Prove $ n equiv s(n) ($mod$ 3)$ using the fact that $ [10^n] = [1]$. [duplicate]
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Why is $9$ special in testing divisiblity by $9$ by summing decimal digits? (casting out nines)
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Prove $ n equiv s(n) (mod 3)$ using the fact that $ [10^n] = [1]$. Let $n = (a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots +(a_1 times 10^1)+ (a_0 times 10^0)$ and $s(n)=(a_k + a_{k-1}+ cdots +a_1+a_0)$.
When trying to solve this question, I combined the information given and found that $$n-s(n) = (a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots +(a_1 times 10^1)+ (a_0 times 10^0)-(a_k + a_{k-1}+ cdots +a_1+a_0)$$
$$=(a_k times 10^k-a_k ) + (a_{k-1} times 10^{k-1}-a_{k-1}) + cdots +(a_1 times 10^1-a_1)+ (a_0 times 10^0-a_0)$$
$$=a_k (10^k-1) + a_{k-1} (10^{k-1}-1) + cdots +a_1 (10^1-1)+ a_0 (1-1)$$
$$=a_k (10^k-1) + a_{k-1} (10^{k-1}-1) + cdots +a_1 (9)$$
I'm not sure where to go from here, I thought maybe I could deduce that since, for the case of $ [10^n] = [1]$ -- since that means that $10^n equiv 1 (mod 3)$ -- I could say that since there exists an integer, call it p, such that $10^n - 1=3p$, and then put that "statement" in the parentheses in the last "=" line that I had above. I have a feeling it doesn't make sense though, and it would be incorrect.
modular-arithmetic
asked Dec 8 '18 at 18:48
ClaireClaire
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Dec 8 '18 at 21:26
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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This question already has an answer here:
Why is $9$ special in testing divisiblity by $9$ by summing decimal digits? (casting out nines)
6 answers
Prove $ n equiv s(n) (mod 3)$ using the fact that $ [10^n] = [1]$. Let $n = (a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots +(a_1 times 10^1)+ (a_0 times 10^0)$ and $s(n)=(a_k + a_{k-1}+ cdots +a_1+a_0)$.
When trying to solve this question, I combined the information given and found that $$n-s(n) = (a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots +(a_1 times 10^1)+ (a_0 times 10^0)-(a_k + a_{k-1}+ cdots +a_1+a_0)$$
$$=(a_k times 10^k-a_k ) + (a_{k-1} times 10^{k-1}-a_{k-1}) + cdots +(a_1 times 10^1-a_1)+ (a_0 times 10^0-a_0)$$
$$=a_k (10^k-1) + a_{k-1} (10^{k-1}-1) + cdots +a_1 (10^1-1)+ a_0 (1-1)$$
$$=a_k (10^k-1) + a_{k-1} (10^{k-1}-1) + cdots +a_1 (9)$$
I'm not sure where to go from here, I thought maybe I could deduce that since, for the case of $ [10^n] = [1]$ -- since that means that $10^n equiv 1 (mod 3)$ -- I could say that since there exists an integer, call it p, such that $10^n - 1=3p$, and then put that "statement" in the parentheses in the last "=" line that I had above. I have a feeling it doesn't make sense though, and it would be incorrect.
modular-arithmetic
asked Dec 8 '18 at 18:48
ClaireClaire
556
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Dec 8 '18 at 21:26
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This question already has an answer here:
Why is $9$ special in testing divisiblity by $9$ by summing decimal digits? (casting out nines)
6 answers
Prove $ n equiv s(n) (mod 3)$ using the fact that $ [10^n] = [1]$. Let $n = (a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots +(a_1 times 10^1)+ (a_0 times 10^0)$ and $s(n)=(a_k + a_{k-1}+ cdots +a_1+a_0)$.
When trying to solve this question, I combined the information given and found that $$n-s(n) = (a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots +(a_1 times 10^1)+ (a_0 times 10^0)-(a_k + a_{k-1}+ cdots +a_1+a_0)$$
$$=(a_k times 10^k-a_k ) + (a_{k-1} times 10^{k-1}-a_{k-1}) + cdots +(a_1 times 10^1-a_1)+ (a_0 times 10^0-a_0)$$
$$=a_k (10^k-1) + a_{k-1} (10^{k-1}-1) + cdots +a_1 (10^1-1)+ a_0 (1-1)$$
$$=a_k (10^k-1) + a_{k-1} (10^{k-1}-1) + cdots +a_1 (9)$$
I'm not sure where to go from here, I thought maybe I could deduce that since, for the case of $ [10^n] = [1]$ -- since that means that $10^n equiv 1 (mod 3)$ -- I could say that since there exists an integer, call it p, such that $10^n - 1=3p$, and then put that "statement" in the parentheses in the last "=" line that I had above. I have a feeling it doesn't make sense though, and it would be incorrect.
modular-arithmetic
asked Dec 8 '18 at 18:48
ClaireClaire
556
$endgroup$
This question already has an answer here:
Why is $9$ special in testing divisiblity by $9$ by summing decimal digits? (casting out nines)
6 answers
Prove $ n equiv s(n) (mod 3)$ using the fact that $ [10^n] = [1]$. Let $n = (a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots +(a_1 times 10^1)+ (a_0 times 10^0)$ and $s(n)=(a_k + a_{k-1}+ cdots +a_1+a_0)$.
