Detemine the unit digit of a number [duplicate]
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This question already has an answer here:
How do I compute $a^b,bmod c$ by hand?
9 answers
Find the unit digit of the number:
$$3^7005 times 6^8000$$
My turn:
$$3^7005times 6^8000=3^7005times 3^8000 times2^8000$$ $$3^13005 times 2^8000$$
But i could not go on ?
algebra-precalculus elementary-number-theory
edited Mar 8 at 14:39
José Carlos Santos
172k23133241
asked Mar 8 at 13:36
Hussien MohamedHussien Mohamed
244
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marked as duplicate by Bill Dubuque algebra-precalculus
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Mar 8 at 15:14
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
How do I compute $a^b,bmod c$ by hand?
9 answers
Find the unit digit of the number:
$$3^7005 times 6^8000$$
My turn:
$$3^7005times 6^8000=3^7005times 3^8000 times2^8000$$ $$3^13005 times 2^8000$$
But i could not go on ?
algebra-precalculus elementary-number-theory
edited Mar 8 at 14:39
José Carlos Santos
172k23133241
asked Mar 8 at 13:36
Hussien MohamedHussien Mohamed
244
$endgroup$
marked as duplicate by Bill Dubuque algebra-precalculus
Users with the algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.
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Mar 8 at 15:14
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Well, you can start by the fact that the last digit of $6^n$ is always $6$ for any positive whole number $n$. Then you only have the $3^7005$ to deal with.
$endgroup$
– Matti P.
Mar 8 at 13:39
add a comment |
$begingroup$
This question already has an answer here:
How do I compute $a^b,bmod c$ by hand?
9 answers
Find the unit digit of the number:
$$3^7005 times 6^8000$$
My turn:
$$3^7005times 6^8000=3^7005times 3^8000 times2^8000$$ $$3^13005 times 2^8000$$
But i could not go on ?
algebra-precalculus elementary-number-theory
edited Mar 8 at 14:39
José Carlos Santos
172k23133241
asked Mar 8 at 13:36
Hussien MohamedHussien Mohamed
244
$endgroup$
This question already has an answer here:
How do I compute $a^b,bmod c$ by hand?
9 answers
Find the unit digit of the number:
$$3^7005 times 6^8000$$
My turn:
$$3^7005times 6^8000=3^7005times 3^8000 times2^8000$$ $$3^13005 times 2^8000$$
But i could not go on ?
This question already has an answer here:
How do I compute $a^b,bmod c$ by hand?
9 answers
algebra-precalculus elementary-number-theory
algebra-precalculus elementary-number-theory
edited Mar 8 at 14:39
José Carlos Santos
172k23133241
asked Mar 8 at 13:36
Hussien MohamedHussien Mohamed
244
edited Mar 8 at 14:39
José Carlos Santos
172k23133241
asked Mar 8 at 13:36
Hussien MohamedHussien Mohamed
244
edited Mar 8 at 14:39
José Carlos Santos
172k23133241
edited Mar 8 at 14:39
José Carlos Santos
172k23133241
edited Mar 8 at 14:39
José Carlos Santos
172k23133241
172k23133241
asked Mar 8 at 13:36
Hussien MohamedHussien Mohamed
244
asked Mar 8 at 13:36
Hussien MohamedHussien Mohamed
244
asked Mar 8 at 13:36
Hussien MohamedHussien Mohamed
244
244
marked as duplicate by Bill Dubuque algebra-precalculus
Users with the algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.
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Mar 8 at 15:14
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Bill Dubuque algebra-precalculus
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Mar 8 at 15:14
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Well, you can start by the fact that the last digit of $6^n$ is always $6$ for any positive whole number $n$. Then you only have the $3^7005$ to deal with.
$endgroup$
– Matti P.
Mar 8 at 13:39
add a comment |
$begingroup$
Well, you can start by the fact that the last digit of $6^n$ is always $6$ for any positive whole number $n$. Then you only have the $3^7005$ to deal with.
$endgroup$
– Matti P.
Mar 8 at 13:39
$begingroup$
Well, you can start by the fact that the last digit of $6^n$ is always $6$ for any positive whole number $n$. Then you only have the $3^7005$ to deal with.
