Mixing
Finding Remainder Using Binomial Theorem
$begingroup$
Find the remainder when $7^{98} $ is divided by $5$.
What I am doing here is expanding ${(5+2)}^{98} $ using binomial theorem and writing it as $5k + 2$, where $k$ is a positive integer but the answer is $4$ and I'm getting $2.$
They are expanding ${(50 - 1)}^{49} $ using binomial theorem and then writing $50k - 1$ and getting $4$ as a reminder.
How those two methods are different?
combinatorics number-theory binomial-theorem
edited Jan 26 at 16:33
Thomas Shelby
4,3042726
asked Jan 26 at 16:21
user638507user638507
142
$endgroup$
add a comment |
$begingroup$
Find the remainder when $7^{98} $ is divided by $5$.
What I am doing here is expanding ${(5+2)}^{98} $ using binomial theorem and writing it as $5k + 2$, where $k$ is a positive integer but the answer is $4$ and I'm getting $2.$
They are expanding ${(50 - 1)}^{49} $ using binomial theorem and then writing $50k - 1$ and getting $4$ as a reminder.
How those two methods are different?
combinatorics number-theory binomial-theorem
edited Jan 26 at 16:33
Thomas Shelby
4,3042726
asked Jan 26 at 16:21
user638507user638507
142
$endgroup$
$begingroup$
$4$ is correct, so you made an error somewhere. If you seek help debugging your calculation then you need to show your work. The point of using BT is that it reduces it to computing $(-1)^{49},,$ which is simpler than computing $,2^{98}pmod 5 $
$endgroup$
– Bill Dubuque
Jan 26 at 16:35
1
$begingroup$
But BT isn't needed: $largebmod 5!:, 7^{large 98}!equiv (7^{large 2})^{large 49}!equiv (-1)^{large 49}!equiv -1equiv 4, $ [or $large, 7equiv 2,$ and $large,2^{large 2}equiv -1$], by using standard congruence arithmetic rules.
$endgroup$
– Bill Dubuque
Jan 26 at 16:47
add a comment |
$begingroup$
Find the remainder when $7^{98} $ is divided by $5$.
What I am doing here is expanding ${(5+2)}^{98} $ using binomial theorem and writing it as $5k + 2$, where $k$ is a positive integer but the answer is $4$ and I'm getting $2.$
They are expanding ${(50 - 1)}^{49} $ using binomial theorem and then writing $50k - 1$ and getting $4$ as a reminder.
How those two methods are different?
combinatorics number-theory binomial-theorem
edited Jan 26 at 16:33
Thomas Shelby
4,3042726
asked Jan 26 at 16:21
user638507user638507
142
$endgroup$
Find the remainder when $7^{98} $ is divided by $5$.
What I am doing here is expanding ${(5+2)}^{98} $ using binomial theorem and writing it as $5k + 2$, where $k$ is a positive integer but the answer is $4$ and I'm getting $2.$
They are expanding ${(50 - 1)}^{49} $ using binomial theorem and then writing $50k - 1$ and getting $4$ as a reminder.
How those two methods are different?
combinatorics number-theory binomial-theorem
combinatorics number-theory binomial-theorem
edited Jan 26 at 16:33
Thomas Shelby
4,3042726
asked Jan 26 at 16:21
user638507user638507
142
edited Jan 26 at 16:33
Thomas Shelby
4,3042726
asked Jan 26 at 16:21
user638507user638507
142
edited Jan 26 at 16:33
Thomas Shelby
4,3042726
edited Jan 26 at 16:33
Thomas Shelby
4,3042726
edited Jan 26 at 16:33
Thomas Shelby
4,3042726
4,3042726
asked Jan 26 at 16:21
user638507user638507
142
asked Jan 26 at 16:21
user638507user638507
142
asked Jan 26 at 16:21
user638507user638507
142
142
$begingroup$
$4$ is correct, so you made an error somewhere. If you seek help debugging your calculation then you need to show your work. The point of using BT is that it reduces it to computing $(-1)^{49},,$ which is simpler than computing $,2^{98}pmod 5 $
$endgroup$
– Bill Dubuque
Jan 26 at 16:35
1
$begingroup$
But BT isn't needed: $largebmod 5!:, 7^{large 98}!equiv (7^{large 2})^{large 49}!equiv (-1)^{large 49}!equiv -1equiv 4, $ [or $large, 7equiv 2,$ and $large,2^{large 2}equiv -1$], by using standard congruence arithmetic rules.
