Trng THCS M Quang: CHUYN BI DNG HC SINH GIIChuyn 2 : DU HIU CHIA
HTPHN I: TM TT L THUYTI. NH NGHA PHP CHIACho 2 s nguyn a v b trong
b 0 ta lun tm c hai s nguyn q v r duy nht sao cho:a = bq + r Vi 0 r
bTrong : a l s b chia, b l s chia, q l thng, r l s d.Khi a chia cho
b c th xy ra bs dr {0; 1; 2; ; b }c bit: r = 0 th a = bq, khi ta ni
a chia ht cho b hay b chia ht a.K hiu: ab hay b\ a Vy:a b C s nguyn
q sao cho a = bqII. CC TNH CHT1. Vi a 0 a a2. Nu a b v b c a c3. Vi
a 0 0 a4. Nu a, b > 0 v a b ; b a a = b5. Nu a b v c bt k ac b6.
Nu a b (t a) (t b)7. Vi a a (t 1)8. Nu a b v c b a tc b9. Nu a b
vcb a tc b10.Nu a + b c v a c b c11.Nu a b vn > 0 an bn12.Nu ac
b v (a, b) =1 c b13.Nu a b, c b v m, n bt k am + cn b14.Nu a b v c
d ac bd15.Tch n s nguyn lin tip chia ht cho n!III. MT S DU HIU CHIA
HTGi N = 0 1 1 n na ...a a a1. Du hiu chia ht cho 2; 5; 4; 25; 8;
125+ N 2 a0 2 a0{0; 2; 4; 6; 8}+ N 5 a0 5 a0{0; 5}+ N 4 (hoc 25) 0
1a a 4 (hoc 25)+ N 8 (hoc 125) 0 1a a a2 8 (hoc 125)2. Du hiu chia
ht cho 3 v 9+ N 3 (hoc 9) a0+a1++an 3 (hoc 9)3. Mt s du hiu khc+ N
11 [(a0+a1+) - (a1+a3+)] 11+ N 101 [(0 1a a+4 5a a+) - (2 3a a+6 7a
a +)]1011Trng THCS M Quang: CHUYN BI DNG HC SINH GII+N 7(hoc13)[(0
1a a a2+ 6 7a a a8+) - [(3 4a a a5+ 9 10a a a11+)11 (hoc 13)+ N 37
(0 1a a a2 + 3 4a a a5+) 37 + N 19 ( a0+2an-1+22an-2++ 2na0) 19IV.
NG D THCa. nh ngha: Cho m l s nguyn dng. Nu hai s nguyn a v b cho
cng s d khi chia cho m th ta ni a ng d vi b theo modun m.K hiu: a b
(modun)Vy: a b (modun) a - b mb. Cc tnh cht1. Vi a a a (modun)2. Nu
a b (modun) b a (modun)3. Nu a b (modun), b c (modun) a c (modun)4.
Nu a b (modun) v c d (modun) a+c b+d (modun)5. Nu a b (modun) v c d
(modun) ac bd (modun)6. Nu a b (modun), d Uc (a, b) v (d, m) =1
dbda(modun)7. Nu a b (modun), d > 0 v d Uc (a, b, m) dbda(modun
dm)V. MT S NH L1. nh l EulerNu m l 1 s nguyn dng (m) l s cc s nguyn
dng nh hn m v nguyn t cng nhau vi m, (a, m) = 1Tha(m)1 (modun)Cng
thc tnh (m)Phn tch m ra tha s nguyn tm = p11 p22 pkkvipi p; i
N*Th(m) = m(1 - ` 11p)(1 - 21p) (1 - kp1)2. nh l FermatNu t l s
nguyn t v a khng chia ht cho p th ap-11 (modp)3. nh l WilsonNu p l
s nguyn t th( P - 1)! + 1 0 (modp)PHN II:CC PHNG PHP GII BI TON
CHIA HT1. Phng php 1: S DNG DU HIU CHIA HTV d 1: Tm cc ch s a, b
sao cho a56b 45GiiTa thy 45 = 5.9 m (5 ; 9) = 1 a56b 45 a56b 5 v
92Trng THCS M Quang: CHUYN BI DNG HC SINH GIIXt a56b 5 b {0 ; 5}Nu
b = 0 ta c s a56b 9 a + 5 + 6 + 0 9a + 11 9 a = 7Nu b = 5 ta c s
a56b 9 a + 5 + 6 + 0 9a + 16 9 a = 2Vy:a = 7 v b = 0 ta c s 7560a =
2 v b = 5 ta c s 2560V d 2: Bit tng cc ch s ca 1 s l khng i khi nhn
