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miercuri, 4 noiembrie 2020

A DIVISIBILITY of TWO AFFINE EXPRESSIONS - Problem S:E19.221

          In "Suplimentul ... " page 7

          English translation from Romanian, thanks to Miss Google:

           "Determine the natural numbers n for which 2n-3 divides 5n+2."

 

ANSWER CiPn=2 and n=11

 Solution CiP

          If 2n-3 \mid 5n+2, then 2n-3 \mid 5n+2-2(2n-3) \Leftrightarrow

(1)                             2n-3 \mid n+8.

From (1) we obtain 2n-3 \mid 2n-3-(n-11), so

(2)                            2n-3 \mid n-11.

If we subtract (1)-(2), we conclude 2n-3 \mid 19 so 2n-3=1 or 2n-3=19.
We find the answer, both values checking the condition.

\blacksquare

           Remark If the statement required n \in \mathbb{Z} instead of n \in \mathbb{N} the answer will be

n \in \{-8, 1, 2, 11\}.



Alternative solution
 
          2n-3 \mid 5n+2 \;\; \Leftrightarrow \;\; 5n+2=(2n-3) \cdot k      for some k \in \mathbb{Z}
(we consider the case more general n \in \mathbb{Z}). The last equation is successively equivalent to
2nk-5n-3k=2 \;\; \Leftrightarrow \;\;4nk-10n-6k=4\;\; \Leftrightarrow \;\;2n(2k-5)-3(2k-5)-15=4\;\; \Leftrightarrow \;\;(2k-5)(2n-3)=19
       The numbers dividing 19 are \pm1 and \pm 19 so 2n-3 \in \{\pm 1, \pm 19 \}. We get the answer again.
\blacksquare
 
 

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