |
$begingroup$
No. First notice that the two numbers you are given are not independent from one another. If $x_1 + ... + x_n = alpha$ then the average is $ frac{x_1+ ... +x_n}{n} = frac{alpha}{n}$ so you only have the information
$$x_1 + ... + x_n = alpha qquad 0<alpha<50, ; alphageq x_i geq 0.$$
An equation which as $binom{alpha + (n-1)}{n-1}$ different solutions.
answered Dec 14 '18 at 18:28
DigitalisDigitalis
528216
$endgroup$
$begingroup$
Are you suggesting that this is the number of required sets? You might be overcounting: ${1,2,3}$ and ${1,3,2}$ are different ordered triples but the same set really
$endgroup$
– Shubham Johri
Dec 14 '18 at 20:23
$begingroup$
You are right I am counting the number of $(x_1,x_2,...,x_n) in mathbb{N}^n$ that verify $x_1 + ... +x_n = alpha$.
$endgroup$
– Digitalis
Dec 15 '18 at 10:16
$begingroup$
Say $x_1+x_2=2$. The possible tuples are $(0,2),(2,0),(1,1); 3$ in total. But ${2,0}, {0,2}$ is the same set. So we only have $2$ distinct sets
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:19
add a comment |
$begingroup$
No. First notice that the two numbers you are given are not independent from one another. If $x_1 + ... + x_n = alpha$ then the average is $ frac{x_1+ ... +x_n}{n} = frac{alpha}{n}$ so you only have the information
$$x_1 + ... + x_n = alpha qquad 0<alpha<50, ; alphageq x_i geq 0.$$
An equation which as $binom{alpha + (n-1)}{n-1}$ different solutions.
answered Dec 14 '18 at 18:28
DigitalisDigitalis
528216
$endgroup$
$begingroup$
Are you suggesting that this is the number of required sets? You might be overcounting: ${1,2,3}$ and ${1,3,2}$ are different ordered triples but the same set really
$endgroup$
– Shubham Johri
Dec 14 '18 at 20:23
$begingroup$
You are right I am counting the number of $(x_1,x_2,...,x_n) in mathbb{N}^n$ that verify $x_1 + ... +x_n = alpha$.
$endgroup$
– Digitalis
Dec 15 '18 at 10:16
$begingroup$
Say $x_1+x_2=2$. The possible tuples are $(0,2),(2,0),(1,1); 3$ in total. But ${2,0}, {0,2}$ is the same set. So we only have $2$ distinct sets
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:19
add a comment |
$begingroup$
No. First notice that the two numbers you are given are not independent from one another. If $x_1 + ... + x_n = alpha$ then the average is $ frac{x_1+ ... +x_n}{n} = frac{alpha}{n}$ so you only have the information
$$x_1 + ... + x_n = alpha qquad 0<alpha<50, ; alphageq x_i geq 0.$$
An equation which as $binom{alpha + (n-1)}{n-1}$ different solutions.
answered Dec 14 '18 at 18:28
DigitalisDigitalis
528216
$endgroup$
No. First notice that the two numbers you are given are not independent from one another. If $x_1 + ... + x_n = alpha$ then the average is $ frac{x_1+ ... +x_n}{n} = frac{alpha}{n}$ so you only have the information
$$x_1 + ... + x_n = alpha qquad 0<alpha<50, ; alphageq x_i geq 0.$$
An equation which as $binom{alpha + (n-1)}{n-1}$ different solutions.
answered Dec 14 '18 at 18:28
DigitalisDigitalis
528216
answered Dec 14 '18 at 18:28
DigitalisDigitalis
528216
answered Dec 14 '18 at 18:28
DigitalisDigitalis
528216
answered Dec 14 '18 at 18:28
DigitalisDigitalis
528216
528216
$begingroup$
Are you suggesting that this is the number of required sets? You might be overcounting: ${1,2,3}$ and ${1,3,2}$ are different ordered triples but the same set really
$endgroup$
– Shubham Johri
Dec 14 '18 at 20:23
$begingroup$
You are right I am counting the number of $(x_1,x_2,...,x_n) in mathbb{N}^n$ that verify $x_1 + ... +x_n = alpha$.
