$3×7×11×dots×2003$. Find last three digits of product [closed]

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What are the last three digits of $3×7×11×dots×2003$?




$$prod_k=0^500(4k+3)bmod1000$$
I have tried a lot and spent a lot of time to but don't seem to be getting anywhere. I tried creating groups for solving separately but that didn't help.




number-theory contest-math




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closed as off-topic by Alex Francisco, Adrian Keister, Isaac Browne, Trần Thúc Minh Trí, Claude Leibovici Jul 17 at 9:56


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." – Alex Francisco, Adrian Keister, Trần Thúc Minh Trí, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Please use the body of your Question to give a full statement of the problem you want help with. The title alone is scarcely adequate to give the problem statement, and here the setup of the problem is lacking. What are the integers being multiplied? If they are exactly of the form $4k+3$, you should say that.
    – hardmath
    Jul 16 at 17:08










  • Perhaps you were able to get the units' place digit? The more evidence of your own efforts, the more accurately a Reader will be able to respond.
    – hardmath
    Jul 16 at 17:10














up vote
0
down vote

favorite
1













What are the last three digits of $3×7×11×dots×2003$?




$$prod_k=0^500(4k+3)bmod1000$$
I have tried a lot and spent a lot of time to but don't seem to be getting anywhere. I tried creating groups for solving separately but that didn't help.




number-theory contest-math




share|cite|improve this question













closed as off-topic by Alex Francisco, Adrian Keister, Isaac Browne, Trần Thúc Minh Trí, Claude Leibovici Jul 17 at 9:56


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." – Alex Francisco, Adrian Keister, Trần Thúc Minh Trí, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Please use the body of your Question to give a full statement of the problem you want help with. The title alone is scarcely adequate to give the problem statement, and here the setup of the problem is lacking. What are the integers being multiplied? If they are exactly of the form $4k+3$, you should say that.
    – hardmath
    Jul 16 at 17:08










  • Perhaps you were able to get the units' place digit? The more evidence of your own efforts, the more accurately a Reader will be able to respond.
    – hardmath
    Jul 16 at 17:10












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1






What are the last three digits of $3×7×11×dots×2003$?




$$prod_k=0^500(4k+3)bmod1000$$
I have tried a lot and spent a lot of time to but don't seem to be getting anywhere. I tried creating groups for solving separately but that didn't help.




number-theory contest-math




share|cite|improve this question














What are the last three digits of $3×7×11×dots×2003$?




$$prod_k=0^500(4k+3)bmod1000$$
I have tried a lot and spent a lot of time to but don't seem to be getting anywhere. I tried creating groups for solving separately but that didn't help.





number-theory contest-math





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 16 at 17:14









Parcly Taxel

33.6k136588




33.6k136588









asked Jul 16 at 17:03









Harsh Katara

458




458




closed as off-topic by Alex Francisco, Adrian Keister, Isaac Browne, Trần Thúc Minh Trí, Claude Leibovici Jul 17 at 9:56


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." – Alex Francisco, Adrian Keister, Trần Thúc Minh Trí, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Alex Francisco, Adrian Keister, Isaac Browne, Trần Thúc Minh Trí, Claude Leibovici Jul 17 at 9:56


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." – Alex Francisco, Adrian Keister, Trần Thúc Minh Trí, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.











  • Please use the body of your Question to give a full statement of the problem you want help with. The title alone is scarcely adequate to give the problem statement, and here the setup of the problem is lacking. What are the integers being multiplied? If they are exactly of the form $4k+3$, you should say that.
    – hardmath
    Jul 16 at 17:08










  • Perhaps you were able to get the units' place digit? The more evidence of your own efforts, the more accurately a Reader will be able to respond.
    – hardmath
    Jul 16 at 17:10
















  • Please use the body of your Question to give a full statement of the problem you want help with. The title alone is scarcely adequate to give the problem statement, and here the setup of the problem is lacking. What are the integers being multiplied? If they are exactly of the form $4k+3$, you should say that.
    – hardmath
    Jul 16 at 17:08










