Mixing
Writing down consecutive natural numbers until a certain number of digits $k$ is reached.
$begingroup$
A person starts writing consecutive natural numbers from $5$ until $k$ digits are reached. For some values of $k$, this will be impossible, for example $6$ or $8$ are impossible as then after writing the single digit numbers, we will have $1$ and $3$ left over respectively, but then we can only write double digit numbers.
An example of this would be if the person wanted to write $9$ digits, they write $5, 6, 7, 8, 9, 10, 11$ for a total of $5 + 2(2) = 9$ digits.
Question: Prove that if $A$, $B$, and $n$ are positive integers, then it is not possible to write $An^{2}$ + $B$ digits for each possible $n$ i.e for each pair of $(A,B)$, there exists an $n$ such that the person cannot write $An^{2}$ + $B$ digits.
I proved that the statement is true for $An$ + $B$, and think that it is true in this case as well, but I can't come up with a proof.
Edit: A sketch of the proof for the above case would be to note that as we go from $n$ to $n + 1$, the initial $An + B$ part is identical, and we only add $A$ digits after. Note that the smallest number we use in this "$A$ " digits interval is ever increasing, so there will come a point when even the smallest number in this interval shall contain more than $A$ digits.
Some perhaps useful facts:
$1)$ The differences between consecutive iterations is $a(2n + 1)$, a linear function in $n$.
$2)$ For sufficiently large $n$, if the $n$-th iteration uses numbers with at most $P$ digits, then the next iteration will use numbers with at most $P + 1$ digits.
$3)$ I imagine the proof is something which can be generalized to higher order polynomials as well.
number-theory integer-partitions decimal-expansion natural-numbers
asked Jan 15 at 20:19
Saad Saad
585211
$endgroup$
|
show 9 more comments
$begingroup$
A person starts writing consecutive natural numbers from $5$ until $k$ digits are reached. For some values of $k$, this will be impossible, for example $6$ or $8$ are impossible as then after writing the single digit numbers, we will have $1$ and $3$ left over respectively, but then we can only write double digit numbers.
An example of this would be if the person wanted to write $9$ digits, they write $5, 6, 7, 8, 9, 10, 11$ for a total of $5 + 2(2) = 9$ digits.
Question: Prove that if $A$, $B$, and $n$ are positive integers, then it is not possible to write $An^{2}$ + $B$ digits for each possible $n$ i.e for each pair of $(A,B)$, there exists an $n$ such that the person cannot write $An^{2}$ + $B$ digits.
I proved that the statement is true for $An$ + $B$, and think that it is true in this case as well, but I can't come up with a proof.
Edit: A sketch of the proof for the above case would be to note that as we go from $n$ to $n + 1$, the initial $An + B$ part is identical, and we only add $A$ digits after. Note that the smallest number we use in this "$A$ " digits interval is ever increasing, so there will come a point when even the smallest number in this interval shall contain more than $A$ digits.
Some perhaps useful facts:
$1)$ The differences between consecutive iterations is $a(2n + 1)$, a linear function in $n$.
$2)$ For sufficiently large $n$, if the $n$-th iteration uses numbers with at most $P$ digits, then the next iteration will use numbers with at most $P + 1$ digits.
$3)$ I imagine the proof is something which can be generalized to higher order polynomials as well.
number-theory integer-partitions decimal-expansion natural-numbers
asked Jan 15 at 20:19
Saad Saad
585211
$endgroup$
$begingroup$
This is not clear. What does "until $k$ digits are reached" mean? I assumed you meant "natural numbers of length $k$" but that's not consistent with the rest of what you wrote.
$endgroup$
– lulu
Jan 15 at 20:35
$begingroup$
For example, until $9$ digits would mean writing 5, 6, 7, 8, 9, 10, 11. So the total number of digits is 5 + 2(2) = 9.
$endgroup$
– Saad
Jan 15 at 20:37
$begingroup$
I don't think that interpretation would occur to very many people. Please edit your post for clarity.
$endgroup$
– lulu
Jan 15 at 20:39
1
$begingroup$
No, @lulu. The question is to show that for each pair $(A,B)$, you cannot write $A(n^{2}) + B$ digits for each $n$. For instance, if $A$ = 2, $B$ = 3, the question is whether you can write out consecutive natural numbers to hit $5, 11, 21, 35, ..$ digits
$endgroup$
– Saad
Jan 15 at 21:05
1
$begingroup$
Sure...but if I could change the start each time, isn't it obvious that I could achieve whatever digit count I liked? I understand it if the starting point is fixed, but you indicated that you thought that irrelevant.
$endgroup$
– lulu
Jan 15 at 21:06
|
show 9 more comments
$begingroup$
A person starts writing consecutive natural numbers from $5$ until $k$ digits are reached. For some values of $k$, this will be impossible, for example $6$ or $8$ are impossible as then after writing the single digit numbers, we will have $1$ and $3$ left over respectively, but then we can only write double digit numbers.