s vi 5. Chng minh rng s chia ht cho 9.GiiGi s cho l aTa c: a v 5a
khi chia cho 9 cng c 1 s d 5a - a 9 4a 9m (4 ; 9) = 1a 9 (pcm)V d
3: CMR s1 s 81111 111 81GiiTa thy: 111111111 9C 1 s 81111 111=
111111111(1072 + 1063 + + 109 + 1)M tng 1072 + 1063 + + 109 + 1 c
tng cc ch s bng 9 91072 + 1063 + + 109 + 1 9 Vy: 1 s 81111 111 81
(pcm)BI TP TNG TBi 1: Tm cc ch s x, y sao cho a.34x5y 4 v 9b. 2x78
17Bi 2: Cho s N = dcba CMRa. N 4 (a + 2b) 4b. N 16 (a + 2b + 4c +
8d) 16 vi b chnc. N 29 (d + 2c + 9b + 27a) 29Bi 3: Tm tt c cc s c 2
ch s sao cho mi s gp 2 ln tch cc ch s ca s .Bi 4: Vit lin tip tt c
cc s c 2 ch s t 19 n 80 ta c s A = 1920217980. Hi s A c chia ht cho
1980 khng ? V sao?Bi 5: Tng ca 46 s t nhin lin tip c chia ht cho 46
khng? V sao?Bi 6: Chng t rng s 1 s 10011 11 2 s 10022 22 l tch ca 2
s t nhin lin tip.HNG DN - P SBi 1:a. x =v y = 2x = v y = 6b. 2x78=
17 (122 + 6x) + 2(2-x)17 x = 23Trng THCS M Quang: CHUYN BI DNG HC
SINH GIIBi 2:a. N4 ab4 10b + a4 8b + (2b + a) 4 a + 2b4b. N16 1000d
+ 100c + 10b + a16 (992d + 96c + 8b) + (8d + 4c + 2b + a) 16 a + 2b
+ 4c + 8d16 vi b chnc. C 100(d + 3c + 9b + 27a) - dbca29m (1000,
29) =1
dbca29(d + 3c + 9b + 27a) 29Bi 3: Gi ab l s c 2 ch s Theo bi ra
ta c:ab= 10a + b = 2ab (1) ab2 b {0; 2; 4; 6; 8} Thay vo (1) a = 3;
b = 6Bi 4: C 1980 = 22.32.5.11 V 2 ch s tn cng ca a l 80 4 v 5A 4 v
5Tng cc s hng l 1+(2+3++7).10+8 = 279Tng cc s hng chn
9+(0+1++9).6+0 = 279C 279 + 279 = 558 9 A 9 279-279 = 0 11A 11Bi 5:
Tng 2 s t nhin lin tip l 1 s l nn khng chia ht cho 2. C 46 s t nhin
lin tip c 23 cp s mi cp c tng l 1 s l tng 23 cp khng chia ht cho 2.
Vy tng ca 46 s t nhin lin tip khng chia ht cho 46.Bi 6: C 100
111...11so123 100 222...22so1 2 3=100 111...11so12399
0100...02so142 43M 99 0100...02so142 43= 3. 99 333...34so12 3100
111...11so123 100 222...22so1 2 3=100 333...33so12399 333...34so12
3 (pcm)2. Phng php 2: S DNG TNH CHT CHIA HT* Ch : Trong n s nguyn
lin tip c 1 v ch 1 s chia ht cho n.CMR: Gi n l s nguyn lin tipm +
1;m + 2; m + nvim Z,n N*Ly n s nguyn lin tip trn chia cho n th ta c
tp hp s d l: {0; 1; 2; n - 1}* Nu tn ti 1 s d l 0: gi sm + i = nqi
;i =n 1,m + i n* Nu khng tn ti s d l 0 khng c s nguyn no trong dy
chia ht cho n phi c t nht 2 s d trng nhau.Gi s: '+ + + +r qjnj mn j
i; 1rnqi i mi - j= n(qi - qj) n i - jnm |i - j|< n i - j= 0 i =
jm + i = m + jVy trong n s c 1 s v ch 1 s chia ht cho n4Trng THCS M
Quang: CHUYN BI DNG HC SINH GIIV d 1:CMR: a. Tch ca 2 s nguyn lin
tip lun chia ht cho 2 b. Tch ca 3 s nguyn lin tip chia ht cho
6.Giia. Trong 2 s nguyn lin tip bao gi cng c 1 s chnS chn chia ht
cho 2.Vy tch ca 2 s nguyn lin tip lun chia ht cho 2.Tch 2 s nguyn