  • Perhaps you were able to get the units' place digit? The more evidence of your own efforts, the more accurately a Reader will be able to respond.
    – hardmath
    Jul 16 at 17:10















Please use the body of your Question to give a full statement of the problem you want help with. The title alone is scarcely adequate to give the problem statement, and here the setup of the problem is lacking. What are the integers being multiplied? If they are exactly of the form $4k+3$, you should say that.
– hardmath
Jul 16 at 17:08




Please use the body of your Question to give a full statement of the problem you want help with. The title alone is scarcely adequate to give the problem statement, and here the setup of the problem is lacking. What are the integers being multiplied? If they are exactly of the form $4k+3$, you should say that.
– hardmath
Jul 16 at 17:08












Perhaps you were able to get the units' place digit? The more evidence of your own efforts, the more accurately a Reader will be able to respond.
– hardmath
Jul 16 at 17:10




Perhaps you were able to get the units' place digit? The more evidence of your own efforts, the more accurately a Reader will be able to respond.
– hardmath
Jul 16 at 17:10










1 Answer
1






active

oldest

votes

















up vote
6
down vote













Compute modulo $125$ and $8$ separately and then use the Chinese Remainder Theorem.



Modulo $125$ is quickly taken care of by $k=3, 8, 13$.



And there are a lot of repeated factors modulo $8$.






share|cite|improve this answer





















  • It has some simpler solution as Chinese Remainder Theorem is not a requirement for this question.
    – Harsh Katara
    Jul 16 at 17:12










  • @HarshKatara: Which kind of contest specifies that a so basic tool as CRT is not allowed?
    – Henning Makholm
    Jul 16 at 17:14










  • Its not the contest but its the exercise but no problems as it is a valid solution.
    – Harsh Katara
    Jul 16 at 17:16










  • @HarshKatara: The edit log for the question looks a lot like you added the contest-math tag to the question originally.
    – Henning Makholm
    Jul 16 at 17:17











  • Yes this question is for contest math but is from a book known as excursion in mathematics and the exercise doesn't tell us about CRT.
    – Harsh Katara
    Jul 16 at 17:19

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
6
down vote













Compute modulo $125$ and $8$ separately and then use the Chinese Remainder Theorem.



Modulo $125$ is quickly taken care of by $k=3, 8, 13$.



And there are a lot of repeated factors modulo $8$.






share|cite|improve this answer





















  • It has some simpler solution as Chinese Remainder Theorem is not a requirement for this question.
    – Harsh Katara
    Jul 16 at 17:12










  • @HarshKatara: Which kind of contest specifies that a so basic tool as CRT is not allowed?
    – Henning Makholm
    Jul 16 at 17:14










  • Its not the contest but its the exercise but no problems as it is a valid solution.
    – Harsh Katara
    Jul 16 at 17:16










  • @HarshKatara: The edit log for the question looks a lot like you added the contest-math tag to the question originally.
    – Henning Makholm
    Jul 16 at 17:17











  • Yes this question is for contest math but is from a book known as excursion in mathematics and the exercise doesn't tell us about CRT.
    – Harsh Katara
    Jul 16 at 17:19














up vote
6
down vote













Compute modulo $125$ and $8$ separately and then use the Chinese Remainder Theorem.



Modulo $125$ is quickly taken care of by $k=3, 8, 13$.