$endgroup$
– Digitalis
Dec 15 '18 at 10:16
$begingroup$
Say $x_1+x_2=2$. The possible tuples are $(0,2),(2,0),(1,1); 3$ in total. But ${2,0}, {0,2}$ is the same set. So we only have $2$ distinct sets
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:19
add a comment |
$begingroup$
Are you suggesting that this is the number of required sets? You might be overcounting: ${1,2,3}$ and ${1,3,2}$ are different ordered triples but the same set really
$endgroup$
– Shubham Johri
Dec 14 '18 at 20:23
$begingroup$
You are right I am counting the number of $(x_1,x_2,...,x_n) in mathbb{N}^n$ that verify $x_1 + ... +x_n = alpha$.
$endgroup$
– Digitalis
Dec 15 '18 at 10:16
$begingroup$
Say $x_1+x_2=2$. The possible tuples are $(0,2),(2,0),(1,1); 3$ in total. But ${2,0}, {0,2}$ is the same set. So we only have $2$ distinct sets
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:19
$begingroup$
Are you suggesting that this is the number of required sets? You might be overcounting: ${1,2,3}$ and ${1,3,2}$ are different ordered triples but the same set really
$endgroup$
– Shubham Johri
Dec 14 '18 at 20:23
$begingroup$
Are you suggesting that this is the number of required sets? You might be overcounting: ${1,2,3}$ and ${1,3,2}$ are different ordered triples but the same set really
$endgroup$
– Shubham Johri
Dec 14 '18 at 20:23
$begingroup$
You are right I am counting the number of $(x_1,x_2,...,x_n) in mathbb{N}^n$ that verify $x_1 + ... +x_n = alpha$.
$endgroup$
– Digitalis
Dec 15 '18 at 10:16
$begingroup$
You are right I am counting the number of $(x_1,x_2,...,x_n) in mathbb{N}^n$ that verify $x_1 + ... +x_n = alpha$.
$endgroup$
– Digitalis
Dec 15 '18 at 10:16
$begingroup$
Say $x_1+x_2=2$. The possible tuples are $(0,2),(2,0),(1,1); 3$ in total. But ${2,0}, {0,2}$ is the same set. So we only have $2$ distinct sets
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:19
$begingroup$
Say $x_1+x_2=2$. The possible tuples are $(0,2),(2,0),(1,1); 3$ in total. But ${2,0}, {0,2}$ is the same set. So we only have $2$ distinct sets
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:19
add a comment |
$begingroup$
If we have $n$ distinct numbers in ${a,a+1,...,b-1,b}$ then $ngeq a+(a+1)+...+(a+n-1)=an+n(n-1)/2$ and $nleq b+(b-1)+...+(b-n+1)=bn-n(n-1)/2$. We find that a set can be found if, and only if, $an+n(n-1)/2leq nleq bn+n(n-1)/2$. The set will be unique if one of the inequalities is an equality, or off-by-one, or if $n=1$.
answered Dec 14 '18 at 18:30
SmileyCraftSmileyCraft
3,376516
$endgroup$
add a comment |
$begingroup$
If we have $n$ distinct numbers in ${a,a+1,...,b-1,b}$ then $ngeq a+(a+1)+...+(a+n-1)=an+n(n-1)/2$ and $nleq b+(b-1)+...+(b-n+1)=bn-n(n-1)/2$. We find that a set can be found if, and only if, $an+n(n-1)/2leq nleq bn+n(n-1)/2$. The set will be unique if one of the inequalities is an equality, or off-by-one, or if $n=1$.