lin tip lun chia ht cho 2 nn tch ca 3 s nguyn lin tip lun chia ht
cho 2b. Trong 3 s nguyn lin tip bao gi cng c 1 s chia ht cho 3.Tch
3 s chia ht cho 3 m (1; 3) = 1.Vy tch ca 3 s nguyn lin tip lun chia
ht cho 6.V d 2:CMR: Tng lp phng ca 3 s nguyn lin tip lun chia ht
cho 9.GiiGi 3 s nguyn lin tip ln lt l: n - 1 ,n ,n+1Ta c: A = (n -
1)3 + n3 + (n + 1)3= 3n3 - 3n + 18n + 9n2 + 9= 3(n - 1)n (n+1) +
9(n2 + 1) + 18nTa thy (n - 1)n (n+ 1) 3 (CM V d 1)3(n - 1)n (n + 1)
9m ' +9 189 ) 1 ( 92nnA 9 (PCM)V d 3: CMR: n4 - 4n3 - 4n2 +16n 3 84
vi n chn, n 4GiiV n chn, n 4 ta t n = 2k, k 2Ta c n4 -4n3 - 4n2 +
16n = 16k4 - 32k3 - 16k2 + 32k = t 16k(k3 - 2k2 -k + 2) = t 16k(k -
2) (k - 1)(k + 1)Vi k 2 nn k - 2, k - 1, k + 1, k l 4 s t nhin lin
tip nn trong 4 s c 1 s chia htcho 2 v 1 s chia ht cho 4. (k - 2)(k
- 1)(k + 1)k 8M (k - 2) (k - 1)k 3;(3,8)=1(k - 2) (k - 1) (k + 1)k
2416(k - 2) (k - 1) (k + 1)k (16,24)Vyn4 - 4n3 - 4n2 +16n 384 vi n
chn, n 4BI TP TNG TBi 1: CMR: a. n(n + 1) (2n + 1) 6b. n5 - 5n3 +
4n 120 Vi n NBi 2: CMR: n4 + 6n3 + 11n2 + 6n 24 Vi n ZBi 3: CMR: Vi
n l tha. n2 + 4n + 3 8b. n3 + 3n2 -n - 3 48c. n12 - n8 -n4 + 1
512Bi 4: Vi p l s nguyn t p > 3CMR: p2 - 1 24Bi 5: CMR: Trong
1900 s t nhin lin tip c 1 s c tng cc ch s chia ht cho 27.HNG DN - P
SBi 1: a. n(n + 1)(2n + 1) = n(n + 1) [(n + 1) + (n + 2)]= n(n + 1)
(n - 1) + n(n + 1) (n + 2) 65Trng THCS M Quang: CHUYN BI DNG HC
SINH GIIb. n5 - 5n3 + 4n = (n4 - 5n2 + 4)n= n(n2 - 1) (n2 - 4)= n(n
+ 1) (n - 1) (n + 2) (n - 2) 120Bi 2: n4 + 6n3 + 6n + 11n2= n(n3 +
6n2 + 6 + 11n)= n(n + 1) (n + 2) (n + 3) 24Bi 3: a. n2 + 4n + 3 =
(n + 1) (n + 3) 8b. n3 + 3n2 - n - 3 = n2(n + 3) - (n + 3)= (n2 -
1) (n + 3)= (n + 1) (n - 1) (n + 3)= (2k + 4) (2k + 2) (2k vi n =
2k + 1, k N)= 8k(k + 1) (k +2) 48c. n12 - n8 - n4 + 1 = n8 (n4 - 1)
- (n4 - 1)= (n4 - 1) (n8 - 1)= (n4 - 1)2 (n4 + 1) = (n2 - 1)2 (n2 -
1)2 (n4 + 1)= 16[k(k + 1)2 (n2 + 1)2 (n4 + 1)Vi n = 2k + 1 n2 + 1 v
n4 + 1 l nhng s chn (n2 + 1)2 2 n4 + 1 2n12 - n8 -n4 + 1 (24.22.
22. 1 . 21)Vy n12 - n8 - n4 + 1 512Bi 4: C p2 - 1 = (p - 1) (p + 1)
v p l s nguyn t p > 3p 3 ta c: (p - 1) (p + 1) 8v p = 3k + 1 hoc
p = 3k + 2 (k N)(p - 1) (p + 1) 3Vy p2 - 1 24Bi 5: Gi s 1900 s t
nhin lin tip l n, n +1; n + 2; ; n + 1989 (1)trong 1000 t nhin lin
tip n, n + 1; n + 2; ; n + 999 c 1 s chia ht cho 1000 gi s n0, khi
n0 c tn cng l 3 ch s 0 gi s tng cc ch s ca n0 l s khi 27 s n0, n0 +
9; n0 + 19; n0 + 29; n0 + 39; ; n0 + 99; n0 + 199; n0 + 899 (2)C
tng cc ch s ln lt l: s; s + 1 ; s + 26C 1 s chia ht cho 27 (PCM)*
Ch :n + 899 n + 999 + 899 < n + 1989Cc s (2) nm trong dy (1)3.
