Find the remainder from the division of $3^{2017}-1$ into $3^{403}-1$
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Here is an interesting problem:
Find the remainder from the division of $3^{2017}-1$ into $3^{403}-1$
elementary-number-theory arithmetic divisibility
edited Nov 20 at 19:19
Martin Sleziak
44.5k7115268
asked Nov 19 at 18:31
ten1o
1335
add a comment |
up vote
0
down vote
favorite
Here is an interesting problem:
Find the remainder from the division of $3^{2017}-1$ into $3^{403}-1$
elementary-number-theory arithmetic divisibility
edited Nov 20 at 19:19
Martin Sleziak
44.5k7115268
asked Nov 19 at 18:31
ten1o
1335
Since you are a fairly new user, I would first of all like to welcome you to our stack exchange! Secondly, I suggest that you put some effort, thoughts or even any knowledge you have over the subject of your questions. This way, your posts will be way better well taken !
– Rebellos
Nov 19 at 19:23
Hint $bmod n-1!: nequiv 1,Rightarrow, f(n)equiv f(1) $ for any polynomial $f(x)$ with integer coefs
– Bill Dubuque
Nov 19 at 20:32
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Here is an interesting problem:
Find the remainder from the division of $3^{2017}-1$ into $3^{403}-1$
elementary-number-theory arithmetic divisibility
edited Nov 20 at 19:19
Martin Sleziak
44.5k7115268
asked Nov 19 at 18:31
ten1o
1335
Here is an interesting problem:
Find the remainder from the division of $3^{2017}-1$ into $3^{403}-1$
elementary-number-theory arithmetic divisibility
elementary-number-theory arithmetic divisibility
edited Nov 20 at 19:19
Martin Sleziak
44.5k7115268
asked Nov 19 at 18:31
ten1o
1335
edited Nov 20 at 19:19
Martin Sleziak
44.5k7115268
asked Nov 19 at 18:31
ten1o
1335
edited Nov 20 at 19:19
Martin Sleziak
44.5k7115268
edited Nov 20 at 19:19
Martin Sleziak
44.5k7115268
edited Nov 20 at 19:19
Martin Sleziak
44.5k7115268
44.5k7115268
asked Nov 19 at 18:31
ten1o
1335
asked Nov 19 at 18:31
ten1o
1335
asked Nov 19 at 18:31
ten1o
1335
1335
Since you are a fairly new user, I would first of all like to welcome you to our stack exchange! Secondly, I suggest that you put some effort, thoughts or even any knowledge you have over the subject of your questions. This way, your posts will be way better well taken !
– Rebellos
Nov 19 at 19:23
Hint $bmod n-1!: nequiv 1,Rightarrow, f(n)equiv f(1) $ for any polynomial $f(x)$ with integer coefs
– Bill Dubuque
Nov 19 at 20:32
add a comment |
Since you are a fairly new user, I would first of all like to welcome you to our stack exchange! Secondly, I suggest that you put some effort, thoughts or even any knowledge you have over the subject of your questions. This way, your posts will be way better well taken !
– Rebellos
Nov 19 at 19:23
Hint $bmod n-1!: nequiv 1,Rightarrow, f(n)equiv f(1) $ for any polynomial $f(x)$ with integer coefs
– Bill Dubuque
Nov 19 at 20:32
Since you are a fairly new user, I would first of all like to welcome you to our stack exchange! Secondly, I suggest that you put some effort, thoughts or even any knowledge you have over the subject of your questions. This way, your posts will be way better well taken !
– Rebellos
Nov 19 at 19:23
Since you are a fairly new user, I would first of all like to welcome you to our stack exchange! Secondly, I suggest that you put some effort, thoughts or even any knowledge you have over the subject of your questions. This way, your posts will be way better well taken !
– Rebellos
Nov 19 at 19:23
Hint $bmod n-1!: nequiv 1,Rightarrow, f(n)equiv f(1) $ for any polynomial $f(x)$ with integer coefs
– Bill Dubuque
Nov 19 at 20:32
Hint $bmod n-1!: nequiv 1,Rightarrow, f(n)equiv f(1) $ for any polynomial $f(x)$ with integer coefs
– Bill Dubuque
Nov 19 at 20:32
add a comment |
3 Answers
3
active
oldest
votes
up vote
4
down vote
Hint :
$$3^{2017} = Big(3^{403}Big)^5 cdot 3^2$$
edited Nov 19 at 18:41
J.G.
19.8k21932
answered Nov 19 at 18:39
Rebellos
12.8k21041
add a comment |
up vote
3
down vote
$$3^{403}equiv 1pmod{3^{403}-1}$$
Raise to the power 5. You get,
$$3^{2015}equiv 1pmod{3^{403}-1}$$
$$3^{2017}equiv 9pmod{3^{403}-1}$$
$$3^{2017}-1equiv 8pmod{3^{403}-1}$$
So the remainder is 8.
add a comment |
up vote
2
down vote
If $n=3^{403}$, you are dividing $9n^5-1$ by $n-1$.
