Remainder on division with $22$
$begingroup$
What is the remainder obtained when $14^{16}$ is divided with $22$?
Is there a general method for this, without using number theory? I wish to solve this question using binomial theorem only - maybe expressing the numerator as a summation in which most terms are divisible by $22$, except the remainder?
How should I proceed?
number-theory elementary-number-theory modular-arithmetic divisibility binomial-theorem
edited Jan 12 at 6:10
Abcd
3,02821235
asked Jul 3 '18 at 10:32
arya_starkarya_stark
676218
$endgroup$
add a comment |
$begingroup$
What is the remainder obtained when $14^{16}$ is divided with $22$?
Is there a general method for this, without using number theory? I wish to solve this question using binomial theorem only - maybe expressing the numerator as a summation in which most terms are divisible by $22$, except the remainder?
How should I proceed?
number-theory elementary-number-theory modular-arithmetic divisibility binomial-theorem
edited Jan 12 at 6:10
Abcd
3,02821235
asked Jul 3 '18 at 10:32
arya_starkarya_stark
676218
$endgroup$
$begingroup$
$14^{16} = (22 - 8)^{16}$
$endgroup$
– ab123
Jul 3 '18 at 10:37
$begingroup$
$8^{16} = 64^8 = (66 - 2)^8$
$endgroup$
– ab123
Jul 3 '18 at 10:41
7
$begingroup$
@ab123 Why not make an answer of that?
$endgroup$
– Arthur
Jul 3 '18 at 10:41
add a comment |
$begingroup$
What is the remainder obtained when $14^{16}$ is divided with $22$?
Is there a general method for this, without using number theory? I wish to solve this question using binomial theorem only - maybe expressing the numerator as a summation in which most terms are divisible by $22$, except the remainder?
How should I proceed?
number-theory elementary-number-theory modular-arithmetic divisibility binomial-theorem
edited Jan 12 at 6:10
Abcd
3,02821235
asked Jul 3 '18 at 10:32
arya_starkarya_stark
676218
$endgroup$
What is the remainder obtained when $14^{16}$ is divided with $22$?
Is there a general method for this, without using number theory? I wish to solve this question using binomial theorem only - maybe expressing the numerator as a summation in which most terms are divisible by $22$, except the remainder?
How should I proceed?
number-theory elementary-number-theory modular-arithmetic divisibility binomial-theorem
number-theory elementary-number-theory modular-arithmetic divisibility binomial-theorem
edited Jan 12 at 6:10
Abcd
3,02821235
asked Jul 3 '18 at 10:32
arya_starkarya_stark
676218
edited Jan 12 at 6:10
Abcd
3,02821235
asked Jul 3 '18 at 10:32
arya_starkarya_stark
676218
edited Jan 12 at 6:10
Abcd
3,02821235
edited Jan 12 at 6:10
Abcd
3,02821235
edited Jan 12 at 6:10
Abcd
3,02821235
3,02821235
asked Jul 3 '18 at 10:32
arya_starkarya_stark
676218
asked Jul 3 '18 at 10:32
arya_starkarya_stark
676218
asked Jul 3 '18 at 10:32
arya_starkarya_stark
676218
676218
$begingroup$
$14^{16} = (22 - 8)^{16}$
$endgroup$
– ab123
Jul 3 '18 at 10:37
$begingroup$
$8^{16} = 64^8 = (66 - 2)^8$
$endgroup$
– ab123
Jul 3 '18 at 10:41
7
$begingroup$
@ab123 Why not make an answer of that?
$endgroup$
– Arthur
Jul 3 '18 at 10:41
add a comment |
$begingroup$
$14^{16} = (22 - 8)^{16}$
$endgroup$
– ab123
Jul 3 '18 at 10:37
$begingroup$
$8^{16} = 64^8 = (66 - 2)^8$
$endgroup$
– ab123
Jul 3 '18 at 10:41
7
$begingroup$
@ab123 Why not make an answer of that?
$endgroup$
– Arthur
Jul 3 '18 at 10:41
$begingroup$
$14^{16} = (22 - 8)^{16}$
$endgroup$
– ab123
Jul 3 '18 at 10:37
$begingroup$
$14^{16} = (22 - 8)^{16}$
$endgroup$
– ab123
Jul 3 '18 at 10:37
$begingroup$
$8^{16} = 64^8 = (66 - 2)^8$
$endgroup$
– ab123
Jul 3 '18 at 10:41
$begingroup$
$8^{16} = 64^8 = (66 - 2)^8$
$endgroup$
– ab123
Jul 3 '18 at 10:41
7
7
$begingroup$
@ab123 Why not make an answer of that?