answered Dec 14 '18 at 18:30
SmileyCraftSmileyCraft
3,376516
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add a comment |
$begingroup$
If we have $n$ distinct numbers in ${a,a+1,...,b-1,b}$ then $ngeq a+(a+1)+...+(a+n-1)=an+n(n-1)/2$ and $nleq b+(b-1)+...+(b-n+1)=bn-n(n-1)/2$. We find that a set can be found if, and only if, $an+n(n-1)/2leq nleq bn+n(n-1)/2$. The set will be unique if one of the inequalities is an equality, or off-by-one, or if $n=1$.
answered Dec 14 '18 at 18:30
SmileyCraftSmileyCraft
3,376516
$endgroup$
If we have $n$ distinct numbers in ${a,a+1,...,b-1,b}$ then $ngeq a+(a+1)+...+(a+n-1)=an+n(n-1)/2$ and $nleq b+(b-1)+...+(b-n+1)=bn-n(n-1)/2$. We find that a set can be found if, and only if, $an+n(n-1)/2leq nleq bn+n(n-1)/2$. The set will be unique if one of the inequalities is an equality, or off-by-one, or if $n=1$.
answered Dec 14 '18 at 18:30
SmileyCraftSmileyCraft
3,376516
answered Dec 14 '18 at 18:30
SmileyCraftSmileyCraft
3,376516
answered Dec 14 '18 at 18:30
SmileyCraftSmileyCraft
3,376516
answered Dec 14 '18 at 18:30
SmileyCraftSmileyCraft
3,376516
3,376516
add a comment |
add a comment |
$begingroup$
No. To start, if you know the sum and the amount of digits then the average doesn’t give you any more information since the average is the sum over the amount of digits (given that they are all distinct). So you could imagine different sets of $ n $ numbers that add up to a given number $ S $. Ex.:
10=1+2+7=1+4+5.
answered Dec 14 '18 at 18:38
KirtpoleKirtpole
1012
$endgroup$
add a comment |
$begingroup$
No. To start, if you know the sum and the amount of digits then the average doesn’t give you any more information since the average is the sum over the amount of digits (given that they are all distinct). So you could imagine different sets of $ n $ numbers that add up to a given number $ S $. Ex.:
10=1+2+7=1+4+5.
answered Dec 14 '18 at 18:38
KirtpoleKirtpole
1012
$endgroup$
add a comment |
$begingroup$
No. To start, if you know the sum and the amount of digits then the average doesn’t give you any more information since the average is the sum over the amount of digits (given that they are all distinct). So you could imagine different sets of $ n $ numbers that add up to a given number $ S $. Ex.:
10=1+2+7=1+4+5.
answered Dec 14 '18 at 18:38
KirtpoleKirtpole
1012
$endgroup$
No. To start, if you know the sum and the amount of digits then the average doesn’t give you any more information since the average is the sum over the amount of digits (given that they are all distinct). So you could imagine different sets of $ n $ numbers that add up to a given number $ S $. Ex.:
10=1+2+7=1+4+5.
answered Dec 14 '18 at 18:38
KirtpoleKirtpole
1012
answered Dec 14 '18 at 18:38
KirtpoleKirtpole
1012
answered Dec 14 '18 at 18:38
KirtpoleKirtpole
1012
answered Dec 14 '18 at 18:38
KirtpoleKirtpole
1012
1012
add a comment |
add a comment |
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$begingroup$
Are they $n$ distinct natural numbers? E.g. (if you don't count $0$ as a natural number) the only set of natural numbers with sum $6$ and average $2$ is ${1,2,3}$.
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– Robert Israel
Dec 14 '18 at 18:19
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Let's say n is equal to 5. It looks like the solution is not unique, but is there any way to functionally determine the set, or is it just guess and check?
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– MathematicalMemester
Dec 14 '18 at 18:21
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@RobertIsrael And yes, they should be distinct natural numbers.
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– MathematicalMemester
Dec 14 '18 at 18:24
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@MathematicalMemester I improved your post (question) by including information you posted in the copies. Ideally, this is the kind of information you should include in a question post, so it does not read like you copied it from an exercise, and shows some thought on your part.
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– amWhy
Dec 14 '18 at 18:32