$endgroup$
– Bill Dubuque
Jan 26 at 16:47
add a comment |
$begingroup$
$4$ is correct, so you made an error somewhere. If you seek help debugging your calculation then you need to show your work. The point of using BT is that it reduces it to computing $(-1)^{49},,$ which is simpler than computing $,2^{98}pmod 5 $
$endgroup$
– Bill Dubuque
Jan 26 at 16:35
1
$begingroup$
But BT isn't needed: $largebmod 5!:, 7^{large 98}!equiv (7^{large 2})^{large 49}!equiv (-1)^{large 49}!equiv -1equiv 4, $ [or $large, 7equiv 2,$ and $large,2^{large 2}equiv -1$], by using standard congruence arithmetic rules.
$endgroup$
– Bill Dubuque
Jan 26 at 16:47
$begingroup$
$4$ is correct, so you made an error somewhere. If you seek help debugging your calculation then you need to show your work. The point of using BT is that it reduces it to computing $(-1)^{49},,$ which is simpler than computing $,2^{98}pmod 5 $
$endgroup$
– Bill Dubuque
Jan 26 at 16:35
$begingroup$
$4$ is correct, so you made an error somewhere. If you seek help debugging your calculation then you need to show your work. The point of using BT is that it reduces it to computing $(-1)^{49},,$ which is simpler than computing $,2^{98}pmod 5 $
$endgroup$
– Bill Dubuque
Jan 26 at 16:35
1
1
$begingroup$
But BT isn't needed: $largebmod 5!:, 7^{large 98}!equiv (7^{large 2})^{large 49}!equiv (-1)^{large 49}!equiv -1equiv 4, $ [or $large, 7equiv 2,$ and $large,2^{large 2}equiv -1$], by using standard congruence arithmetic rules.
$endgroup$
– Bill Dubuque
Jan 26 at 16:47
$begingroup$
But BT isn't needed: $largebmod 5!:, 7^{large 98}!equiv (7^{large 2})^{large 49}!equiv (-1)^{large 49}!equiv -1equiv 4, $ [or $large, 7equiv 2,$ and $large,2^{large 2}equiv -1$], by using standard congruence arithmetic rules.
$endgroup$
– Bill Dubuque
Jan 26 at 16:47
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The binomial theorem tells you
$$(5+2)^{98} = binom{98}{0}cdot 5^{98}+binom{98}{1}cdot 5^{91}2^1+ binom{98}{2}cdot 5^{90}2^2+cdots + binom{98}{97}cdot 5cdot 2^{97}+ binom{98}{98}cdot 2^{98} = 5k+2^{98}.$$
So you need to find the remainder when $2^{98}$ is divided by $5$, not the remainder when $2$ is divided by $5$.
The method using $(50-1)^{49}$ is a little easier because $(-1)^{49}=-1$.
answered Jan 26 at 16:27
kccukccu
10.6k11229
$endgroup$
1
$begingroup$
Also $2^4 equiv 1 pmod 5$ and $2^{98}=2^{4 cdot 24+2}$.
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:30
add a comment |
$begingroup$
With your method, you have to check the divison by 5 of $2^{98} = 4cdot (2^4)^{24} = 4cdot (3cdot 5 +1)^{24}$ and, using the Binomial theorem again, you end up with a rest after division of $4 cdot 1^{24} = 4$ which is also the "other" result.
answered Jan 26 at 16:29
AndreasAndreas
8,3511137
$endgroup$
add a comment |
$begingroup$
When you expand ${(5+2)}^{98}$, you will get $5k+2^{98}$, for some $kinBbb N $.
