Prove $4|n$ if and only if $4|(10a_1+a_0)$ [duplicate]
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This question already has an answer here:
Why doesn't the last digit method work for divisibility by 4?
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Prove $4|n$ if and only if $4|(10a_1+a_0)$. I know there may be "duplicates" out there, but most of them are for the case of $3|n$ and stuff like that. I'm still confused as to how the thought process works though. I have to prove the statement for $n=(a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots + (a_1 times 10^1) + (a_0 times 10^0)$. Before having to prove this in my homework assignment, I had to prove that if $n$ is an integer and $n geq 2$, then $10^n equiv 0 (mod 4)$.
modular-arithmetic
asked Dec 9 '18 at 20:45
ClaireClaire
606
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marked as duplicate by GNUSupporter 8964民主女神 地下教會, Bill Dubuque modular-arithmetic
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Dec 9 '18 at 21:22
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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$begingroup$
This question already has an answer here:
Why doesn't the last digit method work for divisibility by 4?
3 answers
Prove $4|n$ if and only if $4|(10a_1+a_0)$. I know there may be "duplicates" out there, but most of them are for the case of $3|n$ and stuff like that. I'm still confused as to how the thought process works though. I have to prove the statement for $n=(a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots + (a_1 times 10^1) + (a_0 times 10^0)$. Before having to prove this in my homework assignment, I had to prove that if $n$ is an integer and $n geq 2$, then $10^n equiv 0 (mod 4)$.
modular-arithmetic
asked Dec 9 '18 at 20:45
ClaireClaire
606
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marked as duplicate by GNUSupporter 8964民主女神 地下教會, Bill Dubuque modular-arithmetic
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Dec 9 '18 at 21:22
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.
1
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What are your thoughts on the question ?
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– mathcounterexamples.net
Dec 9 '18 at 20:47
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25 times 4 = 100, then what?
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– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 20:48
add a comment |
$begingroup$
This question already has an answer here:
Why doesn't the last digit method work for divisibility by 4?
3 answers
Prove $4|n$ if and only if $4|(10a_1+a_0)$. I know there may be "duplicates" out there, but most of them are for the case of $3|n$ and stuff like that. I'm still confused as to how the thought process works though. I have to prove the statement for $n=(a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots + (a_1 times 10^1) + (a_0 times 10^0)$. Before having to prove this in my homework assignment, I had to prove that if $n$ is an integer and $n geq 2$, then $10^n equiv 0 (mod 4)$.
modular-arithmetic
asked Dec 9 '18 at 20:45
ClaireClaire
606
$endgroup$
This question already has an answer here:
Why doesn't the last digit method work for divisibility by 4?
3 answers
Prove $4|n$ if and only if $4|(10a_1+a_0)$. I know there may be "duplicates" out there, but most of them are for the case of $3|n$ and stuff like that. I'm still confused as to how the thought process works though. I have to prove the statement for $n=(a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots + (a_1 times 10^1) + (a_0 times 10^0)$. Before having to prove this in my homework assignment, I had to prove that if $n$ is an integer and $n geq 2$, then $10^n equiv 0 (mod 4)$.
This question already has an answer here:
Why doesn't the last digit method work for divisibility by 4?
3 answers
modular-arithmetic
modular-arithmetic
asked Dec 9 '18 at 20:45
ClaireClaire
606
asked Dec 9 '18 at 20:45
ClaireClaire
606
asked Dec 9 '18 at 20:45
ClaireClaire
606
asked Dec 9 '18 at 20:45
ClaireClaire
606
asked Dec 9 '18 at 20:45
ClaireClaire
606
606
marked as duplicate by GNUSupporter 8964民主女神 地下教會, Bill Dubuque modular-arithmetic
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Dec 9 '18 at 21:22
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Dec 9 '18 at 21:22
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.
1
$begingroup$
What are your thoughts on the question ?
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:47
$begingroup$
25 times 4 = 100, then what?
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 20:48
add a comment |
1
$begingroup$
What are your thoughts on the question ?
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:47
$begingroup$
25 times 4 = 100, then what?
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 20:48
1
1
$begingroup$
What are your thoughts on the question ?
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:47
$begingroup$
What are your thoughts on the question ?
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:47
$begingroup$
25 times 4 = 100, then what?
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 20:48
$begingroup$
25 times 4 = 100, then what?
