Modular arithmetic modulo $10^2011.$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to solve $ 227x equiv 1 ~ (textmod ~ 2011) $?Large exponential modularmodular arithmetic with exponentsModular Arithmetic - Pirate ProblemSystem of equations - modular arithmeticmodular arithmetic congruenceSolving basic arithmetic moduloModular Arithmetic involving fractional remainderCongruent modulo in modular arithmeticBeing $2011$ prime, calculate 2009! divided by 2011
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Modular arithmetic modulo $10^2011.$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to solve $ 227x equiv 1 ~ (textmod ~ 2011) $?Large exponential modularmodular arithmetic with exponentsModular Arithmetic - Pirate ProblemSystem of equations - modular arithmeticmodular arithmetic congruenceSolving basic arithmetic moduloModular Arithmetic involving fractional remainderCongruent modulo in modular arithmeticBeing $2011$ prime, calculate 2009! divided by 2011
$begingroup$
Is there a natural number $n$ such that $3^n equiv 7 (text mod 10^2011)$
I can't come up with an idea how to solve it...
number-theory modular-arithmetic
edited Apr 8 at 20:22
Dbchatto67
3,052623
asked Apr 8 at 20:05
spyerspyer
1238
$endgroup$
|
show 1 more comment
$begingroup$
Is there a natural number $n$ such that $3^n equiv 7 (text mod 10^2011)$
I can't come up with an idea how to solve it...
number-theory modular-arithmetic
edited Apr 8 at 20:22
Dbchatto67
3,052623
asked Apr 8 at 20:05
spyerspyer
1238
$endgroup$
$begingroup$
Look up Hensel's lifting lemma
$endgroup$
– Ross Millikan
Apr 8 at 20:31
2
$begingroup$
Look at it modulo eight.
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:03
1
$begingroup$
I will gladly explain how I came up with that. But before I do, I want to recommend you to study our guide for new askers. This looks like a contest training problem. Why don't you give the source? That is one of the alternative ways of providing the kind of context we think all the questions should have.
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:06
1
$begingroup$
Ok. A hint as to how I came up with the solution. There are powers of $3$ with the last two digits $07$. Namely $3^15$ and every twentieth ever after. But, writing their three last digits down (I didn't write down the whole things :-), you will spot something about the third last digit..
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:09
$begingroup$
Thank you very much! I will read the guide! This is not a separate problem from a competition. I was solving modular arithmetic related to the last k digits of a number and came up with a problem which I transformed into this. Thank you again!
$endgroup$
– spyer
Apr 8 at 21:22
|
show 1 more comment
$begingroup$
Is there a natural number $n$ such that $3^n equiv 7 (text mod 10^2011)$
I can't come up with an idea how to solve it...
number-theory modular-arithmetic
edited Apr 8 at 20:22
Dbchatto67
3,052623
asked Apr 8 at 20:05
spyerspyer
1238
$endgroup$
Is there a natural number $n$ such that $3^n equiv 7 (text mod 10^2011)$
I can't come up with an idea how to solve it...
number-theory modular-arithmetic
number-theory modular-arithmetic
edited Apr 8 at 20:22
Dbchatto67
3,052623
asked Apr 8 at 20:05
spyerspyer
1238
edited Apr 8 at 20:22
Dbchatto67
3,052623
asked Apr 8 at 20:05
spyerspyer
1238
edited Apr 8 at 20:22
Dbchatto67
3,052623
edited Apr 8 at 20:22
Dbchatto67
3,052623
edited Apr 8 at 20:22
Dbchatto67
3,052623
3,052623
asked Apr 8 at 20:05
spyerspyer
1238
asked Apr 8 at 20:05
spyerspyer
1238
asked Apr 8 at 20:05
spyerspyer
1238
1238
$begingroup$
Look up Hensel's lifting lemma
$endgroup$
– Ross Millikan
Apr 8 at 20:31
2
$begingroup$
Look at it modulo eight.
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:03
1
$begingroup$
I will gladly explain how I came up with that. But before I do, I want to recommend you to study our guide for new askers. This looks like a contest training problem. Why don't you give the source? That is one of the alternative ways of providing the kind of context we think all the questions should have.
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:06
1
$begingroup$
Ok. A hint as to how I came up with the solution. There are powers of $3$ with the last two digits $07$. Namely $3^15$ and every twentieth ever after. But, writing their three last digits down (I didn't write down the whole things :-), you will spot something about the third last digit..
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:09
$begingroup$
Thank you very much! I will read the guide! This is not a separate problem from a competition. I was solving modular arithmetic related to the last k digits of a number and came up with a problem which I transformed into this. Thank you again!
$endgroup$
– spyer
Apr 8 at 21:22
|
show 1 more comment
$begingroup$
Look up Hensel's lifting lemma
$endgroup$
– Ross Millikan
Apr 8 at 20:31
2
$begingroup$
Look at it modulo eight.
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:03
1
$begingroup$
I will gladly explain how I came up with that. But before I do, I want to recommend you to study our guide for new askers. This looks like a contest training problem. Why don't you give the source? That is one of the alternative ways of providing the kind of context we think all the questions should have.
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:06
1
$begingroup$
Ok. A hint as to how I came up with the solution. There are powers of $3$ with the last two digits $07$. Namely $3^15$ and every twentieth ever after. But, writing their three last digits down (I didn't write down the whole things :-), you will spot something about the third last digit..