Phng php 3: XT TP HP S D TRONG PHP CHIAV d 1: CMR: Vi n NTh A(n) =
n(2n + 7) (7n + 7) chia ht cho 6GiiTa thy 1 trong 2 tha s n v 7n +
1 l s chn. Vi n N A(n) 2Ta chng minh A(n) 3Ly n chia cho 3 ta c n =
3k + 1 (k N)Vi r {0; 1; 2}Vi r = 0 n = 3k n 3 A(n) 3Vi r = 1 n = 3k
+ 1 2n + 7 = 6k + 9 3 A(n) 3 Vi r = 2 n = 3k + 2 7n + 1 = 21k + 15
3 A(n) 36Trng THCS M Quang: CHUYN BI DNG HC SINH GIIA(n) 3 vi n m
(2, 3) = 1Vy A(n) 6 vi n NV d 2: CMR: Nu n 3 th A(n) = 32n + 3n + 1
13 Vi n NGiiV n 3 n = 3k + r (k N); r {1; 2; 3}A(n) = 32(3k + r) +
33k+r + 1 = 32r(36k - 1) + 3r (33k - 1) + 32r + 3r + 1ta thy 36k -
1 = (33)2k - 1 = (33 - 1)M= 26M 1333k - 1 = (33 - 1)N = 26N 13vi r
= 1 32n + 3n + 1 = 32 + 3 +1 = 13 1332n + 3n + 1 13vi r = 2 32n +
3n + 1 = 34 + 32 + 1 = 91 1332n + 3n + 1Vy vi n 3 th A(n) = 32n +
3n + 1 13 Vi n NV d 3: Tm tt c cc s t nhin n 2n - 1 7GiiLy n chia
cho 3 ta c n = 3k + 1 (k N); r {0; 1; 2}Vi r = 0 n = 3k ta c 2n - 1
= 23k - 1 = 8k - 1 = (8 - 1)M = 7M 7vi r =1 n = 3k + 1 ta c:2n - 1
= 28k +1 - 1 = 2.23k - 1 = 2(23k - 1) + 1m 23k - 1 7 2n - 1 chia
cho 7 d 1vi r = 2 n = 3k + 2 ta c :2n - 1 = 23k + 2 - 1 = 4(23k -
1) + 3 m 23k - 1 7 2n - 1 chia cho 7 d 3Vy 23k - 1 7 n = 3k (k N)BI
TP TNG TBi 1: CMR: An = n(n2 + 1)(n2 + 4) 5 Vi n ZBi 2: Cho A = a1
+ a2 + + anB = a51 + a52 + + a5nBi 3: CMR: Nu (n, 6) =1 th n2 - 1
24 Vi n ZBi 4: Tm s t nhin W 22n + 2n + 1 7Bi 5: Cho 2 s t nhin m,
n tho mn 24m4 + 1 = n2CMR: mn 55HNG DN - P SBi 1: + A(n) 6+ Ly n
chia cho 5 n = 5q + rr {0; 1; 2; 3; 4}r = 0 n 5 A(n) 5r = 1, 4 n2 +
4 5 A(n) 5r = 2; 3 n2 + 1 5 A(n) 5A(n) 5 A(n) 30Bi 2: Xt hiu B - A
= (a51 - a1) + + (a5n - an) Ch chng minh: a5i - ai 30 l Bi 3: V (n,
6) =1 n = 6k + 1(k N)Vi r {t 1}r = t 1n2 - 1 24Bi 4: Xt n = 3k + r
(k N) Vi r {0; 1; 2}7Trng THCS M Quang: CHUYN BI DNG HC SINH GIITa
c: 22n + 2n + 1 = 22r(26k - 1) + 2r(23k - 1) + 22n + 2n + 1Lm tng t
VD3Bi 5: C 24m4 + 1 = n2 = 25m4 - (m4 - 1)Khi m 5 mn 5Khi m 5 th
(m, 5) = 1 m4 - 1 5(V m5 - m 5 (m4 - 1) 5 m4 - 1 5)n2 5 ni5Vy mn
54. Phng php 4: S DNG PHNG PHP PHN TCH THNH NHN TGi s chng minh an
kTa c th phn tch an cha tha s k hoc phn tch thnh cc tha s m cc tha
s chia ht cho cc tha s ca k.V d 1: CMR: 36n - 26n 35 Vi n NGiiTa c
36n - 26n = (36)n - (26)n = (36 - 26)M= (33 + 23) (33 - 23)M=
35.19M 35 Vy 36n - 26n 35 Vi n NV d 2: CMR: Vi n l s t nhin chn th
biu thc A = 20n + 16n - 3n - 1 232GiiTa thy 232 = 17.19 m (17;19) =
1 ta chng minhA 17 v A 19 ta c A = (20n - 3n) + (16n - 1) c 20n -
3n = (20 - 3)M 17M16n - 1 = (16 + 1)M = 17N 17 (n chn)A 17 (1)ta c:
A = (20n - 1) + (16n - 3n) c 20n - 1 = (20 - 1)p = 19p 19 c 16n -
3n = (16 + 3)Q = 19Q 19 (n chn)A 19 (2)T (1) v (2) A 232V d 3: CMR:
nn - n2 + n - 1 (n - 1)2 Vi n >1GiiVi n = 2 nn - n2 + n - 1 = 1
v (n - 1)2 = (2 - 1)2 = 1nn - n2 + n - 1 (n - 1)2vi n > 2 t A =
nn - n2 + n - 1 ta c A = (nn - n2) + (n - 1)= n2(nn-2 - 1) + (n -
1)= n2(n - 1) (nn-3 + nn-4 + + 1) + (n - 1)= (n - 1) (nn-1 + nn-2 +
+ n2 +1) = (n - 1) [(nn-1 - 1) + +( n2 - 1) + (n - 1)]= (n - 1)2M
(n - 1)2Vy A (n - 1)2 (PCM)BI TP TNG TBi 1: CMR: a. 32n +1 + 22n +2
7 b. mn(m4 - n4) 30Bi 2: CMR: A(n) = 3n + 63 72 vi n chn n N, n 2Bi
3: Cho a v b l 2 s chnh phng l lin tipCMR: a. (a - 1) (b - 1)
1928Trng THCS M Quang: CHUYN BI DNG HC SINH GIIBi 4: CMR: Vi p l 1
s nguyn t p > 5 th p4 - 1 240Bi 5: Cho 3 s nguyn dng a, b, c v
tho mn a2 = b2 + c2CMR: abc 60HNG DN - P SBi 1: a. 32n +1 + 22n +2
= 3.32n + 2.2n= 3.9n + 4.2n= 3(7 + 2)n + 4.2n = 7M + 7.2n 7b. mn(m4
-n4) = mn(m2 - 1)(m2 + 1) -mn(n2 - 1) (n2 + 1) 30Bi 3: C 72 = 9.8 m
(8, 9) = 1 v n = 2k (k N) c 3n + 63 = 32k + 63= (32k - 1) + 64 A(n)
8Bi 4: t a = (2k - 1)2; b = (2k - 1)2 (k N) Ta c (a - 1)(b - 1) =
16k(k + 1)(k - 1) 64 v 3Bi 5: C 60 = 3.4.5t M = abcNu a, b, c u
khng chia ht cho 3 a2, b2 v c2 chia ht cho 3 u d 1 a2 b2 + c2. Do c
t nht 1 s chia ht cho 3. Vy M 3Nu a, b, c u khng chia ht cho 5 a2,
b2 v c2 chia 5 d 1 hoc 4 b2 + c2 chia 5 th d 2; 0 hoc 3.a2 b2 + c2.
Do c t nht 1 s chia ht cho 5. Vy M 5Nu a, b, c l cc s l b2 v c2
chia ht cho 4 d 1.b2 + c2 (mod 4) a2 b2 + c2Do 1 trong 2 s a, b phi
l s chn.Gi s b l s chnNu C l s chn M 4Nu C l s l m a2 = b2 + c2 a l
s l b2 = (a - c) (a + b) ,_
,_
+
,_
2 2 22c a c a b2b chn b 4 m 4Vy M = abc 3.4.5 = 605. Phng php 5:
BIN I BIU THC CN CHNG MINH V DNG TNGGi s chng minh A(n) k ta bin i
A(n) v dng tng ca nhiu hng t v chng minh mi hng t u chia ht cho k.V
d 1: CMR: n3 + 11n 6 vi n z.GiiTa c n3 + 11n = n3 - n + 12n = n(n2
- 1) + 12n= n(n + 1) (n - 1) + 12nV n, n - 1; n + 1 l 3 s nguyn lin
tipn(n + 1) (n - 1) 6 v 12n 6Vy n3 + 11n 6V d 2: Cho a, b z tho mn
(16a +17b) (17a +16b) 11CMR: (16a +17b) (17a +16b) 121Gii9Trng THCS
M Quang: CHUYN BI DNG HC SINH GIIC 11 s nguyn t m (16a +17b) (17a
+16b) 11
++11 16b 17a11 17b 16a(1)C 16a +17b + 17a +16b = 33(a + b) 11
(2)T (1) v (2)
++11 16b 17a11 17b 16aVy (16a +17b) (17a +16b) 121V d 3: Tm n N
sao cho P = (n + 5)(n + 6) 6n.GiiTa c P = (n + 5)(n + 6) = n2 + 11n
+ 30= 12n + n2 - n + 30V 12n 6n nn P 6n n2 - n + 30 6n ''(2) n
30(1) 3 1) - n(n 6n 306 n- n2 T (1) n = 3k hoc n = 3k + 1 (k N) T
(2) n {1; 2; 3; 5; 6; 10; 15; 30}Vy t (1); (2) n {1; 3; 6; 10; 15;
