Mixing
Sum of digits of the multiple of 2003 [closed]
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Can you find the minimum value of $S(2003n)$, with $S(x)$ being the sum of digits in decimal form and n being any positive integer?
(For example, $S(2018)=2+0+1+8=11$ )
number-theory
asked Jul 21 at 14:26
Mathwriter
265
closed as off-topic by user223391, gammatester, Martin R, steven gregory, Batominovski Jul 21 at 14:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, gammatester, Martin R, steven gregory, Batominovski
add a comment |Â
up vote
0
down vote
favorite
Can you find the minimum value of $S(2003n)$, with $S(x)$ being the sum of digits in decimal form and n being any positive integer?
(For example, $S(2018)=2+0+1+8=11$ )
number-theory
asked Jul 21 at 14:26
Mathwriter
265
closed as off-topic by user223391, gammatester, Martin R, steven gregory, Batominovski Jul 21 at 14:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, gammatester, Martin R, steven gregory, Batominovski
Well, looking at the powers of $10pmod 2003$ I see that $10^301equiv -2pmod 2003$. That means that we can get the digit sum $3$. I do not believe that $2$ is possible (but haven't worked hard enough to confirm that). $1$ is clearly not possible.
– lulu
Jul 21 at 14:48
Do you mean to have S(98) = 17 or S(98) = 8?
– steven gregory
Jul 21 at 14:57
Is S(98)=17, sorry for my mistakes.
– Mathwriter
Jul 21 at 15:27
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can you find the minimum value of $S(2003n)$, with $S(x)$ being the sum of digits in decimal form and n being any positive integer?
(For example, $S(2018)=2+0+1+8=11$ )
number-theory
asked Jul 21 at 14:26
Mathwriter
265
Can you find the minimum value of $S(2003n)$, with $S(x)$ being the sum of digits in decimal form and n being any positive integer?
(For example, $S(2018)=2+0+1+8=11$ )
number-theory
asked Jul 21 at 14:26
Mathwriter
265
edited Jul 21 at 15:33
asked Jul 21 at 14:26
Mathwriter
265
asked Jul 21 at 14:26
Mathwriter
265
asked Jul 21 at 14:26
Mathwriter
265
265
closed as off-topic by user223391, gammatester, Martin R, steven gregory, Batominovski Jul 21 at 14:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, gammatester, Martin R, steven gregory, Batominovski
closed as off-topic by user223391, gammatester, Martin R, steven gregory, Batominovski Jul 21 at 14:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, gammatester, Martin R, steven gregory, Batominovski
Well, looking at the powers of $10pmod 2003$ I see that $10^301equiv -2pmod 2003$. That means that we can get the digit sum $3$. I do not believe that $2$ is possible (but haven't worked hard enough to confirm that). $1$ is clearly not possible.
– lulu
Jul 21 at 14:48
Do you mean to have S(98) = 17 or S(98) = 8?
– steven gregory
Jul 21 at 14:57
Is S(98)=17, sorry for my mistakes.
– Mathwriter
Jul 21 at 15:27
add a comment |Â
Well, looking at the powers of $10pmod 2003$ I see that $10^301equiv -2pmod 2003$. That means that we can get the digit sum $3$. I do not believe that $2$ is possible (but haven't worked hard enough to confirm that). $1$ is clearly not possible.
– lulu
Jul 21 at 14:48
Do you mean to have S(98) = 17 or S(98) = 8?
– steven gregory
Jul 21 at 14:57
Is S(98)=17, sorry for my mistakes.
– Mathwriter
Jul 21 at 15:27
Well, looking at the powers of $10pmod 2003$ I see that $10^301equiv -2pmod 2003$. That means that we can get the digit sum $3$. I do not believe that $2$ is possible (but haven't worked hard enough to confirm that). $1$ is clearly not possible.
– lulu
Jul 21 at 14:48
Well, looking at the powers of $10pmod 2003$ I see that $10^301equiv -2pmod 2003$. That means that we can get the digit sum $3$. I do not believe that $2$ is possible (but haven't worked hard enough to confirm that). $1$ is clearly not possible.
– lulu
Jul 21 at 14:48
Do you mean to have S(98) = 17 or S(98) = 8?
– steven gregory
Jul 21 at 14:57
Do you mean to have S(98) = 17 or S(98) = 8?
– steven gregory
Jul 21 at 14:57
Is S(98)=17, sorry for my mistakes.
– Mathwriter
Jul 21 at 15:27
Is S(98)=17, sorry for my mistakes.
– Mathwriter
Jul 21 at 15:27
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Direct computation shows that the order of $10pmod 2003$ is $1001$. Note that this is odd.
Claim $1$: $1$ is impossible.
Pf: The only numbers with digit sum $1$ are powers of $10$ and none of those are divisible by $2003$.
Claim $2$: $2$ is impossible.
Pf: The only numbers with digit sum $2$ are those of the form $10^a(10^b+1)$ for $a,b≥0$. If $b=0$, we have the numbers of the form $2times 10^n$, none of which are divisible by $2003$, so we may assume that $b>0$. In order for such a number to be divisible by $2003$ we'd need $10^bequiv -1pmod 2003$. But this would imply that $10^2bequiv 1pmod 2003$, whence that $1001,|,2b$ which implies that $1001,|,b$. Taking a minimal possible $b$ shows that this is impossible.