$endgroup$
– Matti P.
Mar 8 at 13:39
$begingroup$
Well, you can start by the fact that the last digit of $6^n$ is always $6$ for any positive whole number $n$. Then you only have the $3^7005$ to deal with.
$endgroup$
– Matti P.
Mar 8 at 13:39
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The unit digit of the numbers of the powers of $3$ form a cyclic sequence: $1,3,9,7,1,3,9,7,ldots$ In particular, the unit digit of $3^n$ is $3$ if $n$ is of the form $4k+1$. And the unit digit of any power of $6$ is $6$. Since $7,005$ is of the form $4k+1$, the answer to your question is $8(=3times6pmod10)$.
answered Mar 8 at 13:39
José Carlos SantosJosé Carlos Santos
172k23133241
$endgroup$
$begingroup$
Agreed! I wonder why someone downvoted ...
$endgroup$
– Matti P.
Mar 8 at 13:50
$begingroup$
It happens all the time: :-(
$endgroup$
– José Carlos Santos
Mar 8 at 13:51
$begingroup$
Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
$endgroup$
– B. Goddard
Mar 8 at 14:08
$begingroup$
@B.Goddard Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Mar 8 at 14:15
add a comment |
$begingroup$
We want the unit digit of the number $3^7005 cdot6^8000$. This is equivalent to asking for the residue of the number $3^7005 cdot6^8000$ modulo $10$. Hence, we want some number $n$ with $0 leq n leq 9$ such that
$$
n equiv 3^7005 cdot6^8000 pmod10
$$
Note that $3^4 equiv 1 pmod10$ and $6^n equiv 6 pmod10$ for all $ninmathbbN$. With this result, we have
beginalign
3^7005 cdot6^8000 &= 3^7004cdot 3^1 cdot 6^8000 \
&= left(3^4right)^1751cdot 3^1 cdot 6^8000\
&equiv1^1751cdot 3^1 cdot 6^1 pmod10 \
&= 3 cdot 6 \
&equiv 8 pmod10 \
endalign
Hence, the unit digit of $3^7005 cdot6^8000$ is $8$.
answered Mar 8 at 14:12
BrianBrian
1,268216
$endgroup$
1
$begingroup$
Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
$endgroup$
– Keith Backman
Mar 8 at 14:50
$begingroup$
@KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
$endgroup$
– Brian
Mar 8 at 14:54
$begingroup$
Not to worry -- you should see some of the bloopers I've made!
$endgroup$
– Keith Backman
Mar 8 at 15:56
add a comment |
$begingroup$
If you think about the way you typically do multiplication, the unit digit of a product is the product of the unit digits.
Can you figure out the unit digit of the two terms in the product?
Hint: calculate $3^0, 3^1, 3^2, 3^3, 3^4, 3^5,ldots$ Do you see a pattern in the unit digits? Do you see why the pattern holds?
edited Mar 8 at 13:52
NickD
1,2021512
answered Mar 8 at 13:41
Daniel McLauryDaniel McLaury
16.1k33081
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The unit digit of the numbers of the powers of $3$ form a cyclic sequence: $1,3,9,7,1,3,9,7,ldots$ In particular, the unit digit of $3^n$ is $3$ if $n$ is of the form $4k+1$. And the unit digit of any power of $6$ is $6$. Since $7,005$ is of the form $4k+1$, the answer to your question is $8(=3times6pmod10)$.
answered Mar 8 at 13:39
José Carlos SantosJosé Carlos Santos
172k23133241
$endgroup$
$begingroup$
Agreed! I wonder why someone downvoted ...
$endgroup$
– Matti P.
Mar 8 at 13:50
$begingroup$
It happens all the time: :-(
$endgroup$
– José Carlos Santos
Mar 8 at 13:51
$begingroup$
Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
$endgroup$
– B. Goddard
Mar 8 at 14:08
$begingroup$
@B.Goddard Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Mar 8 at 14:15
add a comment |
$begingroup$
The unit digit of the numbers of the powers of $3$ form a cyclic sequence: $1,3,9,7,1,3,9,7,ldots$ In particular, the unit digit of $3^n$ is $3$ if $n$ is of the form $4k+1$. And the unit digit of any power of $6$ is $6$. Since $7,005$ is of the form $4k+1$, the answer to your question is $8(=3times6pmod10)$.
answered Mar 8 at 13:39
José Carlos SantosJosé Carlos Santos
172k23133241
$endgroup$
$begingroup$
Agreed! I wonder why someone downvoted ...
