Solve a set of congruences
$begingroup$
John is thinking of a number $n$. He's willing to tell us that the number is close to $10000$ and in binary system it ends on $101$. In $7$ and $11$ system it ends on digit $2$ and the last two digits in the ternary numeral system are $21$. What is this number he is thinking of? I understand we need to create a set of congruences and solve it.
I have tried the following: Two of them are $x = 2, (mod 7)$ and $x = 2 (mod 11)$. That should be correct. The first one is $x = 5 (mod 8)$ and the last is $x = 7 (mod 9)$. I can see for this first one that $101$ is $5$ when converted to decimal system, but where do we get modulo $8$, from $2^3$? But why? Thank you for any clarification.
discrete-mathematics modular-arithmetic congruence-relations
edited Jan 18 at 14:29
idriskameni
641319
asked Jan 18 at 12:18
ponikoliponikoli
416
$endgroup$
add a comment |
$begingroup$
John is thinking of a number $n$. He's willing to tell us that the number is close to $10000$ and in binary system it ends on $101$. In $7$ and $11$ system it ends on digit $2$ and the last two digits in the ternary numeral system are $21$. What is this number he is thinking of? I understand we need to create a set of congruences and solve it.
I have tried the following: Two of them are $x = 2, (mod 7)$ and $x = 2 (mod 11)$. That should be correct. The first one is $x = 5 (mod 8)$ and the last is $x = 7 (mod 9)$. I can see for this first one that $101$ is $5$ when converted to decimal system, but where do we get modulo $8$, from $2^3$? But why? Thank you for any clarification.
discrete-mathematics modular-arithmetic congruence-relations
edited Jan 18 at 14:29
idriskameni
641319
asked Jan 18 at 12:18
ponikoliponikoli
416
$endgroup$
$begingroup$
For the same reason that a decimal number that ends in 65 is congruent to 65 modulus 100 as well as being congruenent to 5 modulus 10.
$endgroup$
– William Elliot
Jan 18 at 12:42
add a comment |
$begingroup$
John is thinking of a number $n$. He's willing to tell us that the number is close to $10000$ and in binary system it ends on $101$. In $7$ and $11$ system it ends on digit $2$ and the last two digits in the ternary numeral system are $21$. What is this number he is thinking of? I understand we need to create a set of congruences and solve it.
I have tried the following: Two of them are $x = 2, (mod 7)$ and $x = 2 (mod 11)$. That should be correct. The first one is $x = 5 (mod 8)$ and the last is $x = 7 (mod 9)$. I can see for this first one that $101$ is $5$ when converted to decimal system, but where do we get modulo $8$, from $2^3$? But why? Thank you for any clarification.
discrete-mathematics modular-arithmetic congruence-relations
edited Jan 18 at 14:29
idriskameni
641319
asked Jan 18 at 12:18
ponikoliponikoli
416
$endgroup$
John is thinking of a number $n$. He's willing to tell us that the number is close to $10000$ and in binary system it ends on $101$. In $7$ and $11$ system it ends on digit $2$ and the last two digits in the ternary numeral system are $21$. What is this number he is thinking of? I understand we need to create a set of congruences and solve it.
I have tried the following: Two of them are $x = 2, (mod 7)$ and $x = 2 (mod 11)$. That should be correct. The first one is $x = 5 (mod 8)$ and the last is $x = 7 (mod 9)$. I can see for this first one that $101$ is $5$ when converted to decimal system, but where do we get modulo $8$, from $2^3$? But why? Thank you for any clarification.
discrete-mathematics modular-arithmetic congruence-relations
discrete-mathematics modular-arithmetic congruence-relations
edited Jan 18 at 14:29
idriskameni
641319
asked Jan 18 at 12:18
ponikoliponikoli
416
edited Jan 18 at 14:29
idriskameni
641319
asked Jan 18 at 12:18
ponikoliponikoli
416
edited Jan 18 at 14:29
idriskameni
641319
edited Jan 18 at 14:29
idriskameni
641319
edited Jan 18 at 14:29
idriskameni
641319
641319
asked Jan 18 at 12:18
ponikoliponikoli
416
asked Jan 18 at 12:18
ponikoliponikoli
416
asked Jan 18 at 12:18
ponikoliponikoli
416
416
$begingroup$
For the same reason that a decimal number that ends in 65 is congruent to 65 modulus 100 as well as being congruenent to 5 modulus 10.
$endgroup$
– William Elliot
Jan 18 at 12:42
add a comment |
$begingroup$
For the same reason that a decimal number that ends in 65 is congruent to 65 modulus 100 as well as being congruenent to 5 modulus 10.
$endgroup$
– William Elliot
Jan 18 at 12:42
$begingroup$
For the same reason that a decimal number that ends in 65 is congruent to 65 modulus 100 as well as being congruenent to 5 modulus 10.
$endgroup$
– William Elliot
Jan 18 at 12:42
$begingroup$
For the same reason that a decimal number that ends in 65 is congruent to 65 modulus 100 as well as being congruenent to 5 modulus 10.
