Articles
Register
Sign In
Search
morgane41
@morgane41
January 2021
1
64
Report
bonjour, voici mon dm de spé maths, pouvez vous m'aider ?
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms and service
You must agree before submitting.
Send
Lista de comentários
scoladan
Verified answer
Bonjour,
1) Autant de jetons que de cases. Soit N ce nombre. Le damier a n lignes et n colonnes, donc n² cases.
On peut donc poser N = n²
2) N = n² = 5q + r avec 0 ≤ r < 5
3) Cf. ci-joint
On peut conjecturer que les restes possibles de la division de N par 5 sont 0, 1 ou 4.
4)
a) n = 5p ⇒ N = 25p² = 5 x 5p² + 0 ⇒ r = 0
b) n = 5p + 1
⇒ N = (5p + 1)² = 25p² + 10p + 1 = 5(5p² + 2) + 1 ⇒ r = 1
c) n = 5p + 2
⇒ N = 25p² + 20p + 4 = 5(5p² + 4p) + 4 ⇒ r = 4
d) n = 5p + 3
⇒ N = 25p² + 30p + 9 = 5(5p² + 6p + 1) + 4 ⇒ r = 4
e) n = 5p + 4
⇒ N = 25p² + 40p + 16 = 5(5p² + 8p + 3) + 1 ⇒ r = 1
1 votes
Thanks 0
More Questions From This User
See All
morgane41
January 2021 | 0 Respostas
Responda
morgane41
January 2021 | 0 Respostas
j'ai besoin d'aide pour l'exercice 37 (Sans justification)
Responda
morgane41
January 2021 | 0 Respostas
Responda
morgane41
January 2021 | 0 Respostas
Responda
morgane41
January 2021 | 0 Respostas
Responda
morgane41
January 2021 | 0 Respostas
Responda
Morgane41
May 2019 | 0 Respostas
Responda
Morgane41
May 2019 | 0 Respostas
Responda
×
Report "bonjour, voici mon dm de spé maths, pouvez vous m'aider ?.... Pergunta de ideia de morgane41"
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
Sobre nós
Política de Privacidade
Termos e Condições
direito autoral
Contate-Nos
Helpful Social
Get monthly updates
Submit
Copyright © 2024 ELIBRARY.TIPS - All rights reserved.
Lista de comentários
Verified answer
Bonjour,1) Autant de jetons que de cases. Soit N ce nombre. Le damier a n lignes et n colonnes, donc n² cases.
On peut donc poser N = n²
2) N = n² = 5q + r avec 0 ≤ r < 5
3) Cf. ci-joint
On peut conjecturer que les restes possibles de la division de N par 5 sont 0, 1 ou 4.
4)
a) n = 5p ⇒ N = 25p² = 5 x 5p² + 0 ⇒ r = 0
b) n = 5p + 1
⇒ N = (5p + 1)² = 25p² + 10p + 1 = 5(5p² + 2) + 1 ⇒ r = 1
c) n = 5p + 2
⇒ N = 25p² + 20p + 4 = 5(5p² + 4p) + 4 ⇒ r = 4
d) n = 5p + 3
⇒ N = 25p² + 30p + 9 = 5(5p² + 6p + 1) + 4 ⇒ r = 4
e) n = 5p + 4
⇒ N = 25p² + 40p + 16 = 5(5p² + 8p + 3) + 1 ⇒ r = 1