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Mathematical Reasoning and Divisibility

   

Added on  2020-03-28

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Running head: MATHEMATICAL REASONING Mathematical ReasoningName Institutional Affiliation
Mathematical Reasoning and Divisibility_1

2MATHEMATICAL REASONING Question 1Suppose n=¿. Show that 3n3(dk+dk1+...+d1+d0).For which natural number ¿1other than 3 does this trick work?Proof:We first need to show that indeed the trick works for n=¿Suppose that 3n, where n=¿, and expressing ¿With this notation ¿ and taking base 10 into consideration ¿, 10i can be expressed as ¿ that is10i=¿. Hencedkx10k+dk1x10k1+dk2x10k2+dk3x10k3+...+d3x103+d2x102+d1x101+d0=¿dkx¿¿¿{¿¿{¿¿Since 3¿. Therefore 3n{¿¿. {HINT: if 3|n and 3|k, then 3|n-k}But n={¿¿Implying3{¿¿
Mathematical Reasoning and Divisibility_2

3MATHEMATICAL REASONING That is 3{dk+dk1+dk2+dk3+...d3+d2+d1+d0}Conversely, suppose that3{dk+dk1+dk2+dk3+...d3+d2+d1+d0}. Then 3{¿¿. That is 3¿nHINT: if 3|q and 3|k, then 3|q+kIt can easily be seen the same trick above works when 3 is replaced by 9. That isSince 9¿. Therefore 9n{¿¿. But n={¿¿Implying9{¿¿That is 9{dk+dk1+dk2+dk3+...d3+d2+d1+d0}Conversely, suppose that9{dk+dk1+dk2+dk3+...d3+d2+d1+d0}. Then 9{¿¿. That is 9nThus the same trick works for 9Question 2
Mathematical Reasoning and Divisibility_3

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