Claim $3$: $3$ is possible
Pf: By direct computation, $10^301equiv -2pmod 2003$.
Hence $3$ is the minimum.
answered Jul 21 at 14:59
lulu
35.2k14172
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Direct computation shows that the order of $10pmod 2003$ is $1001$. Note that this is odd.
Claim $1$: $1$ is impossible.
Pf: The only numbers with digit sum $1$ are powers of $10$ and none of those are divisible by $2003$.
Claim $2$: $2$ is impossible.
Pf: The only numbers with digit sum $2$ are those of the form $10^a(10^b+1)$ for $a,b≥0$. If $b=0$, we have the numbers of the form $2times 10^n$, none of which are divisible by $2003$, so we may assume that $b>0$. In order for such a number to be divisible by $2003$ we'd need $10^bequiv -1pmod 2003$. But this would imply that $10^2bequiv 1pmod 2003$, whence that $1001,|,2b$ which implies that $1001,|,b$. Taking a minimal possible $b$ shows that this is impossible.
Claim $3$: $3$ is possible
Pf: By direct computation, $10^301equiv -2pmod 2003$.
Hence $3$ is the minimum.
answered Jul 21 at 14:59
lulu
35.2k14172
add a comment |Â
up vote
4
down vote
accepted
Direct computation shows that the order of $10pmod 2003$ is $1001$. Note that this is odd.
Claim $1$: $1$ is impossible.
Pf: The only numbers with digit sum $1$ are powers of $10$ and none of those are divisible by $2003$.
Claim $2$: $2$ is impossible.
Pf: The only numbers with digit sum $2$ are those of the form $10^a(10^b+1)$ for $a,b≥0$. If $b=0$, we have the numbers of the form $2times 10^n$, none of which are divisible by $2003$, so we may assume that $b>0$. In order for such a number to be divisible by $2003$ we'd need $10^bequiv -1pmod 2003$. But this would imply that $10^2bequiv 1pmod 2003$, whence that $1001,|,2b$ which implies that $1001,|,b$. Taking a minimal possible $b$ shows that this is impossible.
Claim $3$: $3$ is possible
Pf: By direct computation, $10^301equiv -2pmod 2003$.
Hence $3$ is the minimum.
answered Jul 21 at 14:59
lulu
35.2k14172
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Direct computation shows that the order of $10pmod 2003$ is $1001$. Note that this is odd.
Claim $1$: $1$ is impossible.
Pf: The only numbers with digit sum $1$ are powers of $10$ and none of those are divisible by $2003$.
Claim $2$: $2$ is impossible.
Pf: The only numbers with digit sum $2$ are those of the form $10^a(10^b+1)$ for $a,b≥0$. If $b=0$, we have the numbers of the form $2times 10^n$, none of which are divisible by $2003$, so we may assume that $b>0$. In order for such a number to be divisible by $2003$ we'd need $10^bequiv -1pmod 2003$. But this would imply that $10^2bequiv 1pmod 2003$, whence that $1001,|,2b$ which implies that $1001,|,b$. Taking a minimal possible $b$ shows that this is impossible.
Claim $3$: $3$ is possible
Pf: By direct computation, $10^301equiv -2pmod 2003$.
Hence $3$ is the minimum.
answered Jul 21 at 14:59
lulu
35.2k14172
Direct computation shows that the order of $10pmod 2003$ is $1001$. Note that this is odd.
Claim $1$: $1$ is impossible.
Pf: The only numbers with digit sum $1$ are powers of $10$ and none of those are divisible by $2003$.
Claim $2$: $2$ is impossible.
Pf: The only numbers with digit sum $2$ are those of the form $10^a(10^b+1)$ for $a,b≥0$. If $b=0$, we have the numbers of the form $2times 10^n$, none of which are divisible by $2003$, so we may assume that $b>0$. In order for such a number to be divisible by $2003$ we'd need $10^bequiv -1pmod 2003$. But this would imply that $10^2bequiv 1pmod 2003$, whence that $1001,|,2b$ which implies that $1001,|,b$. Taking a minimal possible $b$ shows that this is impossible.
Claim $3$: $3$ is possible
Pf: By direct computation, $10^301equiv -2pmod 2003$.
Hence $3$ is the minimum.
answered Jul 21 at 14:59
lulu
35.2k14172
edited Jul 21 at 15:04
answered Jul 21 at 14:59
lulu
35.2k14172
answered Jul 21 at 14:59
lulu
35.2k14172
answered Jul 21 at 14:59
lulu
35.2k14172
35.2k14172
add a comment |Â
add a comment |Â
Well, looking at the powers of $10pmod 2003$ I see that $10^301equiv -2pmod 2003$. That means that we can get the digit sum $3$. I do not believe that $2$ is possible (but haven't worked hard enough to confirm that). $1$ is clearly not possible.
– lulu
Jul 21 at 14:48
Do you mean to have S(98) = 17 or S(98) = 8?
– steven gregory
Jul 21 at 14:57
Is S(98)=17, sorry for my mistakes.
– Mathwriter
Jul 21 at 15:27