And there are a lot of repeated factors modulo $8$.






share|cite|improve this answer





















  • It has some simpler solution as Chinese Remainder Theorem is not a requirement for this question.
    – Harsh Katara
    Jul 16 at 17:12










  • @HarshKatara: Which kind of contest specifies that a so basic tool as CRT is not allowed?
    – Henning Makholm
    Jul 16 at 17:14










  • Its not the contest but its the exercise but no problems as it is a valid solution.
    – Harsh Katara
    Jul 16 at 17:16










  • @HarshKatara: The edit log for the question looks a lot like you added the contest-math tag to the question originally.
    – Henning Makholm
    Jul 16 at 17:17











  • Yes this question is for contest math but is from a book known as excursion in mathematics and the exercise doesn't tell us about CRT.
    – Harsh Katara
    Jul 16 at 17:19












up vote
6
down vote










up vote
6
down vote









Compute modulo $125$ and $8$ separately and then use the Chinese Remainder Theorem.



Modulo $125$ is quickly taken care of by $k=3, 8, 13$.



And there are a lot of repeated factors modulo $8$.






share|cite|improve this answer













Compute modulo $125$ and $8$ separately and then use the Chinese Remainder Theorem.



Modulo $125$ is quickly taken care of by $k=3, 8, 13$.



And there are a lot of repeated factors modulo $8$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 16 at 17:09









Henning Makholm

226k16291520




226k16291520











  • It has some simpler solution as Chinese Remainder Theorem is not a requirement for this question.
    – Harsh Katara
    Jul 16 at 17:12










  • @HarshKatara: Which kind of contest specifies that a so basic tool as CRT is not allowed?
    – Henning Makholm
    Jul 16 at 17:14










  • Its not the contest but its the exercise but no problems as it is a valid solution.
    – Harsh Katara
    Jul 16 at 17:16










  • @HarshKatara: The edit log for the question looks a lot like you added the contest-math tag to the question originally.
    – Henning Makholm
    Jul 16 at 17:17











  • Yes this question is for contest math but is from a book known as excursion in mathematics and the exercise doesn't tell us about CRT.
    – Harsh Katara
    Jul 16 at 17:19
















  • It has some simpler solution as Chinese Remainder Theorem is not a requirement for this question.
    – Harsh Katara
    Jul 16 at 17:12










  • @HarshKatara: Which kind of contest specifies that a so basic tool as CRT is not allowed?
    – Henning Makholm
    Jul 16 at 17:14










  • Its not the contest but its the exercise but no problems as it is a valid solution.
    – Harsh Katara
    Jul 16 at 17:16










  • @HarshKatara: The edit log for the question looks a lot like you added the contest-math tag to the question originally.
    – Henning Makholm
    Jul 16 at 17:17











  • Yes this question is for contest math but is from a book known as excursion in mathematics and the exercise doesn't tell us about CRT.
    – Harsh Katara
    Jul 16 at 17:19















It has some simpler solution as Chinese Remainder Theorem is not a requirement for this question.
– Harsh Katara
Jul 16 at 17:12




It has some simpler solution as Chinese Remainder Theorem is not a requirement for this question.
– Harsh Katara
Jul 16 at 17:12












@HarshKatara: Which kind of contest specifies that a so basic tool as CRT is not allowed?
– Henning Makholm
Jul 16 at 17:14




@HarshKatara: Which kind of contest specifies that a so basic tool as CRT is not allowed?
– Henning Makholm
Jul 16 at 17:14












Its not the contest but its the exercise but no problems as it is a valid solution.
– Harsh Katara
Jul 16 at 17:16




Its not the contest but its the exercise but no problems as it is a valid solution.
– Harsh Katara
Jul 16 at 17:16












@HarshKatara: The edit log for the question looks a lot like you added the contest-math tag to the question originally.
– Henning Makholm
Jul 16 at 17:17





@HarshKatara: The edit log for the question looks a lot like you added the contest-math tag to the question originally.
– Henning Makholm
Jul 16 at 17:17













Yes this question is for contest math but is from a book known as excursion in mathematics and the exercise doesn't tell us about CRT.
– Harsh Katara
Jul 16 at 17:19




Yes this question is for contest math but is from a book known as excursion in mathematics and the exercise doesn't tell us about CRT.
– Harsh Katara
Jul 16 at 17:19


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