When trying to solve this question, I combined the information given and found that $$n-s(n) = (a_k times 10^k) + (a_{k-1} times 10^{k-1}) + cdots +(a_1 times 10^1)+ (a_0 times 10^0)-(a_k + a_{k-1}+ cdots +a_1+a_0)$$
$$=(a_k times 10^k-a_k ) + (a_{k-1} times 10^{k-1}-a_{k-1}) + cdots +(a_1 times 10^1-a_1)+ (a_0 times 10^0-a_0)$$
$$=a_k (10^k-1) + a_{k-1} (10^{k-1}-1) + cdots +a_1 (10^1-1)+ a_0 (1-1)$$
$$=a_k (10^k-1) + a_{k-1} (10^{k-1}-1) + cdots +a_1 (9)$$
I'm not sure where to go from here, I thought maybe I could deduce that since, for the case of $ [10^n] = [1]$ -- since that means that $10^n equiv 1 (mod 3)$ -- I could say that since there exists an integer, call it p, such that $10^n - 1=3p$, and then put that "statement" in the parentheses in the last "=" line that I had above. I have a feeling it doesn't make sense though, and it would be incorrect.
This question already has an answer here:
Why is $9$ special in testing divisiblity by $9$ by summing decimal digits? (casting out nines)
6 answers
modular-arithmetic
modular-arithmetic
asked Dec 8 '18 at 18:48
ClaireClaire
556
asked Dec 8 '18 at 18:48
ClaireClaire
556
edited Dec 8 '18 at 20:08
Claire
asked Dec 8 '18 at 18:48
ClaireClaire
556
asked Dec 8 '18 at 18:48
ClaireClaire
556
asked Dec 8 '18 at 18:48
ClaireClaire
556
556
marked as duplicate by Bill Dubuque modular-arithmetic
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Dec 8 '18 at 21:26
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1 Answer
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Hint: Let $$z_n=a_n10^n+a_{n-1}10^{n-1}+...+a_1cdot 10+a_0$$ with your hint we get
$$z_nequiv a_n+a_{n-1}+...+a_1+a_0mod 3$$ since $$10^iequiv 1mod 3$$
answered Dec 8 '18 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
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Okay, that makes sense to me now.
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– Claire
Dec 8 '18 at 19:22
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Do you know how I would go about showing that $3|n$ if and only if $3|s(n)$?
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– Claire
Dec 8 '18 at 19:22
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1 Answer
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1 Answer
1
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active
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active
oldest
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$begingroup$
Hint: Let $$z_n=a_n10^n+a_{n-1}10^{n-1}+...+a_1cdot 10+a_0$$ with your hint we get
$$z_nequiv a_n+a_{n-1}+...+a_1+a_0mod 3$$ since $$10^iequiv 1mod 3$$
answered Dec 8 '18 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
$endgroup$
$begingroup$
Okay, that makes sense to me now.
$endgroup$
– Claire
Dec 8 '18 at 19:22
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Do you know how I would go about showing that $3|n$ if and only if $3|s(n)$?
$endgroup$
– Claire
Dec 8 '18 at 19:22
add a comment |
$begingroup$
Hint: Let $$z_n=a_n10^n+a_{n-1}10^{n-1}+...+a_1cdot 10+a_0$$ with your hint we get
$$z_nequiv a_n+a_{n-1}+...+a_1+a_0mod 3$$ since $$10^iequiv 1mod 3$$
answered Dec 8 '18 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
$endgroup$
$begingroup$
Okay, that makes sense to me now.
$endgroup$
– Claire
Dec 8 '18 at 19:22
$begingroup$
Do you know how I would go about showing that $3|n$ if and only if $3|s(n)$?
$endgroup$
– Claire
Dec 8 '18 at 19:22
add a comment |
$begingroup$
Hint: Let $$z_n=a_n10^n+a_{n-1}10^{n-1}+...+a_1cdot 10+a_0$$ with your hint we get
$$z_nequiv a_n+a_{n-1}+...+a_1+a_0mod 3$$ since $$10^iequiv 1mod 3$$
answered Dec 8 '18 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
$endgroup$
Hint: Let $$z_n=a_n10^n+a_{n-1}10^{n-1}+...+a_1cdot 10+a_0$$ with your hint we get
$$z_nequiv a_n+a_{n-1}+...+a_1+a_0mod 3$$ since $$10^iequiv 1mod 3$$
answered Dec 8 '18 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
answered Dec 8 '18 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
answered Dec 8 '18 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
answered Dec 8 '18 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
74.8k42865
$begingroup$
Okay, that makes sense to me now.
$endgroup$
– Claire
Dec 8 '18 at 19:22
$begingroup$
Do you know how I would go about showing that $3|n$ if and only if $3|s(n)$?
$endgroup$
– Claire
Dec 8 '18 at 19:22
add a comment |
$begingroup$
Okay, that makes sense to me now.
$endgroup$
– Claire
Dec 8 '18 at 19:22
$begingroup$
Do you know how I would go about showing that $3|n$ if and only if $3|s(n)$?
$endgroup$
– Claire
Dec 8 '18 at 19:22
$begingroup$
Okay, that makes sense to me now.
$endgroup$
– Claire
Dec 8 '18 at 19:22
$begingroup$
Okay, that makes sense to me now.
$endgroup$
– Claire
Dec 8 '18 at 19:22
$begingroup$
Do you know how I would go about showing that $3|n$ if and only if $3|s(n)$?
$endgroup$
– Claire
Dec 8 '18 at 19:22
$begingroup$
Do you know how I would go about showing that $3|n$ if and only if $3|s(n)$?
$endgroup$
– Claire
Dec 8 '18 at 19:22
add a comment |
