[PDF] Exercices bac -- 2011-2015 -- arithmétique E 1 E 2

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p q p q p-1≡1 (p) (un)ិ n u n=2n+3n+6n-1. n un n un (un) p

6×2p-2≡3 (p)6×3p-2≡2 (p)

6up-2≡0 (p)

11x-7y=5 xy

ិ (u;v)

11u-7v=1

(O;⃗ı,⃗ȷ) D

11x-7y-5=0 C M(x;y)

0⩽x⩽500⩽y⩽50

D C

11x2-7y2=5 xy

(x;y) x2≡2y2 xy x x2 y 2y2 x2 2y2 (x;y) xy xy (x;y) p a p ap-1≡ 1 (p) (un) ិ u

0=1 n,un+1=10un+21.

u1,u2u3 n,

3un=10n+1-7

n un u2 (un) n,un n, 3un≡4-(-1)n(11) n,un 1016≡1(17) k,u16k+8 ab (u;v) ិau+bv=1 a,b,c a bc ab ac pq pq a a≡0 [p]a≡0 [q] a≡0 [pq]

S nិ

n≡9 [17] n≡3 [5] S (u;v) 17u+5v=1 ិ (u;v) n0=3×17u+9×5v n0 S n0 S S n S n-n0≡0 [85] n S n=43+85kk a,b,c,d n (E) : 23x-26y=1,xy ិ (-9 ;-8) (E) (E) a 0⩽a⩽2523a≡1 (mod 26) (x1;x2)x1 x2 (x1;x2) (y1;y2)

S1)

y

1≡11x1+3x2(mod 26)

y

2≡7x1+4x2(mod 26)0⩽y1⩽250⩽y2⩽25.

(y1;y2) |{z} (x1;x2)ិ (S1) ិ

S2)

23x1≡4y1+23y2(mod 26)

23x2≡19y1+11y2(mod 26)

(x1;x2)ិ

S2) ិ

S3)

x

1≡16y1+y2(mod 26)

x

2≡11y1+5y2(mod 26)

(x1;x2)ិ (S3) ិ (S1)

25x-108y=1xy

ិ (13 ; 3) a cg ិ 25g-108c=1 p a p ap-1 p ap-1≡1 [p] x x≡a[7]x≡a[19] x≡a[133] a a6≡1 [7] a108≡1 [7] (a25)g≡a[7] (a25)g≡a[7] a(a25)g≡a[19] (a25)g≡a[133] a 1⩽a⩽26 a r a25≡r[133]

0⩽r<133

rr1r13≡r1[133]0⩽r1<133 ិ r1≡a[133]

128 59

a b c a b a⩾b ឹc ឹa a=13b=4 025
m

9m+526

p p ិ m ឹ p x 9x≡1 [26]

9m+5≡p[26]⇐⇒m≡3p-15 [26].

E 026

A A a→0,b→1,...z→25 a b c d e f g h i j k l m n o p q r s t u v w x y z⋆ a 0,b z 25 ⋆ 26
• xE g

4x+327

xE,g(x) E • g(x) s→18,g(18)=2121→v s v xE g(x)=x g x E y

E y≡4x+3 x≡7y+6

e e e xy xy xy xy x 0ཝ y ཝ ឹxy ិ 8x+15y=1

8x+15y=146xy

nN n

2≡N-1N.

n×n3≡1N k1 5k1≡126 k 0⩽k⩽255k≡126

A=

4 1 , B= 2-1 , X= x1 x

Y=

y1 y 6A-A2

A A-1

A-1=αI+βA αβ

ិ B=5A-1

AX=Y 5X=BY

X=

x1 x x1 x2

X Y=

y1 y Y=AX

Y R=

r1 r r1 y1 r2 y2 r1r2 →X 14 →Y= 76
→R= 24

X Y=

y1 y Y=AX 5x1=2y1-y2

5x2= -3y1+4y2.

