How to prove that if a figure in a number changes, then the rest also changes.
We have a 8 figure number, lets call it $N_1=a_1a_2a_3a_4a_5a_6a_7a_8$. Prove that if we interchange two numbers (for example $N_2=a_1a_2a_4a_3a_5a_6a_7a_8$ and $N_1 neq N_2$) then the rest we get by dividing $N_1$ and $N_2$ with 23 isn't the same.
Then prove that if we change one number (for example: $N_3=a_1a_2ba_4a_5a_6a_7a_8$ and $bneq a_3$) the rest we get by dividing $N_1$ and $N_3$ by 23 is not the same.
Then prove that this is not true if we divide the numbers by 24. I mean, that we can find $N_1, N_2$ and $N_3$ defined as I said before that after dividing by 24 the rest is the same.
I do not even know how to start. Thanks in advance.
arithmetic
asked Nov 25 at 13:54
Andarrkor
306
|
show 4 more comments
We have a 8 figure number, lets call it $N_1=a_1a_2a_3a_4a_5a_6a_7a_8$. Prove that if we interchange two numbers (for example $N_2=a_1a_2a_4a_3a_5a_6a_7a_8$ and $N_1 neq N_2$) then the rest we get by dividing $N_1$ and $N_2$ with 23 isn't the same.
Then prove that if we change one number (for example: $N_3=a_1a_2ba_4a_5a_6a_7a_8$ and $bneq a_3$) the rest we get by dividing $N_1$ and $N_3$ by 23 is not the same.
Then prove that this is not true if we divide the numbers by 24. I mean, that we can find $N_1, N_2$ and $N_3$ defined as I said before that after dividing by 24 the rest is the same.
I do not even know how to start. Thanks in advance.
arithmetic
asked Nov 25 at 13:54
Andarrkor
306
3
Well, the first claim isn't true as stated...since nothing prevents $a_3=a_4$. If you add the condition that $a_ineq a_j$, then $N_1-N_2=(a_i-a_j)times 10^{8-i}-(a_j-a_i)10^{8-j}$. (Note: check the exponents there). Work from there.
– lulu
Nov 25 at 14:04
2
Side note: life (or at least arithmetic) gets easier if you change notation to $N_1=a_7a_6cdots a_0=sum a_i10^i$.
– lulu
Nov 25 at 14:12
@lulu I edited it. Not all the figures have to be different, but the new number $N_2$ can't be the same as $N_1$ (so the interchanged figures can not be the same figure).
– Andarrkor
Nov 25 at 14:49
Sure. As I said, I figured you meant to add that condition.
– lulu
Nov 25 at 15:02
@lulu I do not know what that subtraction has to do with the rest. I don't understand why this changes when the dividing number is 24. I suppose that it is something related to prime numbers, as 23 is prime. Thanks for the answers.
– Andarrkor
Nov 25 at 15:09
|
show 4 more comments
We have a 8 figure number, lets call it $N_1=a_1a_2a_3a_4a_5a_6a_7a_8$. Prove that if we interchange two numbers (for example $N_2=a_1a_2a_4a_3a_5a_6a_7a_8$ and $N_1 neq N_2$) then the rest we get by dividing $N_1$ and $N_2$ with 23 isn't the same.
Then prove that if we change one number (for example: $N_3=a_1a_2ba_4a_5a_6a_7a_8$ and $bneq a_3$) the rest we get by dividing $N_1$ and $N_3$ by 23 is not the same.
Then prove that this is not true if we divide the numbers by 24. I mean, that we can find $N_1, N_2$ and $N_3$ defined as I said before that after dividing by 24 the rest is the same.
I do not even know how to start. Thanks in advance.
arithmetic
asked Nov 25 at 13:54
Andarrkor
306
We have a 8 figure number, lets call it $N_1=a_1a_2a_3a_4a_5a_6a_7a_8$. Prove that if we interchange two numbers (for example $N_2=a_1a_2a_4a_3a_5a_6a_7a_8$ and $N_1 neq N_2$) then the rest we get by dividing $N_1$ and $N_2$ with 23 isn't the same.
Then prove that if we change one number (for example: $N_3=a_1a_2ba_4a_5a_6a_7a_8$ and $bneq a_3$) the rest we get by dividing $N_1$ and $N_3$ by 23 is not the same.
Then prove that this is not true if we divide the numbers by 24. I mean, that we can find $N_1, N_2$ and $N_3$ defined as I said before that after dividing by 24 the rest is the same.
I do not even know how to start. Thanks in advance.
arithmetic
arithmetic
asked Nov 25 at 13:54
Andarrkor
306
asked Nov 25 at 13:54
Andarrkor
306
edited Nov 25 at 14:47
asked Nov 25 at 13:54
Andarrkor
306
asked Nov 25 at 13:54
Andarrkor
306
asked Nov 25 at 13:54
Andarrkor
306
306
3
Well, the first claim isn't true as stated...since nothing prevents $a_3=a_4$. If you add the condition that $a_ineq a_j$, then $N_1-N_2=(a_i-a_j)times 10^{8-i}-(a_j-a_i)10^{8-j}$. (Note: check the exponents there). Work from there.