$endgroup$
– Matti P.
Mar 8 at 13:50
$begingroup$
It happens all the time: :-(
$endgroup$
– José Carlos Santos
Mar 8 at 13:51
$begingroup$
Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
$endgroup$
– B. Goddard
Mar 8 at 14:08
$begingroup$
@B.Goddard Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Mar 8 at 14:15
add a comment |
$begingroup$
The unit digit of the numbers of the powers of $3$ form a cyclic sequence: $1,3,9,7,1,3,9,7,ldots$ In particular, the unit digit of $3^n$ is $3$ if $n$ is of the form $4k+1$. And the unit digit of any power of $6$ is $6$. Since $7,005$ is of the form $4k+1$, the answer to your question is $8(=3times6pmod10)$.
answered Mar 8 at 13:39
José Carlos SantosJosé Carlos Santos
172k23133241
$endgroup$
The unit digit of the numbers of the powers of $3$ form a cyclic sequence: $1,3,9,7,1,3,9,7,ldots$ In particular, the unit digit of $3^n$ is $3$ if $n$ is of the form $4k+1$. And the unit digit of any power of $6$ is $6$. Since $7,005$ is of the form $4k+1$, the answer to your question is $8(=3times6pmod10)$.
answered Mar 8 at 13:39
José Carlos SantosJosé Carlos Santos
172k23133241
edited Mar 8 at 14:14
answered Mar 8 at 13:39
José Carlos SantosJosé Carlos Santos
172k23133241
answered Mar 8 at 13:39
José Carlos SantosJosé Carlos Santos
172k23133241
answered Mar 8 at 13:39
José Carlos SantosJosé Carlos Santos
172k23133241
172k23133241
$begingroup$
Agreed! I wonder why someone downvoted ...
$endgroup$
– Matti P.
Mar 8 at 13:50
$begingroup$
It happens all the time: :-(
$endgroup$
– José Carlos Santos
Mar 8 at 13:51
$begingroup$
Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
$endgroup$
– B. Goddard
Mar 8 at 14:08
$begingroup$
@B.Goddard Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Mar 8 at 14:15
add a comment |
$begingroup$
Agreed! I wonder why someone downvoted ...
$endgroup$
– Matti P.
Mar 8 at 13:50
$begingroup$
It happens all the time: :-(
$endgroup$
– José Carlos Santos
Mar 8 at 13:51
$begingroup$
Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
$endgroup$
– B. Goddard
Mar 8 at 14:08
$begingroup$
@B.Goddard Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Mar 8 at 14:15
$begingroup$
Agreed! I wonder why someone downvoted ...
$endgroup$
– Matti P.
Mar 8 at 13:50
$begingroup$
Agreed! I wonder why someone downvoted ...
$endgroup$
– Matti P.
Mar 8 at 13:50
$begingroup$
It happens all the time: :-(
$endgroup$
– José Carlos Santos
Mar 8 at 13:51
$begingroup$
It happens all the time: :-(
$endgroup$
– José Carlos Santos
Mar 8 at 13:51
$begingroup$
Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
$endgroup$
– B. Goddard
Mar 8 at 14:08
$begingroup$
Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
$endgroup$
– B. Goddard
Mar 8 at 14:08
$begingroup$
@B.Goddard Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Mar 8 at 14:15
$begingroup$
@B.Goddard Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Mar 8 at 14:15
add a comment |
$begingroup$
We want the unit digit of the number $3^7005 cdot6^8000$. This is equivalent to asking for the residue of the number $3^7005 cdot6^8000$ modulo $10$. Hence, we want some number $n$ with $0 leq n leq 9$ such that
$$
n equiv 3^7005 cdot6^8000 pmod10
$$
Note that $3^4 equiv 1 pmod10$ and $6^n equiv 6 pmod10$ for all $ninmathbbN$. With this result, we have
beginalign
3^7005 cdot6^8000 &= 3^7004cdot 3^1 cdot 6^8000 \
&= left(3^4right)^1751cdot 3^1 cdot 6^8000\
&equiv1^1751cdot 3^1 cdot 6^1 pmod10 \
&= 3 cdot 6 \
&equiv 8 pmod10 \
endalign
Hence, the unit digit of $3^7005 cdot6^8000$ is $8$.