$endgroup$
– Arthur
Jul 3 '18 at 10:41
$begingroup$
@ab123 Why not make an answer of that?
$endgroup$
– Arthur
Jul 3 '18 at 10:41
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You can use binomial expansions and see that
$$14^{16} = (22 - 8)^{16}$$
implies that the remainder is just the remainder when $(-8)^{16}( = 8^{16})$ is divided by $22.$
Proceeding similarly,
$8^{16} = 64^8 = (66 - 2)^8 implies 2^8 = 256 text{ divided by } 22 implies text{remainder = 14}$
answered Jul 3 '18 at 10:45
ab123ab123
1,654423
$endgroup$
add a comment |
$begingroup$
Since $$ 14^2 equiv -2 $$ so $$14^{16} equiv (-2)^8 equiv 16^2equiv (-6)^2 equiv 14$$
or
$$ 14^2 = 22k -2 $$ so $$14^{16} = (22k-2)^8 = 22l+2^8=22l+22cdot 11+14$$
answered Jul 3 '18 at 10:39
greedoidgreedoid
39.9k114798
$endgroup$
add a comment |
$begingroup$
A method that uses FLT.
Finding the remainder on dividing $14^{16}$ by $22$ is equivalent to twice that on dividing $7cdot14^{15}$ by $11$.
By FLT, $14^{10}equiv1pmod{11}$ so $$7cdot14^{15}equiv7cdot14^5equiv98cdot196cdot196equiv-1cdot(-2)cdot(-2)equiv-4equiv7pmod{11}$$ hence the required remainder is $7times 2=14$.
answered Jul 3 '18 at 14:12
TheSimpliFireTheSimpliFire
12.4k62460
$endgroup$
add a comment |
$begingroup$
Using the Euclidean Algorithm
Note that
$$
begin{align}
14^{16}&equiv0&pmod2tag1
end{align}
$$
Reducing mod $11$ and using Fermat's Little Theorem, we get
$$
begin{align}
14^{16}
&equiv3^6&pmod{11}\
&equiv3&pmod{11}tag2
end{align}
$$
Solving $11a+2b=1$ using the Euclidean Algorithm as implemented in this answer,
$$
begin{array}{r}
&&5&2\hline
1&0&1&-2\
0&1&-5&11\
11&2&1&0\
end{array}tag3
$$
we have
$$
11(1)+2(-5)=1tag4
$$
from which we get
$$
begin{align}
-10&equiv0&pmod2\
-10&equiv1&pmod{11}
end{align}tag5
$$
Multiplying $(5)$ by $3$ gives
$$
begin{align}
-30&equiv0&pmod2\
-30&equiv3&pmod{11}
end{align}tag6
$$
Using the Chinese Remainder Theorem
The Chinese Remainder Theorem says that the solution to $(6)$ is unique mod ${22}$. Thus, we get the smallest positive solution to be
$$
begin{align}
14&equiv0&pmod2\
14&equiv3&pmod{11}
end{align}tag7
$$
Thus, $(1)$, $(2)$, and $(7)$ yields
$$
begin{align}
14^{16}&equiv14&pmod{22}tag8
end{align}
$$
answered Jul 3 '18 at 15:05
robjohn♦robjohn
266k27306629
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use binomial expansions and see that
$$14^{16} = (22 - 8)^{16}$$
implies that the remainder is just the remainder when $(-8)^{16}( = 8^{16})$ is divided by $22.$
Proceeding similarly,
$8^{16} = 64^8 = (66 - 2)^8 implies 2^8 = 256 text{ divided by } 22 implies text{remainder = 14}$
answered Jul 3 '18 at 10:45
ab123ab123
1,654423
$endgroup$
add a comment |
$begingroup$
You can use binomial expansions and see that
$$14^{16} = (22 - 8)^{16}$$
implies that the remainder is just the remainder when $(-8)^{16}( = 8^{16})$ is divided by $22.$
Proceeding similarly,