30}Thay cc gi tr ca n vo P ta c n {1; 3; 10; 30} l tho mnVy n {1;
3; 10; 15; 30} th P = (n + 5)(n + 6) 6n.BI TP TNG TBi 1: CMR: 13 +
33 + 53 + 73 23Bi 2: CMR: 36n2 + 60n + 24 24Bi 3: CMR: a. 5n+2 +
26.5n + 8 2n+1 59b. 9 2n + 14 5Bi 4: Tm n N sao cho n3 - 8n2 + 2n
n2 + 1HNG DN - P SBi 1: 13 + 33 + 53 + 73 = (13 + 73) + (33 + 53)=
8m + 8N 23Bi 2: 362 + 60n + 24 = 12n(3n + 5) + 24Ta thy n v 3n + 5
khng ng thi cng chn hoc cng l n(3n + 5) 2 PCMBi 3: a. 5n+2 + 26.5n
+ 8 2n+1 = 5n(25 + 26) + 8 2n+1 = 5n(59 - 8) + 8.64 n=5n.59 + 8.59m
59b. 9 2n + 14 = 9 2n - 1 + 15= (81n - 1) + 15= 80m + 15 5Bi 4: C
n3 - 8n2 + 2n = (n2 + 1)(n - 8) + n + 8 (n2 + 1) n + 8 n2 + 1Nu n +
8 = 0 n = -8 (tho mn)Nu n + 8 0 |n + 8| n2 + 1
+ +
+ + +8 0 7 n8 0 9 n8 1 n 8 n 8 1 -n 8 n 2222n nn
nnnViViViVi10Trng THCS M Quang: CHUYN BI DNG HC SINH GIIn {-2; 0;
2} th liVy n {-8; 0; 2} Phng php 6:DNG QUY NP TON HCGi s CM A(n) P
vi n a (1)Bc 1: Ta CM (1) ng vi n = a tc l CM A(n) PBc 2: Gi s (1)
ng vi n = k tc l CM A(k) P vi k aTa CM (1) ng vi n = k + 1 tc l phi
CM A(k+1) PBc 3: Kt lun A(n) P vi n aV d 1: Chng minh A(n) = 16n -
15n - 1 225 vi n N*GiiVi n = 1 A(n) = 225 225 vy n = 1 ngGi s n = k
1 ngha l A(k) = 16k - 15k - 1 225Ta phi CM A(k+1) = 16 k+1 - 15(k +
1) - 1 225Tht vy: A(k+1) = 16 k+1 - 15(k + 1) - 1= 16.16k - 15k -
16= (16k - 15k - 1) + 15.16k - 15= 16k - 15k - 1 + 15.15m= A(k) +
225m A(k) 225 (gi thit quy np) 225m 225Vy A(n) 225V d 2: CMR: vi n
N* v n l s t nhin l ta c 2 22 1+nnm GiiVi n = 1 m2 - 1 = (m + 1)(m
- 1) 8 (v m + 1; m - 1 l 2 s chn lin tip nn tch ca chng chia ht cho
8)Gi s vi n = k ta c 2 22 1+kkm ta phi chng minh3 22 11++kkm Tht vy
2 22 1+kkm ) ( . 2 12 2z q q mkk +1 . 22 2+ +q mkk c ( ) ( ) q q q
m mk k kk k. 2 . 2 1 1 . 2 1 13 2 42222 21+ + ++ + += 3 2 1 32 ) 2
( 2+ + ++k k kq q Vy 2 22 1+nnm vi n 1BI TP TNG TBi 1: CMR: 33n+3 -
26n - 27 29 vi n 1Bi 2: CMR: 42n+2 - 1 15Bi 3: CMR s c thnh lp bi
3n ch s ging nhau th chia ht cho 3n vi n l s nguyn dng.HNG DN - P
SBi 1: Tng t v d 1.11Trng THCS M Quang: CHUYN BI DNG HC SINH GIIBi
2: Tng t v d 1.Bi 3: Ta cn CM sana aa3... 3n (1)Vi n = 1 ta ca aa
...3 111 a Gi s (1) ng vi n = k tc l saka aa3... 3kTa chng minh (1)
ng vi n = k + 1 tc l phi chng minha s13...+ ka aa 3k+1 ta c 3k+1 =
3.3k = 3k + 3k +3kC k k kka a a a a a a aa3 3 33... ... ... ...1+a
skk ka a a aa a aa33 3 . 2... 10 . ... 10 . ... + + ( )1 3 3 . 233
1 10 10 ...++ + kk kka aa 1. Phng php 7: S DNG NG D THCGii bi ton
da vo ng d thc ch yu l s dng nh l Euler v nh l FermatV d 1: CMR:
22225555 + 55552222 7GiiC 2222 - 4 (mod 7) 22225555 + 55552222 (-