– lulu
Nov 25 at 14:04
2
Side note: life (or at least arithmetic) gets easier if you change notation to $N_1=a_7a_6cdots a_0=sum a_i10^i$.
– lulu
Nov 25 at 14:12
@lulu I edited it. Not all the figures have to be different, but the new number $N_2$ can't be the same as $N_1$ (so the interchanged figures can not be the same figure).
– Andarrkor
Nov 25 at 14:49
Sure. As I said, I figured you meant to add that condition.
– lulu
Nov 25 at 15:02
@lulu I do not know what that subtraction has to do with the rest. I don't understand why this changes when the dividing number is 24. I suppose that it is something related to prime numbers, as 23 is prime. Thanks for the answers.
– Andarrkor
Nov 25 at 15:09
|
show 4 more comments
3
Well, the first claim isn't true as stated...since nothing prevents $a_3=a_4$. If you add the condition that $a_ineq a_j$, then $N_1-N_2=(a_i-a_j)times 10^{8-i}-(a_j-a_i)10^{8-j}$. (Note: check the exponents there). Work from there.
– lulu
Nov 25 at 14:04
2
Side note: life (or at least arithmetic) gets easier if you change notation to $N_1=a_7a_6cdots a_0=sum a_i10^i$.
– lulu
Nov 25 at 14:12
@lulu I edited it. Not all the figures have to be different, but the new number $N_2$ can't be the same as $N_1$ (so the interchanged figures can not be the same figure).
– Andarrkor
Nov 25 at 14:49
Sure. As I said, I figured you meant to add that condition.
– lulu
Nov 25 at 15:02
@lulu I do not know what that subtraction has to do with the rest. I don't understand why this changes when the dividing number is 24. I suppose that it is something related to prime numbers, as 23 is prime. Thanks for the answers.
– Andarrkor
Nov 25 at 15:09
3
3
Well, the first claim isn't true as stated...since nothing prevents $a_3=a_4$. If you add the condition that $a_ineq a_j$, then $N_1-N_2=(a_i-a_j)times 10^{8-i}-(a_j-a_i)10^{8-j}$. (Note: check the exponents there). Work from there.
– lulu
Nov 25 at 14:04
Well, the first claim isn't true as stated...since nothing prevents $a_3=a_4$. If you add the condition that $a_ineq a_j$, then $N_1-N_2=(a_i-a_j)times 10^{8-i}-(a_j-a_i)10^{8-j}$. (Note: check the exponents there). Work from there.
– lulu
Nov 25 at 14:04
2
2
Side note: life (or at least arithmetic) gets easier if you change notation to $N_1=a_7a_6cdots a_0=sum a_i10^i$.
– lulu
Nov 25 at 14:12
Side note: life (or at least arithmetic) gets easier if you change notation to $N_1=a_7a_6cdots a_0=sum a_i10^i$.
– lulu
Nov 25 at 14:12
@lulu I edited it. Not all the figures have to be different, but the new number $N_2$ can't be the same as $N_1$ (so the interchanged figures can not be the same figure).
– Andarrkor
Nov 25 at 14:49
@lulu I edited it. Not all the figures have to be different, but the new number $N_2$ can't be the same as $N_1$ (so the interchanged figures can not be the same figure).
– Andarrkor
Nov 25 at 14:49
Sure. As I said, I figured you meant to add that condition.
– lulu
Nov 25 at 15:02
Sure. As I said, I figured you meant to add that condition.
– lulu
Nov 25 at 15:02
@lulu I do not know what that subtraction has to do with the rest. I don't understand why this changes when the dividing number is 24. I suppose that it is something related to prime numbers, as 23 is prime. Thanks for the answers.
– Andarrkor
Nov 25 at 15:09
@lulu I do not know what that subtraction has to do with the rest. I don't understand why this changes when the dividing number is 24. I suppose that it is something related to prime numbers, as 23 is prime. Thanks for the answers.
– Andarrkor
Nov 25 at 15:09
|
show 4 more comments
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3
Well, the first claim isn't true as stated...since nothing prevents $a_3=a_4$. If you add the condition that $a_ineq a_j$, then $N_1-N_2=(a_i-a_j)times 10^{8-i}-(a_j-a_i)10^{8-j}$. (Note: check the exponents there). Work from there.
– lulu
Nov 25 at 14:04
2
Side note: life (or at least arithmetic) gets easier if you change notation to $N_1=a_7a_6cdots a_0=sum a_i10^i$.
– lulu
Nov 25 at 14:12
@lulu I edited it. Not all the figures have to be different, but the new number $N_2$ can't be the same as $N_1$ (so the interchanged figures can not be the same figure).
– Andarrkor
Nov 25 at 14:49
Sure. As I said, I figured you meant to add that condition.
– lulu
Nov 25 at 15:02
@lulu I do not know what that subtraction has to do with the rest. I don't understand why this changes when the dividing number is 24. I suppose that it is something related to prime numbers, as 23 is prime. Thanks for the answers.
– Andarrkor
Nov 25 at 15:09