$8^{16} = 64^8 = (66 - 2)^8 implies 2^8 = 256 text{ divided by } 22 implies text{remainder = 14}$
answered Jul 3 '18 at 10:45
ab123ab123
1,654423
$endgroup$
add a comment |
$begingroup$
You can use binomial expansions and see that
$$14^{16} = (22 - 8)^{16}$$
implies that the remainder is just the remainder when $(-8)^{16}( = 8^{16})$ is divided by $22.$
Proceeding similarly,
$8^{16} = 64^8 = (66 - 2)^8 implies 2^8 = 256 text{ divided by } 22 implies text{remainder = 14}$
answered Jul 3 '18 at 10:45
ab123ab123
1,654423
$endgroup$
You can use binomial expansions and see that
$$14^{16} = (22 - 8)^{16}$$
implies that the remainder is just the remainder when $(-8)^{16}( = 8^{16})$ is divided by $22.$
Proceeding similarly,
$8^{16} = 64^8 = (66 - 2)^8 implies 2^8 = 256 text{ divided by } 22 implies text{remainder = 14}$
answered Jul 3 '18 at 10:45
ab123ab123
1,654423
edited Jul 3 '18 at 10:48
answered Jul 3 '18 at 10:45
ab123ab123
1,654423
answered Jul 3 '18 at 10:45
ab123ab123
1,654423
answered Jul 3 '18 at 10:45
ab123ab123
1,654423
1,654423
add a comment |
add a comment |
$begingroup$
Since $$ 14^2 equiv -2 $$ so $$14^{16} equiv (-2)^8 equiv 16^2equiv (-6)^2 equiv 14$$
or
$$ 14^2 = 22k -2 $$ so $$14^{16} = (22k-2)^8 = 22l+2^8=22l+22cdot 11+14$$
answered Jul 3 '18 at 10:39
greedoidgreedoid
39.9k114798
$endgroup$
add a comment |
$begingroup$
Since $$ 14^2 equiv -2 $$ so $$14^{16} equiv (-2)^8 equiv 16^2equiv (-6)^2 equiv 14$$
or
$$ 14^2 = 22k -2 $$ so $$14^{16} = (22k-2)^8 = 22l+2^8=22l+22cdot 11+14$$
answered Jul 3 '18 at 10:39
greedoidgreedoid
39.9k114798
$endgroup$
add a comment |
$begingroup$
Since $$ 14^2 equiv -2 $$ so $$14^{16} equiv (-2)^8 equiv 16^2equiv (-6)^2 equiv 14$$
or
$$ 14^2 = 22k -2 $$ so $$14^{16} = (22k-2)^8 = 22l+2^8=22l+22cdot 11+14$$
answered Jul 3 '18 at 10:39
greedoidgreedoid
39.9k114798
$endgroup$
Since $$ 14^2 equiv -2 $$ so $$14^{16} equiv (-2)^8 equiv 16^2equiv (-6)^2 equiv 14$$
or
$$ 14^2 = 22k -2 $$ so $$14^{16} = (22k-2)^8 = 22l+2^8=22l+22cdot 11+14$$
answered Jul 3 '18 at 10:39
greedoidgreedoid
39.9k114798
edited Jul 3 '18 at 10:53
answered Jul 3 '18 at 10:39
greedoidgreedoid
39.9k114798
answered Jul 3 '18 at 10:39
greedoidgreedoid
39.9k114798
answered Jul 3 '18 at 10:39
greedoidgreedoid
39.9k114798
39.9k114798
add a comment |
add a comment |
$begingroup$
A method that uses FLT.
Finding the remainder on dividing $14^{16}$ by $22$ is equivalent to twice that on dividing $7cdot14^{15}$ by $11$.
By FLT, $14^{10}equiv1pmod{11}$ so $$7cdot14^{15}equiv7cdot14^5equiv98cdot196cdot196equiv-1cdot(-2)cdot(-2)equiv-4equiv7pmod{11}$$ hence the required remainder is $7times 2=14$.
answered Jul 3 '18 at 14:12
TheSimpliFireTheSimpliFire
12.4k62460
$endgroup$
add a comment |
$begingroup$
A method that uses FLT.
Finding the remainder on dividing $14^{16}$ by $22$ is equivalent to twice that on dividing $7cdot14^{15}$ by $11$.