4)5555 + 45555 (mod 7)Li c: (- 4)5555 + 42222 = - 45555 + 42222 = -
42222 (43333 - 1) = ( ) ( ) 1 4 4 -11113 2222V 43 = 64 (mod 7) ( )
0 1 411113 (mod 7)22225555 + 55552222 0 (mod 7)Vy 22225555 +
55552222 7V d 2: CMR: 22 5 3 31 4 1 43 2 + ++ + n n vi n NGiiTheo
nh l Fermat ta c: 310 1 (mod 11)210 1 (mod 11)Ta tm d trong php
chia l 24n+1 v 34n+1 cho 10C 24n+1 = 2.16n 2 (mod 10)24n+1 = 10q +
2 (q N)C 34n+1 = 3.81n 3 (mod 10)34n+1 = 10k + 3 (k N)Ta c: 3 10 2
10 3 22 3 5 3 31 4 1 4+ ++ + ++ +k qn n= 32.310q + 23.210k + 51+0+1
(mod 2)0 (mod 2) 12Trng THCS M Quang: CHUYN BI DNG HC SINH GIIm (2,
11) = 1Vy 22 5 3 31 4 1 43 2 + ++ + n n vi n NV d 3: CMR: 11 7 21
42 ++ n vi n NGiiTa c: 24 6 (mod) 24n+1 2 (mod 10)24n+1 = 10q + 2
(q N)2 10 22 21 4++qnTheo nh l Fermat ta c: 210 1 (mod 11)210q 1
(mod 11)7 2 7 22 10 21 4+ +++qn4+7 (mod 11) 0 (mod 11) Vy 11 7 21
42 ++ n vi n N (PCM)BI TP TNG TBi 1: CMR 19 3 22 62 ++ n vi n NBi
2: CMR vi n 1 ta c52n-1. 22n-15n+1 + 3n+1 .22n-1 38Bi 3: Cho s p
> 3, p (P)CMR 3p - 2p - 1 42pBi 4: CMR vi mi s nguyn t p u c
dng2n - n (n N) chia ht cho p.HNG DN - P SBi 1: Lm tng t nh VD3Bi
2: Ta thy 52n-1. 22n-15n+1 + 3n+1 .22n-1 2Mt khc 52n-1. 22n-15n+1 +
3n+1 .22n-1 = 2n(52n-1.10 + 9. 6n-1)V 25 6 (mod 19) 5n-1 6n-1 (mod
19)25n-1.10 + 9. 6n-1 6n-1.19 (mod 19) 0 (mod 19)Bi 3: t A = 3p -2p
- 1 (p l)D dng CM A 2 v A 3 A 6Nu p = 7 A = 37 -27 - 1 49 A 7pNu p
7 (p, 7) = 1Theo nh l Fermat ta c:A = (3p - 3) - (2p - 2) pt p = 3q
+ r (q N; r = 1, 2)A = (33q+1 - 3) - (23q+r - 2)= 3r.27q -2r.8q - 1
= 7k + 3r(-1)q - 2r - 1 (k N)vi r = 1, q phi chn (v p l)A = 7k - 9
- 4 - 1 = 7k - 14Vy A 7 m A p, (p, 7) = 1 A 7pM (7, 6) = 1; A 6A
42p.Bi 4: Nu P = 2 22 - 2 = 2 2Nu n > 2 Theo nh l Fermat ta
c:2p-1 1 (mod p)13Trng THCS M Quang: CHUYN BI DNG HC SINH
GII2m(p-1) 1 (mod p) (m N)Xt A = 2m(p-1) + m - mpA p m = kq - 1Nh
vy nu p > 2 p c dng 2n - n trong N = (kp - 1)(p - 1), k N u chia
ht cho p2. Phng php 8: S DNG NGUYN L IRICHLETNu em n + 1 con th nht
vo n lng th c t nht 1 lng cha t 2 con tr ln.V d 1: CMR: Trong n + 1
s nguyn bt k c 2 s c hiu chia ht cho n.GiiLy n + 1 s nguyn cho chia
cho n th c n + 1 s d nhn 1 trong cc s sau: 0; 1; 2; ; n - 1c t nht
2 s d c cng s d khi chia cho n.Gi s ai = nq1 + r0 r < naj = nq2
+ r a1; q2 Naj - aj = n(q1 - q2) nVy trong n +1 s nguyn bt k c 2 s
c hiu chia ht cho n.Nu khng c 1 tng no trong cc tng trn chia ht cho
n nh vy s d khi chia mi tng trn cho n ta c n s d l 1; 2; ; n - 1Vy
theo nguyn l irichlet s tn ti t nht 2 tng m chi cho n c cng s d
(theo VD1) hiu cadr tng ny chia ht cho n (PCM).BI TP TNG TBi 1:
CMR: Tn ti n N sao cho 17n - 1 25Bi 2: CMR: Tn ti 1 bi ca s 1993 ch
cha ton s 1.Bi 3: CMR: Vi 17 s nguyn bt k bao gi cng tn ti 1 tng 5
s chia ht cho 5.Bi 4: C hay khng 1 s c dng.19931993 1993000 00
1994HNG DN - P SBi 1: Xt dy s 17, 172, , 1725 (tng t VD2)Bi 2: Ta c
1994 s nguyn cha ton b s 1 l:1111111 s 199411 111 Khi chia cho 1993
th c 1993 s d theo nguyn l irichlet c t nht 2 s c cng s d.Gi s l ai
= 1993q + r0 r < 1993aj = 1993k + r i > j; q, k Naj - aj =
1993(q - k) ) ( 1993 0 00 11 111 k q 0 s i 1 s 1994 j - i) ( 1993
10 . 11 111 k qj 1 s 1994 j - i14Trng THCS M Quang: CHUYN BI DNG HC
SINH GIIm (10j, 1993) = 1 1 s 199411 111 1993 (PCM)Bi 3: Xt dy s gm
17 s nguyn bt k l a1, a2, , a17Chia cc s cho 5 ta c 17 s d t phi c
5 s d thuc tp hp{0; 1; 2; 3; 4}Nu trong 17 s trn c 5 s khi chia cho
5 c cng s d th tng ca chng s chia ht cho 5.Nu trong 17 s trn khng c
s no c cng s d khi chia cho 5 tn ti 5 s c s d khc nhau tng cc s d
l: 0 + 1 + 2 + 3 + 4 = 10 10Vy tng ca 5 s ny chia ht cho 5.Bi 4: Xt
dy s a1 = 1993, a2 = 19931993, a1994 = 1993 s 19941993 1993 em chia
cho 1994 c 1994 s d thuc tp {1; 2; ; 1993} theo nguyn l irichlet c
t nht 2 s hng c cng s d.Gi s: ai = 1993 1993 (i s 1993)aj = 1993
1993 (j s 1993)aj - aj 19941 i < j 19941993 10 . 1993 1993
ni1993 s i - j3. Phng php 9PHNG PHP PHN CHNG CM A(n) p (hoc A(n) p
)+ Gi s: A(n) p (hoc A(n) p )+ CM trn gi s l sai+ Kt lun: A(n) p
(hoc A(n) p )V d 1: CMR n2 + 3n + 5 121 vi n NGi s tn ti n N sao
cho n2 + 3n + 5 1214n2 + 12n + 20 121 (v (n, 121) = 1)(2n + 3)2 +
11 121 (1)(2n + 3)2 11V 11 l s nguyn t 2n + 3 11(2n + 3)2 121 (2)T
(1) v (2) 11 121 v lVy n2 + 3n + 5 121V d 2: CMR n2 - 1 n vi n
N*GiiXt tp hp s t nhin N*Gi s n 1, n N* sao cho n2 - 1 nGi d l c s
chung nh nht khc 1 ca n d (p) theo nh l Format ta c2d-1 1 (mod d) m
< dta chng minh m\nGi s n = mq + r (0 r < m)15Trng THCS M
Quang: CHUYN BI DNG HC SINH GIITheo gi s n2 - 1 n nmq+r - 1 n2r(nmq
- 1) + (2r - 1) n 2r - 1 d v r < m m m N, m nh nht khc 1 c tnh
cht (1)r = 0 m\n m m < d cng c tnh cht (1) nn iu gi s l sai.Vy
n2 - 1 n vi n N*BI TP TNG TBi 1: C tn ti n N sao cho n2 + n + 2 49
khng?Bi 2: CMR: n2 + n + 1 9 vi n N*Bi 3: CMR: 4n2 - 4n + 18 289 vi
n NHNG DN - P SBi 1: Gi s tn ti n N n2 + n + 2 49 4n2 + 4n + 8
49(2n + 1)2 + 7 49 (1) (2n + 1)2 7V 7 l s nguyn t 2n + 1 7 (2n +
1)2 49 (2)T (1); (2) 7 49 v l.Bi 2: Gi s tn ti n2 + n + 1 9 vi n(n
+ 2)(n - 1) + 3 3 (1)v 3 l s nguyn t
+3 13 2nn (n + 2)(n - 1) 9 (2)T (1) v (2) 3 9 v lBi 3: Gi s n N
4n2 - 4n + 18 289 (2n - 1)2 + 17 172(2n - 1) 1717 l s nguyn t (2n -
1) 17 (2n - 1)2 28917 289 v l.16