If you know Fermat's little theorem, you have $2^4equiv 1pmod 5$. Hence $2^{98}equiv 2^{2}equiv 4pmod 5$. So the remainder is $4$.
answered Jan 26 at 16:28
Thomas ShelbyThomas Shelby
4,3042726
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The binomial theorem tells you
$$(5+2)^{98} = binom{98}{0}cdot 5^{98}+binom{98}{1}cdot 5^{91}2^1+ binom{98}{2}cdot 5^{90}2^2+cdots + binom{98}{97}cdot 5cdot 2^{97}+ binom{98}{98}cdot 2^{98} = 5k+2^{98}.$$
So you need to find the remainder when $2^{98}$ is divided by $5$, not the remainder when $2$ is divided by $5$.
The method using $(50-1)^{49}$ is a little easier because $(-1)^{49}=-1$.
answered Jan 26 at 16:27
kccukccu
10.6k11229
$endgroup$
1
$begingroup$
Also $2^4 equiv 1 pmod 5$ and $2^{98}=2^{4 cdot 24+2}$.
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:30
add a comment |
$begingroup$
The binomial theorem tells you
$$(5+2)^{98} = binom{98}{0}cdot 5^{98}+binom{98}{1}cdot 5^{91}2^1+ binom{98}{2}cdot 5^{90}2^2+cdots + binom{98}{97}cdot 5cdot 2^{97}+ binom{98}{98}cdot 2^{98} = 5k+2^{98}.$$
So you need to find the remainder when $2^{98}$ is divided by $5$, not the remainder when $2$ is divided by $5$.
The method using $(50-1)^{49}$ is a little easier because $(-1)^{49}=-1$.
answered Jan 26 at 16:27
kccukccu
10.6k11229
$endgroup$
1
$begingroup$
Also $2^4 equiv 1 pmod 5$ and $2^{98}=2^{4 cdot 24+2}$.
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:30
add a comment |
$begingroup$
The binomial theorem tells you
$$(5+2)^{98} = binom{98}{0}cdot 5^{98}+binom{98}{1}cdot 5^{91}2^1+ binom{98}{2}cdot 5^{90}2^2+cdots + binom{98}{97}cdot 5cdot 2^{97}+ binom{98}{98}cdot 2^{98} = 5k+2^{98}.$$
So you need to find the remainder when $2^{98}$ is divided by $5$, not the remainder when $2$ is divided by $5$.
The method using $(50-1)^{49}$ is a little easier because $(-1)^{49}=-1$.
answered Jan 26 at 16:27
kccukccu
10.6k11229
$endgroup$
The binomial theorem tells you
$$(5+2)^{98} = binom{98}{0}cdot 5^{98}+binom{98}{1}cdot 5^{91}2^1+ binom{98}{2}cdot 5^{90}2^2+cdots + binom{98}{97}cdot 5cdot 2^{97}+ binom{98}{98}cdot 2^{98} = 5k+2^{98}.$$
So you need to find the remainder when $2^{98}$ is divided by $5$, not the remainder when $2$ is divided by $5$.
The method using $(50-1)^{49}$ is a little easier because $(-1)^{49}=-1$.
answered Jan 26 at 16:27
kccukccu
10.6k11229
answered Jan 26 at 16:27
kccukccu
10.6k11229
answered Jan 26 at 16:27
kccukccu
10.6k11229
answered Jan 26 at 16:27
kccukccu
10.6k11229
10.6k11229
1
$begingroup$
Also $2^4 equiv 1 pmod 5$ and $2^{98}=2^{4 cdot 24+2}$.
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:30
add a comment |
1
$begingroup$
Also $2^4 equiv 1 pmod 5$ and $2^{98}=2^{4 cdot 24+2}$.
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:30
1
1
$begingroup$
Also $2^4 equiv 1 pmod 5$ and $2^{98}=2^{4 cdot 24+2}$.
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:30
$begingroup$
Also $2^4 equiv 1 pmod 5$ and $2^{98}=2^{4 cdot 24+2}$.