An example of this would be if the person wanted to write $9$ digits, they write $5, 6, 7, 8, 9, 10, 11$ for a total of $5 + 2(2) = 9$ digits.
Question: Prove that if $A$, $B$, and $n$ are positive integers, then it is not possible to write $An^{2}$ + $B$ digits for each possible $n$ i.e for each pair of $(A,B)$, there exists an $n$ such that the person cannot write $An^{2}$ + $B$ digits.
I proved that the statement is true for $An$ + $B$, and think that it is true in this case as well, but I can't come up with a proof.
Edit: A sketch of the proof for the above case would be to note that as we go from $n$ to $n + 1$, the initial $An + B$ part is identical, and we only add $A$ digits after. Note that the smallest number we use in this "$A$ " digits interval is ever increasing, so there will come a point when even the smallest number in this interval shall contain more than $A$ digits.
Some perhaps useful facts:
$1)$ The differences between consecutive iterations is $a(2n + 1)$, a linear function in $n$.
$2)$ For sufficiently large $n$, if the $n$-th iteration uses numbers with at most $P$ digits, then the next iteration will use numbers with at most $P + 1$ digits.
$3)$ I imagine the proof is something which can be generalized to higher order polynomials as well.
number-theory integer-partitions decimal-expansion natural-numbers
asked Jan 15 at 20:19
Saad Saad
585211
$endgroup$
A person starts writing consecutive natural numbers from $5$ until $k$ digits are reached. For some values of $k$, this will be impossible, for example $6$ or $8$ are impossible as then after writing the single digit numbers, we will have $1$ and $3$ left over respectively, but then we can only write double digit numbers.
An example of this would be if the person wanted to write $9$ digits, they write $5, 6, 7, 8, 9, 10, 11$ for a total of $5 + 2(2) = 9$ digits.
Question: Prove that if $A$, $B$, and $n$ are positive integers, then it is not possible to write $An^{2}$ + $B$ digits for each possible $n$ i.e for each pair of $(A,B)$, there exists an $n$ such that the person cannot write $An^{2}$ + $B$ digits.
I proved that the statement is true for $An$ + $B$, and think that it is true in this case as well, but I can't come up with a proof.
Edit: A sketch of the proof for the above case would be to note that as we go from $n$ to $n + 1$, the initial $An + B$ part is identical, and we only add $A$ digits after. Note that the smallest number we use in this "$A$ " digits interval is ever increasing, so there will come a point when even the smallest number in this interval shall contain more than $A$ digits.
Some perhaps useful facts:
$1)$ The differences between consecutive iterations is $a(2n + 1)$, a linear function in $n$.
$2)$ For sufficiently large $n$, if the $n$-th iteration uses numbers with at most $P$ digits, then the next iteration will use numbers with at most $P + 1$ digits.
$3)$ I imagine the proof is something which can be generalized to higher order polynomials as well.
number-theory integer-partitions decimal-expansion natural-numbers
number-theory integer-partitions decimal-expansion natural-numbers
asked Jan 15 at 20:19
Saad Saad
585211
asked Jan 15 at 20:19
Saad Saad
585211
edited Jan 15 at 21:24
Saad
asked Jan 15 at 20:19
Saad Saad
585211
asked Jan 15 at 20:19
Saad Saad
585211
asked Jan 15 at 20:19
Saad Saad
585211
585211
$begingroup$
This is not clear. What does "until $k$ digits are reached" mean? I assumed you meant "natural numbers of length $k$" but that's not consistent with the rest of what you wrote.
$endgroup$
– lulu
Jan 15 at 20:35
$begingroup$
For example, until $9$ digits would mean writing 5, 6, 7, 8, 9, 10, 11. So the total number of digits is 5 + 2(2) = 9.
$endgroup$
– Saad
Jan 15 at 20:37
$begingroup$
I don't think that interpretation would occur to very many people. Please edit your post for clarity.
$endgroup$
– lulu
Jan 15 at 20:39
1
$begingroup$
No, @lulu. The question is to show that for each pair $(A,B)$, you cannot write $A(n^{2}) + B$ digits for each $n$. For instance, if $A$ = 2, $B$ = 3, the question is whether you can write out consecutive natural numbers to hit $5, 11, 21, 35, ..$ digits
$endgroup$
– Saad
Jan 15 at 21:05
1
$begingroup$
Sure...but if I could change the start each time, isn't it obvious that I could achieve whatever digit count I liked? I understand it if the starting point is fixed, but you indicated that you thought that irrelevant.
$endgroup$
– lulu
Jan 15 at 21:06
|
show 9 more comments
$begingroup$
This is not clear. What does "until $k$ digits are reached" mean? I assumed you meant "natural numbers of length $k$" but that's not consistent with the rest of what you wrote.
