1 / 4

Question 4 – p92

Question 4 – p92. X 2 1 (mod p) for p >= 3 be prime X 2 – 1 0 (mod p) (x – 1) (x + 1) 0 (mod p) Hence, either (x – 1) 0 (mod p) or (x + 1) 0 (mod p) Equivalent to, x 1 (mod p) or x -1(mod p). Question 6 – p92. Let x = the total number of people,

sitara
Télécharger la présentation

Question 4 – p92

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Question 4 – p92 X2 1 (mod p) for p >= 3 be prime X2 – 1 0 (mod p) (x – 1) (x + 1) 0 (mod p) Hence, either (x – 1) 0 (mod p) or (x + 1) 0 (mod p) Equivalent to, x 1 (mod p) or x -1(mod p)

  2. Question 6 – p92 Let x = the total number of people, Line up 3 rows, 1 left over : x 1 (mod 3) Line up 4 rows, 2 left over : x 2 (mod 4) Line up 5 rows, 3 left over : x 3 (mod 5) Find the smallest x that satisfy 3 conditions.

  3. Question 6 – p92 x 1 (mod 3) Let x = 3n + 1 (for some n is positive integer) x 2 (mod 4) 3n + 1 2 (mod 4) 3n 1 (mod 4) n 3 (mod 4) n = 4m + 3 (for some m is positive integer)

  4. Question 6 – p92 Now, x = 12m + 10 x 3 (mod 5) 12m + 10 3 (mod 5) 12m -7 -2 3 (mod 5) 4m 1 (mod 5) m 4 (mod 5) m = 5o + 4 (for some o is positive integer) Hence x = 60o + 58 Possible x : 58, 118, 178, 238, 298, 358 ….. Notice : 12 = lcm(3,4) 60 = lcm(3,4,5)

More Related