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:09
$begingroup$
Thank you very much! I will read the guide! This is not a separate problem from a competition. I was solving modular arithmetic related to the last k digits of a number and came up with a problem which I transformed into this. Thank you again!
$endgroup$
– spyer
Apr 8 at 21:22
$begingroup$
Look up Hensel's lifting lemma
$endgroup$
– Ross Millikan
Apr 8 at 20:31
$begingroup$
Look up Hensel's lifting lemma
$endgroup$
– Ross Millikan
Apr 8 at 20:31
2
2
$begingroup$
Look at it modulo eight.
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:03
$begingroup$
Look at it modulo eight.
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:03
1
1
$begingroup$
I will gladly explain how I came up with that. But before I do, I want to recommend you to study our guide for new askers. This looks like a contest training problem. Why don't you give the source? That is one of the alternative ways of providing the kind of context we think all the questions should have.
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:06
$begingroup$
I will gladly explain how I came up with that. But before I do, I want to recommend you to study our guide for new askers. This looks like a contest training problem. Why don't you give the source? That is one of the alternative ways of providing the kind of context we think all the questions should have.
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:06
1
1
$begingroup$
Ok. A hint as to how I came up with the solution. There are powers of $3$ with the last two digits $07$. Namely $3^15$ and every twentieth ever after. But, writing their three last digits down (I didn't write down the whole things :-), you will spot something about the third last digit..
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:09
$begingroup$
Ok. A hint as to how I came up with the solution. There are powers of $3$ with the last two digits $07$. Namely $3^15$ and every twentieth ever after. But, writing their three last digits down (I didn't write down the whole things :-), you will spot something about the third last digit..
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:09
$begingroup$
Thank you very much! I will read the guide! This is not a separate problem from a competition. I was solving modular arithmetic related to the last k digits of a number and came up with a problem which I transformed into this. Thank you again!
$endgroup$
– spyer
Apr 8 at 21:22
$begingroup$
Thank you very much! I will read the guide! This is not a separate problem from a competition. I was solving modular arithmetic related to the last k digits of a number and came up with a problem which I transformed into this. Thank you again!
$endgroup$
– spyer
Apr 8 at 21:22
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Simply to get this out of the unanswered queue. No. the original congruence can be rewritten as $$10^2011y+7=3^n$$ which works for any divisor of $10^2011$ 8 is such a divisor. The problem is no power of 3, is 7 mod 8. It therefore fails to be true.
answered Apr 8 at 23:57
Roddy MacPheeRoddy MacPhee
780118
$endgroup$
add a comment |
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$begingroup$
Simply to get this out of the unanswered queue. No. the original congruence can be rewritten as $$10^2011y+7=3^n$$ which works for any divisor of $10^2011$ 8 is such a divisor. The problem is no power of 3, is 7 mod 8. It therefore fails to be true.
answered Apr 8 at 23:57
Roddy MacPheeRoddy MacPhee
780118
$endgroup$
add a comment |
$begingroup$
Simply to get this out of the unanswered queue. No. the original congruence can be rewritten as $$10^2011y+7=3^n$$ which works for any divisor of $10^2011$ 8 is such a divisor. The problem is no power of 3, is 7 mod 8. It therefore fails to be true.
answered Apr 8 at 23:57
Roddy MacPheeRoddy MacPhee
780118
$endgroup$
add a comment |
$begingroup$
Simply to get this out of the unanswered queue. No. the original congruence can be rewritten as $$10^2011y+7=3^n$$ which works for any divisor of $10^2011$ 8 is such a divisor. The problem is no power of 3, is 7 mod 8. It therefore fails to be true.
answered Apr 8 at 23:57
Roddy MacPheeRoddy MacPhee
780118
$endgroup$
Simply to get this out of the unanswered queue. No. the original congruence can be rewritten as $$10^2011y+7=3^n$$ which works for any divisor of $10^2011$ 8 is such a divisor. The problem is no power of 3, is 7 mod 8. It therefore fails to be true.
answered Apr 8 at 23:57
Roddy MacPheeRoddy MacPhee
780118
edited 2 days ago
answered Apr 8 at 23:57
Roddy MacPheeRoddy MacPhee
780118
answered Apr 8 at 23:57
Roddy MacPheeRoddy MacPhee
780118
answered Apr 8 at 23:57
Roddy MacPheeRoddy MacPhee
780118
780118
add a comment |
add a comment |
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$begingroup$
Look up Hensel's lifting lemma
$endgroup$
– Ross Millikan
Apr 8 at 20:31
2
$begingroup$
Look at it modulo eight.
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:03
1
$begingroup$
I will gladly explain how I came up with that. But before I do, I want to recommend you to study our guide for new askers. This looks like a contest training problem. Why don't you give the source? That is one of the alternative ways of providing the kind of context we think all the questions should have.
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:06
1
$begingroup$
Ok. A hint as to how I came up with the solution. There are powers of $3$ with the last two digits $07$. Namely $3^15$ and every twentieth ever after. But, writing their three last digits down (I didn't write down the whole things :-), you will spot something about the third last digit..
$endgroup$
– Jyrki Lahtonen
Apr 8 at 21:09
$begingroup$
Thank you very much! I will read the guide! This is not a separate problem from a competition. I was solving modular arithmetic related to the last k digits of a number and came up with a problem which I transformed into this. Thank you again!
$endgroup$
– spyer
Apr 8 at 21:22