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:30
add a comment |
$begingroup$
With your method, you have to check the divison by 5 of $2^{98} = 4cdot (2^4)^{24} = 4cdot (3cdot 5 +1)^{24}$ and, using the Binomial theorem again, you end up with a rest after division of $4 cdot 1^{24} = 4$ which is also the "other" result.
answered Jan 26 at 16:29
AndreasAndreas
8,3511137
$endgroup$
add a comment |
$begingroup$
With your method, you have to check the divison by 5 of $2^{98} = 4cdot (2^4)^{24} = 4cdot (3cdot 5 +1)^{24}$ and, using the Binomial theorem again, you end up with a rest after division of $4 cdot 1^{24} = 4$ which is also the "other" result.
answered Jan 26 at 16:29
AndreasAndreas
8,3511137
$endgroup$
add a comment |
$begingroup$
With your method, you have to check the divison by 5 of $2^{98} = 4cdot (2^4)^{24} = 4cdot (3cdot 5 +1)^{24}$ and, using the Binomial theorem again, you end up with a rest after division of $4 cdot 1^{24} = 4$ which is also the "other" result.
answered Jan 26 at 16:29
AndreasAndreas
8,3511137
$endgroup$
With your method, you have to check the divison by 5 of $2^{98} = 4cdot (2^4)^{24} = 4cdot (3cdot 5 +1)^{24}$ and, using the Binomial theorem again, you end up with a rest after division of $4 cdot 1^{24} = 4$ which is also the "other" result.
answered Jan 26 at 16:29
AndreasAndreas
8,3511137
answered Jan 26 at 16:29
AndreasAndreas
8,3511137
answered Jan 26 at 16:29
AndreasAndreas
8,3511137
answered Jan 26 at 16:29
AndreasAndreas
8,3511137
8,3511137
add a comment |
add a comment |
$begingroup$
When you expand ${(5+2)}^{98}$, you will get $5k+2^{98}$, for some $kinBbb N $.
If you know Fermat's little theorem, you have $2^4equiv 1pmod 5$. Hence $2^{98}equiv 2^{2}equiv 4pmod 5$. So the remainder is $4$.
answered Jan 26 at 16:28
Thomas ShelbyThomas Shelby
4,3042726
$endgroup$
add a comment |
$begingroup$
When you expand ${(5+2)}^{98}$, you will get $5k+2^{98}$, for some $kinBbb N $.
If you know Fermat's little theorem, you have $2^4equiv 1pmod 5$. Hence $2^{98}equiv 2^{2}equiv 4pmod 5$. So the remainder is $4$.
answered Jan 26 at 16:28
Thomas ShelbyThomas Shelby
4,3042726
$endgroup$
add a comment |
$begingroup$
When you expand ${(5+2)}^{98}$, you will get $5k+2^{98}$, for some $kinBbb N $.
If you know Fermat's little theorem, you have $2^4equiv 1pmod 5$. Hence $2^{98}equiv 2^{2}equiv 4pmod 5$. So the remainder is $4$.
answered Jan 26 at 16:28
Thomas ShelbyThomas Shelby
4,3042726
$endgroup$
When you expand ${(5+2)}^{98}$, you will get $5k+2^{98}$, for some $kinBbb N $.
If you know Fermat's little theorem, you have $2^4equiv 1pmod 5$. Hence $2^{98}equiv 2^{2}equiv 4pmod 5$. So the remainder is $4$.
answered Jan 26 at 16:28
Thomas ShelbyThomas Shelby
4,3042726
edited Jan 26 at 16:35
answered Jan 26 at 16:28
Thomas ShelbyThomas Shelby
4,3042726
answered Jan 26 at 16:28
Thomas ShelbyThomas Shelby
4,3042726
answered Jan 26 at 16:28
Thomas ShelbyThomas Shelby
4,3042726
4,3042726
add a comment |
add a comment |
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$begingroup$
$4$ is correct, so you made an error somewhere. If you seek help debugging your calculation then you need to show your work. The point of using BT is that it reduces it to computing $(-1)^{49},,$ which is simpler than computing $,2^{98}pmod 5 $
$endgroup$
– Bill Dubuque
Jan 26 at 16:35
1
$begingroup$
But BT isn't needed: $largebmod 5!:, 7^{large 98}!equiv (7^{large 2})^{large 49}!equiv (-1)^{large 49}!equiv -1equiv 4, $ [or $large, 7equiv 2,$ and $large,2^{large 2}equiv -1$], by using standard congruence arithmetic rules.
$endgroup$
– Bill Dubuque
Jan 26 at 16:47