Construct a positive decimal number of length $n$ or less that is divisible by $2^n-1$ for n in $Z^+_0$












0












$begingroup$


I got this question on a discrete mathematics test and I no idea on how to solve it:




Create a method for constructing a positive decimal number of length $n$ or less with digits from set ${0,1,8,9}$ that is divisible by $2^n-1$, where n is in the nonnegative integers.




I would appreciate any help. I spent so long thinking about it and nothing clicks.






combinatorics number-theory elementary-number-theory algorithms







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try thinking in binary. What is $2^n - 1$ in binary? What does it look like in binary when you create a decimal number with ${0,1,8,9}$?
    $endgroup$
    – Larry B.
    Dec 18 '18 at 23:54
















0












$begingroup$


I got this question on a discrete mathematics test and I no idea on how to solve it:




Create a method for constructing a positive decimal number of length $n$ or less with digits from set ${0,1,8,9}$ that is divisible by $2^n-1$, where n is in the nonnegative integers.




I would appreciate any help. I spent so long thinking about it and nothing clicks.






combinatorics number-theory elementary-number-theory algorithms







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try thinking in binary. What is $2^n - 1$ in binary? What does it look like in binary when you create a decimal number with ${0,1,8,9}$?
    $endgroup$
    – Larry B.
    Dec 18 '18 at 23:54














0












0








0





$begingroup$


I got this question on a discrete mathematics test and I no idea on how to solve it:




Create a method for constructing a positive decimal number of length $n$ or less with digits from set ${0,1,8,9}$ that is divisible by $2^n-1$, where n is in the nonnegative integers.




I would appreciate any help. I spent so long thinking about it and nothing clicks.






combinatorics number-theory elementary-number-theory algorithms







share|cite|improve this question











$endgroup$




I got this question on a discrete mathematics test and I no idea on how to solve it:




Create a method for constructing a positive decimal number of length $n$ or less with digits from set ${0,1,8,9}$ that is divisible by $2^n-1$, where n is in the nonnegative integers.




I would appreciate any help. I spent so long thinking about it and nothing clicks.






combinatorics number-theory elementary-number-theory algorithms




combinatorics number-theory elementary-number-theory algorithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 23:50









Lucas Henrique

1,036414




1,036414










asked Dec 18 '18 at 23:44









TheGodProjectTheGodProject

32




32








  • 1




    $begingroup$
    Try thinking in binary. What is $2^n - 1$ in binary? What does it look like in binary when you create a decimal number with ${0,1,8,9}$?
    $endgroup$
    – Larry B.
    Dec 18 '18 at 23:54














  • 1




    $begingroup$
    Try thinking in binary. What is $2^n - 1$ in binary? What does it look like in binary when you create a decimal number with ${0,1,8,9}$?
    $endgroup$
    – Larry B.
    Dec 18 '18 at 23:54








1




1




$begingroup$
Try thinking in binary. What is $2^n - 1$ in binary? What does it look like in binary when you create a decimal number with ${0,1,8,9}$?
$endgroup$
– Larry B.
Dec 18 '18 at 23:54




$begingroup$
Try thinking in binary. What is $2^n - 1$ in binary? What does it look like in binary when you create a decimal number with ${0,1,8,9}$?
$endgroup$
– Larry B.
Dec 18 '18 at 23:54










1 Answer
1






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0












$begingroup$

We consider two points:



First:Note that the sum of given digits is $18$, that indicates the required number is divisible by $9$.



Second: Any number of form $(d_1d_2d_3...0d_1d_2d_3)$ is divisible by 11 iff number of digits $d_1, d_2, d_3 . . .$ is even, for example $23023=11times2093$ or $472904729=11times42991339$ etc.



Considering these points, we suppose $n=11$; with a four digit number like $abcd$; which is not divisible by 11 by given digits (i.e. 0,1,8 and 9), for $a=8$ , $b=0$ and $c=9$ we get:



$abcd=8091$



The required number must also be divisible by $11$, so the solution can be $809108091$ and we have:



$2^{11-1}-1=2^{10}-1=1023$



And:



$809108091=1023times790917$



Numbers $8091080910$ and $80910809100$ can also be solutions.
Other arrangements of digits does not give the solution.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    It needn't be divisible by 9; 181 would be valid but not divisible by 9
    $endgroup$
    – Benedict Randall Shaw
    Dec 20 '18 at 16:39











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0












$begingroup$

We consider two points:



First:Note that the sum of given digits is $18$, that indicates the required number is divisible by $9$.



