Sujet 2013 - Mathématiques – Académie de Besançon

n= 24 2 + 4 = 6 24
6
364
n n
110 111 112
A= 30 ×31 ×32 ×33 A
98 208 030 98 208 031 98 208 032 98 208 033
i0 8
p i 9
p p + 2
p p + 2
22

300 160
ˆıˆr
ˆı= ˆr
[MN]
[P O]N
[P O]
[NO]
[NO]
[MN]

σN∗n
σ(21) = 32 21 1 3 7 21 1 + 3 + 7 + 21 = 32
p1p
2
72 = 23×32
n
1 + 2 + . . . +n=n(n+ 1)
2.
x6= 1 n
1 + x+x2+. . . +xn=xn+1 −1
x−1.
72 = 23×3272
1 ; 2 ; 22; 23; 3 ; 32; 2 ×3 ; 22×3 ; 23×3 ; 2 ×32; 22×32; 23×32.
350
σ(350)
σ(n)n

n
σ0
k . . . . . .
n k 0
σ . . .
σ
n k 0
σ
σ(n)n= 1 n= 2 n= 3 n= 4 n= 5 n= 6
p σ(p)
n>2σ(n)>n+ 1
n σ(n)6n(n+ 1)
2
n
28 28 28 1 2 4 7 14
28=1+2+4+7+14
n σ(n)=2n
10
n
•28 = 2n2n+1 −1
•2n+1 −1
2n2n+1 −1
n2n+1 −1
n2n+1 −1
2n2n+1 −1
n p 2
2np n p
n2n+1 −1
σ2n2n+1 −1n2n2n+1 −1
n p 2
2np p = 2n+1 −1
n σ(n)=2n−1
16
k2k