By FLT, $14^{10}equiv1pmod{11}$ so $$7cdot14^{15}equiv7cdot14^5equiv98cdot196cdot196equiv-1cdot(-2)cdot(-2)equiv-4equiv7pmod{11}$$ hence the required remainder is $7times 2=14$.
answered Jul 3 '18 at 14:12
TheSimpliFireTheSimpliFire
12.4k62460
$endgroup$
add a comment |
$begingroup$
A method that uses FLT.
Finding the remainder on dividing $14^{16}$ by $22$ is equivalent to twice that on dividing $7cdot14^{15}$ by $11$.
By FLT, $14^{10}equiv1pmod{11}$ so $$7cdot14^{15}equiv7cdot14^5equiv98cdot196cdot196equiv-1cdot(-2)cdot(-2)equiv-4equiv7pmod{11}$$ hence the required remainder is $7times 2=14$.
answered Jul 3 '18 at 14:12
TheSimpliFireTheSimpliFire
12.4k62460
$endgroup$
A method that uses FLT.
Finding the remainder on dividing $14^{16}$ by $22$ is equivalent to twice that on dividing $7cdot14^{15}$ by $11$.
By FLT, $14^{10}equiv1pmod{11}$ so $$7cdot14^{15}equiv7cdot14^5equiv98cdot196cdot196equiv-1cdot(-2)cdot(-2)equiv-4equiv7pmod{11}$$ hence the required remainder is $7times 2=14$.
answered Jul 3 '18 at 14:12
TheSimpliFireTheSimpliFire
12.4k62460
answered Jul 3 '18 at 14:12
TheSimpliFireTheSimpliFire
12.4k62460
answered Jul 3 '18 at 14:12
TheSimpliFireTheSimpliFire
12.4k62460
answered Jul 3 '18 at 14:12
TheSimpliFireTheSimpliFire
12.4k62460
12.4k62460
add a comment |
add a comment |
$begingroup$
Using the Euclidean Algorithm
Note that
$$
begin{align}
14^{16}&equiv0&pmod2tag1
end{align}
$$
Reducing mod $11$ and using Fermat's Little Theorem, we get
$$
begin{align}
14^{16}
&equiv3^6&pmod{11}\
&equiv3&pmod{11}tag2
end{align}
$$
Solving $11a+2b=1$ using the Euclidean Algorithm as implemented in this answer,
$$
begin{array}{r}
&&5&2\hline
1&0&1&-2\
0&1&-5&11\
11&2&1&0\
end{array}tag3
$$
we have
$$
11(1)+2(-5)=1tag4
$$
from which we get
$$
begin{align}
-10&equiv0&pmod2\
-10&equiv1&pmod{11}
end{align}tag5
$$
Multiplying $(5)$ by $3$ gives
$$
begin{align}
-30&equiv0&pmod2\
-30&equiv3&pmod{11}
end{align}tag6
$$
Using the Chinese Remainder Theorem
The Chinese Remainder Theorem says that the solution to $(6)$ is unique mod ${22}$. Thus, we get the smallest positive solution to be
$$
begin{align}
14&equiv0&pmod2\
14&equiv3&pmod{11}
end{align}tag7
$$
Thus, $(1)$, $(2)$, and $(7)$ yields
$$
begin{align}
14^{16}&equiv14&pmod{22}tag8
end{align}
$$
answered Jul 3 '18 at 15:05
robjohn♦robjohn
266k27306629
$endgroup$
add a comment |
$begingroup$
Using the Euclidean Algorithm
Note that
$$
begin{align}
14^{16}&equiv0&pmod2tag1
end{align}
$$
Reducing mod $11$ and using Fermat's Little Theorem, we get
$$
begin{align}
14^{16}
&equiv3^6&pmod{11}\
&equiv3&pmod{11}tag2
end{align}
$$
Solving $11a+2b=1$ using the Euclidean Algorithm as implemented in this answer,
$$
begin{array}{r}
&&5&2\hline
1&0&1&-2\
0&1&-5&11\
11&2&1&0\
end{array}tag3
$$
we have
$$
11(1)+2(-5)=1tag4
$$
from which we get
$$
begin{align}
-10&equiv0&pmod2\
-10&equiv1&pmod{11}
end{align}tag5
$$
Multiplying $(5)$ by $3$ gives
$$
begin{align}
-30&equiv0&pmod2\
-30&equiv3&pmod{11}
end{align}tag6
$$