$endgroup$
– William Elliot
Jan 18 at 12:42
add a comment |
1 Answer
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If your binary and ternary numerals are $ b_nb_{n-1}dots b_2b_1b_0 $ and $ t_mt_{m-1}dots t_1t_0 $, respectively, with $b_2=1, b_1=0, b_0=1, t_1=2 $ and $ t_0=1 $, then, by definition, the numbers they represent are:
begin{eqnarray}
B &=& b_n,2^n + b_{n-1},2^{n-1} + dots + b_3,2^3 + 1times2^2 + 0times2 + 1\
&=& 2^3,left(,b_n,2^{n-3}+b_{n-1},2^{n-4} + dots + b_3, right) + 5 mbox{and}\
T &=& t_m,3^m + t_{m-1},3^{m-1} + dots + t_2,3^2 + 2times3 + 1\
&=& 3^2,left(,t_m,3^{m-2} + t_{m-1},3^{m-3} + dots + t_2,right) + 7 ,
end{eqnarray}
from which you should be able to see that if you subtract $ 5 $ from $ B $ the result is divisible by $ 8 $, and if you subtract $ 7 $ from $ T $ the result is divisible by 9.
answered Jan 18 at 15:33
lonza leggieralonza leggiera
70117
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
If your binary and ternary numerals are $ b_nb_{n-1}dots b_2b_1b_0 $ and $ t_mt_{m-1}dots t_1t_0 $, respectively, with $b_2=1, b_1=0, b_0=1, t_1=2 $ and $ t_0=1 $, then, by definition, the numbers they represent are:
begin{eqnarray}
B &=& b_n,2^n + b_{n-1},2^{n-1} + dots + b_3,2^3 + 1times2^2 + 0times2 + 1\
&=& 2^3,left(,b_n,2^{n-3}+b_{n-1},2^{n-4} + dots + b_3, right) + 5 mbox{and}\
T &=& t_m,3^m + t_{m-1},3^{m-1} + dots + t_2,3^2 + 2times3 + 1\
&=& 3^2,left(,t_m,3^{m-2} + t_{m-1},3^{m-3} + dots + t_2,right) + 7 ,
end{eqnarray}
from which you should be able to see that if you subtract $ 5 $ from $ B $ the result is divisible by $ 8 $, and if you subtract $ 7 $ from $ T $ the result is divisible by 9.
answered Jan 18 at 15:33
lonza leggieralonza leggiera
70117
$endgroup$
add a comment |
$begingroup$
If your binary and ternary numerals are $ b_nb_{n-1}dots b_2b_1b_0 $ and $ t_mt_{m-1}dots t_1t_0 $, respectively, with $b_2=1, b_1=0, b_0=1, t_1=2 $ and $ t_0=1 $, then, by definition, the numbers they represent are:
begin{eqnarray}
B &=& b_n,2^n + b_{n-1},2^{n-1} + dots + b_3,2^3 + 1times2^2 + 0times2 + 1\
&=& 2^3,left(,b_n,2^{n-3}+b_{n-1},2^{n-4} + dots + b_3, right) + 5 mbox{and}\
T &=& t_m,3^m + t_{m-1},3^{m-1} + dots + t_2,3^2 + 2times3 + 1\
&=& 3^2,left(,t_m,3^{m-2} + t_{m-1},3^{m-3} + dots + t_2,right) + 7 ,
end{eqnarray}
from which you should be able to see that if you subtract $ 5 $ from $ B $ the result is divisible by $ 8 $, and if you subtract $ 7 $ from $ T $ the result is divisible by 9.
answered Jan 18 at 15:33
lonza leggieralonza leggiera
70117
$endgroup$
add a comment |
$begingroup$
If your binary and ternary numerals are $ b_nb_{n-1}dots b_2b_1b_0 $ and $ t_mt_{m-1}dots t_1t_0 $, respectively, with $b_2=1, b_1=0, b_0=1, t_1=2 $ and $ t_0=1 $, then, by definition, the numbers they represent are:
begin{eqnarray}
B &=& b_n,2^n + b_{n-1},2^{n-1} + dots + b_3,2^3 + 1times2^2 + 0times2 + 1\
&=& 2^3,left(,b_n,2^{n-3}+b_{n-1},2^{n-4} + dots + b_3, right) + 5 mbox{and}\
T &=& t_m,3^m + t_{m-1},3^{m-1} + dots + t_2,3^2 + 2times3 + 1\
&=& 3^2,left(,t_m,3^{m-2} + t_{m-1},3^{m-3} + dots + t_2,right) + 7 ,
end{eqnarray}
from which you should be able to see that if you subtract $ 5 $ from $ B $ the result is divisible by $ 8 $, and if you subtract $ 7 $ from $ T $ the result is divisible by 9.
answered Jan 18 at 15:33
lonza leggieralonza leggiera
70117
$endgroup$
If your binary and ternary numerals are $ b_nb_{n-1}dots b_2b_1b_0 $ and $ t_mt_{m-1}dots t_1t_0 $, respectively, with $b_2=1, b_1=0, b_0=1, t_1=2 $ and $ t_0=1 $, then, by definition, the numbers they represent are:
begin{eqnarray}
B &=& b_n,2^n + b_{n-1},2^{n-1} + dots + b_3,2^3 + 1times2^2 + 0times2 + 1\
&=& 2^3,left(,b_n,2^{n-3}+b_{n-1},2^{n-4} + dots + b_3, right) + 5 mbox{and}\
T &=& t_m,3^m + t_{m-1},3^{m-1} + dots + t_2,3^2 + 2times3 + 1\
&=& 3^2,left(,t_m,3^{m-2} + t_{m-1},3^{m-3} + dots + t_2,right) + 7 ,
end{eqnarray}
from which you should be able to see that if you subtract $ 5 $ from $ B $ the result is divisible by $ 8 $, and if you subtract $ 7 $ from $ T $ the result is divisible by 9.
answered Jan 18 at 15:33
lonza leggieralonza leggiera
70117
answered Jan 18 at 15:33
lonza leggieralonza leggiera
70117
answered Jan 18 at 15:33
lonza leggieralonza leggiera
70117
answered Jan 18 at 15:33
lonza leggieralonza leggiera
70117
70117
add a comment |
add a comment |
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$begingroup$
For the same reason that a decimal number that ends in 65 is congruent to 65 modulus 100 as well as being congruenent to 5 modulus 10.
$endgroup$
– William Elliot
Jan 18 at 12:42