$endgroup$
– lulu
Jan 15 at 20:35
$begingroup$
For example, until $9$ digits would mean writing 5, 6, 7, 8, 9, 10, 11. So the total number of digits is 5 + 2(2) = 9.
$endgroup$
– Saad
Jan 15 at 20:37
$begingroup$
I don't think that interpretation would occur to very many people. Please edit your post for clarity.
$endgroup$
– lulu
Jan 15 at 20:39
1
$begingroup$
No, @lulu. The question is to show that for each pair $(A,B)$, you cannot write $A(n^{2}) + B$ digits for each $n$. For instance, if $A$ = 2, $B$ = 3, the question is whether you can write out consecutive natural numbers to hit $5, 11, 21, 35, ..$ digits
$endgroup$
– Saad
Jan 15 at 21:05
1
$begingroup$
Sure...but if I could change the start each time, isn't it obvious that I could achieve whatever digit count I liked? I understand it if the starting point is fixed, but you indicated that you thought that irrelevant.
$endgroup$
– lulu
Jan 15 at 21:06
$begingroup$
This is not clear. What does "until $k$ digits are reached" mean? I assumed you meant "natural numbers of length $k$" but that's not consistent with the rest of what you wrote.
$endgroup$
– lulu
Jan 15 at 20:35
$begingroup$
This is not clear. What does "until $k$ digits are reached" mean? I assumed you meant "natural numbers of length $k$" but that's not consistent with the rest of what you wrote.
$endgroup$
– lulu
Jan 15 at 20:35
$begingroup$
For example, until $9$ digits would mean writing 5, 6, 7, 8, 9, 10, 11. So the total number of digits is 5 + 2(2) = 9.
$endgroup$
– Saad
Jan 15 at 20:37
$begingroup$
For example, until $9$ digits would mean writing 5, 6, 7, 8, 9, 10, 11. So the total number of digits is 5 + 2(2) = 9.
$endgroup$
– Saad
Jan 15 at 20:37
$begingroup$
I don't think that interpretation would occur to very many people. Please edit your post for clarity.
$endgroup$
– lulu
Jan 15 at 20:39
$begingroup$
I don't think that interpretation would occur to very many people. Please edit your post for clarity.
$endgroup$
– lulu
Jan 15 at 20:39
1
1
$begingroup$
No, @lulu. The question is to show that for each pair $(A,B)$, you cannot write $A(n^{2}) + B$ digits for each $n$. For instance, if $A$ = 2, $B$ = 3, the question is whether you can write out consecutive natural numbers to hit $5, 11, 21, 35, ..$ digits
$endgroup$
– Saad
Jan 15 at 21:05
$begingroup$
No, @lulu. The question is to show that for each pair $(A,B)$, you cannot write $A(n^{2}) + B$ digits for each $n$. For instance, if $A$ = 2, $B$ = 3, the question is whether you can write out consecutive natural numbers to hit $5, 11, 21, 35, ..$ digits
$endgroup$
– Saad
Jan 15 at 21:05
1
1
$begingroup$
Sure...but if I could change the start each time, isn't it obvious that I could achieve whatever digit count I liked? I understand it if the starting point is fixed, but you indicated that you thought that irrelevant.
$endgroup$
– lulu
Jan 15 at 21:06
$begingroup$
Sure...but if I could change the start each time, isn't it obvious that I could achieve whatever digit count I liked? I understand it if the starting point is fixed, but you indicated that you thought that irrelevant.
$endgroup$
– lulu
Jan 15 at 21:06
|
show 9 more comments
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$begingroup$
This is not clear. What does "until $k$ digits are reached" mean? I assumed you meant "natural numbers of length $k$" but that's not consistent with the rest of what you wrote.
$endgroup$
– lulu
Jan 15 at 20:35
$begingroup$
For example, until $9$ digits would mean writing 5, 6, 7, 8, 9, 10, 11. So the total number of digits is 5 + 2(2) = 9.
$endgroup$
– Saad
Jan 15 at 20:37
$begingroup$
I don't think that interpretation would occur to very many people. Please edit your post for clarity.
$endgroup$
– lulu
Jan 15 at 20:39
1
$begingroup$
No, @lulu. The question is to show that for each pair $(A,B)$, you cannot write $A(n^{2}) + B$ digits for each $n$. For instance, if $A$ = 2, $B$ = 3, the question is whether you can write out consecutive natural numbers to hit $5, 11, 21, 35, ..$ digits
$endgroup$
– Saad
Jan 15 at 21:05
1
$begingroup$
Sure...but if I could change the start each time, isn't it obvious that I could achieve whatever digit count I liked? I understand it if the starting point is fixed, but you indicated that you thought that irrelevant.
$endgroup$
– lulu
Jan 15 at 21:06