But $9n^5-1=9(n^5-1)+8$ and the remainder is $8$.
answered Nov 19 at 18:47
Yves Daoust
122k668218
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Hint :
$$3^{2017} = Big(3^{403}Big)^5 cdot 3^2$$
edited Nov 19 at 18:41
J.G.
19.8k21932
answered Nov 19 at 18:39
Rebellos
12.8k21041
add a comment |
up vote
4
down vote
Hint :
$$3^{2017} = Big(3^{403}Big)^5 cdot 3^2$$
edited Nov 19 at 18:41
J.G.
19.8k21932
answered Nov 19 at 18:39
Rebellos
12.8k21041
add a comment |
up vote
4
down vote
up vote
4
down vote
Hint :
$$3^{2017} = Big(3^{403}Big)^5 cdot 3^2$$
edited Nov 19 at 18:41
J.G.
19.8k21932
answered Nov 19 at 18:39
Rebellos
12.8k21041
Hint :
$$3^{2017} = Big(3^{403}Big)^5 cdot 3^2$$
edited Nov 19 at 18:41
J.G.
19.8k21932
answered Nov 19 at 18:39
Rebellos
12.8k21041
edited Nov 19 at 18:41
J.G.
19.8k21932
edited Nov 19 at 18:41
J.G.
19.8k21932
edited Nov 19 at 18:41
J.G.
19.8k21932
19.8k21932
answered Nov 19 at 18:39
Rebellos
12.8k21041
answered Nov 19 at 18:39
Rebellos
12.8k21041
answered Nov 19 at 18:39
Rebellos
12.8k21041
12.8k21041
add a comment |
add a comment |
up vote
3
down vote
$$3^{403}equiv 1pmod{3^{403}-1}$$
Raise to the power 5. You get,
$$3^{2015}equiv 1pmod{3^{403}-1}$$
$$3^{2017}equiv 9pmod{3^{403}-1}$$
$$3^{2017}-1equiv 8pmod{3^{403}-1}$$
So the remainder is 8.
add a comment |
up vote
3
down vote
$$3^{403}equiv 1pmod{3^{403}-1}$$
Raise to the power 5. You get,
$$3^{2015}equiv 1pmod{3^{403}-1}$$
$$3^{2017}equiv 9pmod{3^{403}-1}$$
$$3^{2017}-1equiv 8pmod{3^{403}-1}$$
So the remainder is 8.
add a comment |
up vote
3
down vote
up vote
3
down vote
$$3^{403}equiv 1pmod{3^{403}-1}$$
Raise to the power 5. You get,
$$3^{2015}equiv 1pmod{3^{403}-1}$$
$$3^{2017}equiv 9pmod{3^{403}-1}$$
$$3^{2017}-1equiv 8pmod{3^{403}-1}$$
So the remainder is 8.
$$3^{403}equiv 1pmod{3^{403}-1}$$
Raise to the power 5. You get,
$$3^{2015}equiv 1pmod{3^{403}-1}$$
$$3^{2017}equiv 9pmod{3^{403}-1}$$
$$3^{2017}-1equiv 8pmod{3^{403}-1}$$
So the remainder is 8.
answered Nov 19 at 18:46
user612946
add a comment |
add a comment |
up vote
2
down vote
If $n=3^{403}$, you are dividing $9n^5-1$ by $n-1$.
But $9n^5-1=9(n^5-1)+8$ and the remainder is $8$.
answered Nov 19 at 18:47
Yves Daoust
122k668218
add a comment |
up vote
2
down vote
If $n=3^{403}$, you are dividing $9n^5-1$ by $n-1$.
But $9n^5-1=9(n^5-1)+8$ and the remainder is $8$.
answered Nov 19 at 18:47
Yves Daoust
122k668218
add a comment |
up vote
2
down vote
up vote
2
down vote
If $n=3^{403}$, you are dividing $9n^5-1$ by $n-1$.
But $9n^5-1=9(n^5-1)+8$ and the remainder is $8$.
answered Nov 19 at 18:47
Yves Daoust
122k668218
If $n=3^{403}$, you are dividing $9n^5-1$ by $n-1$.
But $9n^5-1=9(n^5-1)+8$ and the remainder is $8$.
answered Nov 19 at 18:47
Yves Daoust
122k668218
answered Nov 19 at 18:47
Yves Daoust
122k668218
answered Nov 19 at 18:47
Yves Daoust
122k668218
answered Nov 19 at 18:47
Yves Daoust
122k668218
122k668218
add a comment |
add a comment |
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Since you are a fairly new user, I would first of all like to welcome you to our stack exchange! Secondly, I suggest that you put some effort, thoughts or even any knowledge you have over the subject of your questions. This way, your posts will be way better well taken !
– Rebellos
Nov 19 at 19:23
Hint $bmod n-1!: nequiv 1,Rightarrow, f(n)equiv f(1) $ for any polynomial $f(x)$ with integer coefs
– Bill Dubuque
Nov 19 at 20:32