answered Mar 8 at 14:12
BrianBrian
1,268216
$endgroup$
1
$begingroup$
Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
$endgroup$
– Keith Backman
Mar 8 at 14:50
$begingroup$
@KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
$endgroup$
– Brian
Mar 8 at 14:54
$begingroup$
Not to worry -- you should see some of the bloopers I've made!
$endgroup$
– Keith Backman
Mar 8 at 15:56
add a comment |
$begingroup$
We want the unit digit of the number $3^7005 cdot6^8000$. This is equivalent to asking for the residue of the number $3^7005 cdot6^8000$ modulo $10$. Hence, we want some number $n$ with $0 leq n leq 9$ such that
$$
n equiv 3^7005 cdot6^8000 pmod10
$$
Note that $3^4 equiv 1 pmod10$ and $6^n equiv 6 pmod10$ for all $ninmathbbN$. With this result, we have
beginalign
3^7005 cdot6^8000 &= 3^7004cdot 3^1 cdot 6^8000 \
&= left(3^4right)^1751cdot 3^1 cdot 6^8000\
&equiv1^1751cdot 3^1 cdot 6^1 pmod10 \
&= 3 cdot 6 \
&equiv 8 pmod10 \
endalign
Hence, the unit digit of $3^7005 cdot6^8000$ is $8$.
answered Mar 8 at 14:12
BrianBrian
1,268216
$endgroup$
1
$begingroup$
Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
$endgroup$
– Keith Backman
Mar 8 at 14:50
$begingroup$
@KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
$endgroup$
– Brian
Mar 8 at 14:54
$begingroup$
Not to worry -- you should see some of the bloopers I've made!
$endgroup$
– Keith Backman
Mar 8 at 15:56
add a comment |
$begingroup$
We want the unit digit of the number $3^7005 cdot6^8000$. This is equivalent to asking for the residue of the number $3^7005 cdot6^8000$ modulo $10$. Hence, we want some number $n$ with $0 leq n leq 9$ such that
$$
n equiv 3^7005 cdot6^8000 pmod10
$$
Note that $3^4 equiv 1 pmod10$ and $6^n equiv 6 pmod10$ for all $ninmathbbN$. With this result, we have
beginalign
3^7005 cdot6^8000 &= 3^7004cdot 3^1 cdot 6^8000 \
&= left(3^4right)^1751cdot 3^1 cdot 6^8000\
&equiv1^1751cdot 3^1 cdot 6^1 pmod10 \
&= 3 cdot 6 \
&equiv 8 pmod10 \
endalign
Hence, the unit digit of $3^7005 cdot6^8000$ is $8$.
answered Mar 8 at 14:12
BrianBrian
1,268216
$endgroup$
We want the unit digit of the number $3^7005 cdot6^8000$. This is equivalent to asking for the residue of the number $3^7005 cdot6^8000$ modulo $10$. Hence, we want some number $n$ with $0 leq n leq 9$ such that
$$
n equiv 3^7005 cdot6^8000 pmod10
$$
Note that $3^4 equiv 1 pmod10$ and $6^n equiv 6 pmod10$ for all $ninmathbbN$. With this result, we have
beginalign
3^7005 cdot6^8000 &= 3^7004cdot 3^1 cdot 6^8000 \
&= left(3^4right)^1751cdot 3^1 cdot 6^8000\
&equiv1^1751cdot 3^1 cdot 6^1 pmod10 \
&= 3 cdot 6 \
&equiv 8 pmod10 \
endalign
Hence, the unit digit of $3^7005 cdot6^8000$ is $8$.
answered Mar 8 at 14:12
BrianBrian
1,268216
edited Mar 8 at 14:53
answered Mar 8 at 14:12
BrianBrian
1,268216
answered Mar 8 at 14:12
BrianBrian
1,268216
answered Mar 8 at 14:12
BrianBrian
1,268216
1,268216
1
$begingroup$
Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
$endgroup$
– Keith Backman
Mar 8 at 14:50
$begingroup$
@KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
$endgroup$
– Brian
Mar 8 at 14:54
$begingroup$
Not to worry -- you should see some of the bloopers I've made!