Second: Any number of form $(d_1d_2d_3...0d_1d_2d_3)$ is divisible by 11 iff number of digits $d_1, d_2, d_3 . . .$ is even, for example $23023=11times2093$ or $472904729=11times42991339$ etc.



Considering these points, we suppose $n=11$; with a four digit number like $abcd$; which is not divisible by 11 by given digits (i.e. 0,1,8 and 9), for $a=8$ , $b=0$ and $c=9$ we get:



$abcd=8091$



The required number must also be divisible by $11$, so the solution can be $809108091$ and we have:



$2^{11-1}-1=2^{10}-1=1023$



And:



$809108091=1023times790917$



Numbers $8091080910$ and $80910809100$ can also be solutions.
Other arrangements of digits does not give the solution.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    It needn't be divisible by 9; 181 would be valid but not divisible by 9
    $endgroup$
    – Benedict Randall Shaw
    Dec 20 '18 at 16:39
















0












$begingroup$

We consider two points:



First:Note that the sum of given digits is $18$, that indicates the required number is divisible by $9$.



Second: Any number of form $(d_1d_2d_3...0d_1d_2d_3)$ is divisible by 11 iff number of digits $d_1, d_2, d_3 . . .$ is even, for example $23023=11times2093$ or $472904729=11times42991339$ etc.



Considering these points, we suppose $n=11$; with a four digit number like $abcd$; which is not divisible by 11 by given digits (i.e. 0,1,8 and 9), for $a=8$ , $b=0$ and $c=9$ we get:



$abcd=8091$



The required number must also be divisible by $11$, so the solution can be $809108091$ and we have:



$2^{11-1}-1=2^{10}-1=1023$



And:



$809108091=1023times790917$



Numbers $8091080910$ and $80910809100$ can also be solutions.
Other arrangements of digits does not give the solution.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    It needn't be divisible by 9; 181 would be valid but not divisible by 9
    $endgroup$
    – Benedict Randall Shaw
    Dec 20 '18 at 16:39














0












0








0





$begingroup$

We consider two points:



First:Note that the sum of given digits is $18$, that indicates the required number is divisible by $9$.



Second: Any number of form $(d_1d_2d_3...0d_1d_2d_3)$ is divisible by 11 iff number of digits $d_1, d_2, d_3 . . .$ is even, for example $23023=11times2093$ or $472904729=11times42991339$ etc.



Considering these points, we suppose $n=11$; with a four digit number like $abcd$; which is not divisible by 11 by given digits (i.e. 0,1,8 and 9), for $a=8$ , $b=0$ and $c=9$ we get:



$abcd=8091$



The required number must also be divisible by $11$, so the solution can be $809108091$ and we have:



$2^{11-1}-1=2^{10}-1=1023$



And:



$809108091=1023times790917$



Numbers $8091080910$ and $80910809100$ can also be solutions.
Other arrangements of digits does not give the solution.






share|cite|improve this answer











$endgroup$



We consider two points:



First:Note that the sum of given digits is $18$, that indicates the required number is divisible by $9$.



Second: Any number of form $(d_1d_2d_3...0d_1d_2d_3)$ is divisible by 11 iff number of digits $d_1, d_2, d_3 . . .$ is even, for example $23023=11times2093$ or $472904729=11times42991339$ etc.



Considering these points, we suppose $n=11$; with a four digit number like $abcd$; which is not divisible by 11 by given digits (i.e. 0,1,8 and 9), for $a=8$ , $b=0$ and $c=9$ we get:



$abcd=8091$



The required number must also be divisible by $11$, so the solution can be $809108091$ and we have:



$2^{11-1}-1=2^{10}-1=1023$



And:



$809108091=1023times790917$



Numbers $8091080910$ and $80910809100$ can also be solutions.
Other arrangements of digits does not give the solution.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '18 at 9:53

























answered Dec 20 '18 at 16:35









siroussirous

1,6711514




1,6711514








  • 1




    $begingroup$
    It needn't be divisible by 9; 181 would be valid but not divisible by 9
    $endgroup$
    – Benedict Randall Shaw
    Dec 20 '18 at 16:39














  • 1




    $begingroup$
    It needn't be divisible by 9; 181 would be valid but not divisible by 9
    $endgroup$
    – Benedict Randall Shaw
    Dec 20 '18 at 16:39








1




1




$begingroup$
It needn't be divisible by 9; 181 would be valid but not divisible by 9
$endgroup$
– Benedict Randall Shaw
Dec 20 '18 at 16:39




$begingroup$
It needn't be divisible by 9; 181 would be valid but not divisible by 9
$endgroup$
– Benedict Randall Shaw
Dec 20 '18 at 16:39


















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