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naeln
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Bunuel
naeln
I am on shaky grounds with remainder questions. For me, solving this type is always time consuming, so I was wondering whether there is an efficient method to solving this type of questions.

Theory on remainders problems: remainders-144665.html
Tips on remainders: remainders-tips-and-hints-175000.html

Units digits, exponents, remainders problems: new-units-digits-exponents-remainders-problems-168569.html

All DS remainders problems to practice: search.php?search_id=tag&tag_id=198
All PS remainders problems to practice: search.php?search_id=tag&tag_id=199

Hope this helps.

Thank you Bunuel for the links and the clear explanation. Truly appreciate it :-D
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There is 1 variable (m) in the original condition. In order to match the number of variables and the number of equations, we need 1 equation. Since con 1) and con 2) each has 1 equation, there is a high chance that D is the correct answer. We can substitute the other parts.
In case of con 1), if we divide m=28p+5=5, 33, 61,.... by 14, the remainder is always 5. The answer is unique and the condition is sufficient.
In case of con 2), the number that satisfies m=2q+1=7n+5 is 5, which is the remainder of dividing m by 14. The answer is unique and the condition is sufficient. Hence, the correct answer is D.

- For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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