x

1≡16y1+5y2

x

2≡15y1+6y226

AB A

AB A D AB D>0 B A A D B>A

D B-A

D A-B

ឹA A=12B=14 AB

A=221B=331 ឹ

ិ (x;y) () 221x-331y=1. ិ (3 ; 2) (x;y) (un)(vn)ិ n u n=2+221n v 0=3 v n+1=vn+331 vn n (p;q) u p=vq, 0⩽p⩽5000⩽q⩽500 ab r(a,b) ab c ab a b c̸=0 ឹb a=26b=9 abc ab ិ ab pq 025 x x′ិ x x′ p=9q=2 ឹ uv

9u+26v=1 ិ (u,v)

x′≡9x+2 [26] x≡3x′+20 [26] q=2p p p p=13q=2 (x,y,z) x

2+y2=z2.

32+42=9+16=25=52

(x,y,z) p (px,py,pz) (x,y,z) xyz n n=2α×kα k n=2α×k n

9=20×9,

120=23×15

192
xz x=2α×k z=2β×m 2x2z2

2x2 z2

(x,z) 2x2=z2 xyz x,y (x,y,z) xyz x4032=169×961

(x,z) z2-x2=4032 x<403 (x, 2015,z)

15x-26k=mxk

m ិ (u;v)

15u-26v=1

(x0;k0) (x;k) 15 (x-x0)-26(k-k0)=0 (x;k) x=26q+7m k=15q+4mq∈?. x xy 15x+7 26
y x y

15x+726 k 15x-26k=y-7

x=7y+326 n S(n) n ិ S(6)=12 S(7) n S(n)⩾1+n n S(n)=1+n np×qpq

S(n)=(1+p)(1+q)

nm

S(n×m)=S(n)×S(m)

npk p k n

S(n)=1-pk+1

1-p np13×q7 pq m mn st

0⩽s⩽130⩽t⩽7 m=ps×qt

S(n)=1-p14

1-p×1-q8

1-q 2n-1n a,bc (b;c)=1 baca bca 233-1
(233-1)÷3 (233-1)÷4 (233-1)÷12 (233-1) (233-1) (233-1) (233-1)

2≡-1 [3] 233-1

S=1+23+(23)2+(23)3+···+(23)10

233-1

27-1 ិ

(N,k) Nk n k n (2n-1,k)̸=0k⩽p 2 n-1 ງk k>p 2 n-1 ឹ n=33 n=7 kឹ

M=

1 1 1 1-1 1 I= 1 0 0 0 1 0

M2 M3=

20 10 11

12 2 9

ិ M3=M2+8M+6I

M M-1=1

6 (M2-M-8I) abc y=ax2+bx+c (-1 ;-1) abc M a b 1 -1 abc ិ abcpqr (O;⃗ı,⃗ȷ) (1 ;p) (-1 ;q) (2 ;r) pqr y=ax2+bx+c a b =M-1 p q  abc -3p+q+2r≡0[6]

3p-3q≡0[6]

6p+2q-2r≡0[6]

q-r≡0[3] p-q≡0[2] q-r≡0[3] p-q≡0[2] abc y=ax2+bx+c

2r+q-3p=0

p=7 qrabc y=ax2+bx+c a,bc a bc ab k bc=ka ab uv au+bv=1 c acu+bcv=c bcka acu+kav=c⇐⇒a(cu+kv)=c ac u1=2+3+6-1=10 u

2=4+9+36-1=48

u

3=8+27+216-1=250

u

4=16+81+1296-1=1392

u

5=32+243+7776-1=8050

u

6=64+729+46656-1=47448

2≡0 mod 2=⇒2n≡0 mod 2

3≡1 mod 2=⇒3n≡1 mod 2

6≡0 mod 2=⇒6n≡0 mod 2un=2n+3n+6n-1≡0+1+0-1≡0 mod 2

u n u n n k∈?∗ n=2k

4≡0 mod 4, 4k≡0 mod 4

4 k×9k≡0 mod 4

9≡1 mod 4 9k≡1 mod 4

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