Using the Chinese Remainder Theorem
The Chinese Remainder Theorem says that the solution to $(6)$ is unique mod ${22}$. Thus, we get the smallest positive solution to be
$$
begin{align}
14&equiv0&pmod2\
14&equiv3&pmod{11}
end{align}tag7
$$
Thus, $(1)$, $(2)$, and $(7)$ yields
$$
begin{align}
14^{16}&equiv14&pmod{22}tag8
end{align}
$$
answered Jul 3 '18 at 15:05
robjohn♦robjohn
266k27306629
$endgroup$
add a comment |
$begingroup$
Using the Euclidean Algorithm
Note that
$$
begin{align}
14^{16}&equiv0&pmod2tag1
end{align}
$$
Reducing mod $11$ and using Fermat's Little Theorem, we get
$$
begin{align}
14^{16}
&equiv3^6&pmod{11}\
&equiv3&pmod{11}tag2
end{align}
$$
Solving $11a+2b=1$ using the Euclidean Algorithm as implemented in this answer,
$$
begin{array}{r}
&&5&2\hline
1&0&1&-2\
0&1&-5&11\
11&2&1&0\
end{array}tag3
$$
we have
$$
11(1)+2(-5)=1tag4
$$
from which we get
$$
begin{align}
-10&equiv0&pmod2\
-10&equiv1&pmod{11}
end{align}tag5
$$
Multiplying $(5)$ by $3$ gives
$$
begin{align}
-30&equiv0&pmod2\
-30&equiv3&pmod{11}
end{align}tag6
$$
Using the Chinese Remainder Theorem
The Chinese Remainder Theorem says that the solution to $(6)$ is unique mod ${22}$. Thus, we get the smallest positive solution to be
$$
begin{align}
14&equiv0&pmod2\
14&equiv3&pmod{11}
end{align}tag7
$$
Thus, $(1)$, $(2)$, and $(7)$ yields
$$
begin{align}
14^{16}&equiv14&pmod{22}tag8
end{align}
$$
answered Jul 3 '18 at 15:05
robjohn♦robjohn
266k27306629
$endgroup$
Using the Euclidean Algorithm
Note that
$$
begin{align}
14^{16}&equiv0&pmod2tag1
end{align}
$$
Reducing mod $11$ and using Fermat's Little Theorem, we get
$$
begin{align}
14^{16}
&equiv3^6&pmod{11}\
&equiv3&pmod{11}tag2
end{align}
$$
Solving $11a+2b=1$ using the Euclidean Algorithm as implemented in this answer,
$$
begin{array}{r}
&&5&2\hline
1&0&1&-2\
0&1&-5&11\
11&2&1&0\
end{array}tag3
$$
we have
$$
11(1)+2(-5)=1tag4
$$
from which we get
$$
begin{align}
-10&equiv0&pmod2\
-10&equiv1&pmod{11}
end{align}tag5
$$
Multiplying $(5)$ by $3$ gives
$$
begin{align}
-30&equiv0&pmod2\
-30&equiv3&pmod{11}
end{align}tag6
$$
Using the Chinese Remainder Theorem
The Chinese Remainder Theorem says that the solution to $(6)$ is unique mod ${22}$. Thus, we get the smallest positive solution to be
$$
begin{align}
14&equiv0&pmod2\
14&equiv3&pmod{11}
end{align}tag7
$$
Thus, $(1)$, $(2)$, and $(7)$ yields
$$
begin{align}
14^{16}&equiv14&pmod{22}tag8
end{align}
$$
answered Jul 3 '18 at 15:05
robjohn♦robjohn
266k27306629
answered Jul 3 '18 at 15:05
robjohn♦robjohn
266k27306629
answered Jul 3 '18 at 15:05
robjohn♦robjohn
266k27306629
answered Jul 3 '18 at 15:05
robjohn♦robjohn
266k27306629
266k27306629
add a comment |
add a comment |
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$begingroup$
$14^{16} = (22 - 8)^{16}$
$endgroup$
– ab123
Jul 3 '18 at 10:37
$begingroup$
$8^{16} = 64^8 = (66 - 2)^8$
$endgroup$
– ab123
Jul 3 '18 at 10:41
7
$begingroup$
@ab123 Why not make an answer of that?
$endgroup$
– Arthur
Jul 3 '18 at 10:41