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 20:48
add a comment |
2 Answers
2
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oldest
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Your idea is good. If we can show that $10^nequiv0$ modulo $4$ for "most" $n$ ($n>1$), then the problem is done already. Now notice that $4times 25=100=10^2$, and $10^{n}=10^210^{n-2}$ for $ngeq 2$. Can you continue?
answered Dec 9 '18 at 20:49
YiFanYiFan
3,2591424
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$begingroup$
begin{align}
n&=(a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots + a_1times 10^2+(a_1 times 10) + a_0\
&= underbrace{(a_k times 10^{k-2}) + (a_{k-1} times 10^{k-3}) + cdots + a_1)times 10^2}_{m} + (a_1 times 10) + a_0
end{align}
so $;n=m+10a_1+a_0$, and $m$ is divisible by $4$. Can you proceed?
answered Dec 9 '18 at 20:53
BernardBernard
120k740114
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your idea is good. If we can show that $10^nequiv0$ modulo $4$ for "most" $n$ ($n>1$), then the problem is done already. Now notice that $4times 25=100=10^2$, and $10^{n}=10^210^{n-2}$ for $ngeq 2$. Can you continue?
answered Dec 9 '18 at 20:49
YiFanYiFan
3,2591424
$endgroup$
add a comment |
$begingroup$
Your idea is good. If we can show that $10^nequiv0$ modulo $4$ for "most" $n$ ($n>1$), then the problem is done already. Now notice that $4times 25=100=10^2$, and $10^{n}=10^210^{n-2}$ for $ngeq 2$. Can you continue?
answered Dec 9 '18 at 20:49
YiFanYiFan
3,2591424
$endgroup$
add a comment |
$begingroup$
Your idea is good. If we can show that $10^nequiv0$ modulo $4$ for "most" $n$ ($n>1$), then the problem is done already. Now notice that $4times 25=100=10^2$, and $10^{n}=10^210^{n-2}$ for $ngeq 2$. Can you continue?
answered Dec 9 '18 at 20:49
YiFanYiFan
3,2591424
$endgroup$
Your idea is good. If we can show that $10^nequiv0$ modulo $4$ for "most" $n$ ($n>1$), then the problem is done already. Now notice that $4times 25=100=10^2$, and $10^{n}=10^210^{n-2}$ for $ngeq 2$. Can you continue?
answered Dec 9 '18 at 20:49
YiFanYiFan
3,2591424
answered Dec 9 '18 at 20:49
YiFanYiFan
3,2591424
answered Dec 9 '18 at 20:49
YiFanYiFan
3,2591424
answered Dec 9 '18 at 20:49
YiFanYiFan
3,2591424
3,2591424
add a comment |
add a comment |
$begingroup$
begin{align}
n&=(a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots + a_1times 10^2+(a_1 times 10) + a_0\
&= underbrace{(a_k times 10^{k-2}) + (a_{k-1} times 10^{k-3}) + cdots + a_1)times 10^2}_{m} + (a_1 times 10) + a_0
end{align}
so $;n=m+10a_1+a_0$, and $m$ is divisible by $4$. Can you proceed?
answered Dec 9 '18 at 20:53
BernardBernard
120k740114
$endgroup$
add a comment |
$begingroup$
begin{align}
n&=(a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots + a_1times 10^2+(a_1 times 10) + a_0\
&= underbrace{(a_k times 10^{k-2}) + (a_{k-1} times 10^{k-3}) + cdots + a_1)times 10^2}_{m} + (a_1 times 10) + a_0
end{align}
so $;n=m+10a_1+a_0$, and $m$ is divisible by $4$. Can you proceed?
answered Dec 9 '18 at 20:53
BernardBernard
120k740114
$endgroup$
add a comment |
$begingroup$
begin{align}
n&=(a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots + a_1times 10^2+(a_1 times 10) + a_0\
&= underbrace{(a_k times 10^{k-2}) + (a_{k-1} times 10^{k-3}) + cdots + a_1)times 10^2}_{m} + (a_1 times 10) + a_0
end{align}
so $;n=m+10a_1+a_0$, and $m$ is divisible by $4$. Can you proceed?
answered Dec 9 '18 at 20:53
BernardBernard
120k740114
$endgroup$
begin{align}
n&=(a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots + a_1times 10^2+(a_1 times 10) + a_0\
&= underbrace{(a_k times 10^{k-2}) + (a_{k-1} times 10^{k-3}) + cdots + a_1)times 10^2}_{m} + (a_1 times 10) + a_0
end{align}
so $;n=m+10a_1+a_0$, and $m$ is divisible by $4$. Can you proceed?
answered Dec 9 '18 at 20:53
BernardBernard
120k740114
answered Dec 9 '18 at 20:53
BernardBernard
120k740114
answered Dec 9 '18 at 20:53
BernardBernard
120k740114
answered Dec 9 '18 at 20:53
BernardBernard
120k740114
120k740114
add a comment |
add a comment |
1
$begingroup$
What are your thoughts on the question ?
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:47
$begingroup$
25 times 4 = 100, then what?
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 20:48