$endgroup$
– Keith Backman
Mar 8 at 15:56
add a comment |
1
$begingroup$
Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
$endgroup$
– Keith Backman
Mar 8 at 14:50
$begingroup$
@KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
$endgroup$
– Brian
Mar 8 at 14:54
$begingroup$
Not to worry -- you should see some of the bloopers I've made!
$endgroup$
– Keith Backman
Mar 8 at 15:56
1
1
$begingroup$
Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
$endgroup$
– Keith Backman
Mar 8 at 14:50
$begingroup$
Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
$endgroup$
– Keith Backman
Mar 8 at 14:50
$begingroup$
@KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
$endgroup$
– Brian
Mar 8 at 14:54
$begingroup$
@KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
$endgroup$
– Brian
Mar 8 at 14:54
$begingroup$
Not to worry -- you should see some of the bloopers I've made!
$endgroup$
– Keith Backman
Mar 8 at 15:56
$begingroup$
Not to worry -- you should see some of the bloopers I've made!
$endgroup$
– Keith Backman
Mar 8 at 15:56
add a comment |
$begingroup$
If you think about the way you typically do multiplication, the unit digit of a product is the product of the unit digits.
Can you figure out the unit digit of the two terms in the product?
Hint: calculate $3^0, 3^1, 3^2, 3^3, 3^4, 3^5,ldots$ Do you see a pattern in the unit digits? Do you see why the pattern holds?
edited Mar 8 at 13:52
NickD
1,2021512
answered Mar 8 at 13:41
Daniel McLauryDaniel McLaury
16.1k33081
$endgroup$
add a comment |
$begingroup$
If you think about the way you typically do multiplication, the unit digit of a product is the product of the unit digits.
Can you figure out the unit digit of the two terms in the product?
Hint: calculate $3^0, 3^1, 3^2, 3^3, 3^4, 3^5,ldots$ Do you see a pattern in the unit digits? Do you see why the pattern holds?
edited Mar 8 at 13:52
NickD
1,2021512
answered Mar 8 at 13:41
Daniel McLauryDaniel McLaury
16.1k33081
$endgroup$
add a comment |
$begingroup$
If you think about the way you typically do multiplication, the unit digit of a product is the product of the unit digits.
Can you figure out the unit digit of the two terms in the product?
Hint: calculate $3^0, 3^1, 3^2, 3^3, 3^4, 3^5,ldots$ Do you see a pattern in the unit digits? Do you see why the pattern holds?
edited Mar 8 at 13:52
NickD
1,2021512
answered Mar 8 at 13:41
Daniel McLauryDaniel McLaury
16.1k33081
$endgroup$
If you think about the way you typically do multiplication, the unit digit of a product is the product of the unit digits.
Can you figure out the unit digit of the two terms in the product?
Hint: calculate $3^0, 3^1, 3^2, 3^3, 3^4, 3^5,ldots$ Do you see a pattern in the unit digits? Do you see why the pattern holds?
edited Mar 8 at 13:52
NickD
1,2021512
answered Mar 8 at 13:41
Daniel McLauryDaniel McLaury
16.1k33081
edited Mar 8 at 13:52
NickD
1,2021512
edited Mar 8 at 13:52
NickD
1,2021512
edited Mar 8 at 13:52
NickD
1,2021512
1,2021512
answered Mar 8 at 13:41
Daniel McLauryDaniel McLaury
16.1k33081
answered Mar 8 at 13:41
Daniel McLauryDaniel McLaury
16.1k33081
answered Mar 8 at 13:41
Daniel McLauryDaniel McLaury
16.1k33081
16.1k33081
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Well, you can start by the fact that the last digit of $6^n$ is always $6$ for any positive whole number $n$. Then you only have the $3^7005$ to deal with.
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– Matti P.
Mar 8 at 13:39