Chuyn Bi Dng HSG Ton 6S CHNH PHNGI. NH NGHA:S chnh phng l s bng bnh phng ng ca mt s nguyn.II. TNH CHT:1. S chnh phng ch c th c ch s tn cng bng 0, 1, 4, 5, 6, 9 ; khng th c ch s tn cng bng 2, 3, 7, 8.2. Khi phn tch ra tha s nguyn t, s chnh phng ch cha cc tha s nguyn t vi s m chn.3. S chnh phng ch c th c mt trong hai dng 4n hoc 4n + 1. Khng c s chnh phng no c dng 4n + 2 hoc 4n + 3 (n N).4. S chnh phng ch c th c mt trong hai dng 3n hoc 3n + 1. Khng c s chnh phng no c dng 3n + 2 (n N).5. S chnh phngtn cng bng 1 hoc 9 th ch s hng chc l ch s chn.S chnh phng tn cng bng 5 th ch s hng chc l 2S chnh phng tn cng bng 4 th ch s hng chc l ch s chn.S chnh phng tn cng bng 6 th ch s hng chc l ch s l.6. S chnh phng chia ht cho 2 th chia ht cho 4. S chnh phng chia ht cho 3 th chia ht cho 9. S chnh phng chia ht cho 5 th chia ht cho 25. S chnh phng chia ht cho 8 th chia ht cho 16.III. MT S DNG BI TP V S CHNH PHNGA. DNG1 : CHNG MINH MT S L S CHNH PHNGBi 1: Chng minh rng vi mi s nguyn x, y th A = (x + y)(x + 2y)(x + 3y)(x + 4y) + y4 l s chnh phng.Ta c A = (x + y)(x + 2y)(x + 3y)(x + 4y) + y4 = (x2 + 5xy + 4y2)( x2 + 5xy + 6y2) + y4 tx2 + 5xy + 5y2 = t ( t Z)thA = (t - y2)( t + y2) + y4 = t2 y4 + y4 = t2 = (x2 + 5xy + 5y2)2 V x, y, z Z nn x2 Z,5xy Z,5y2 Z x2 + 5xy + 5y2 ZVy A l s chnh phng.Bi 2: Chng minh tch ca 4 s t nhin lin tip cng 1 lun l s chnh phng.Gi 4 s t nhin, lin tip l n, n + 1, n+ 2, n + 3(n N). Ta cTrang 1Chuyn Bi Dng HSG Ton 6n(n + 1)(n + 2)(n + 3) + 1 = n.(n + 3(n + 1)(n + 2) + 1 = (n2 + 3n)( n2 + 3n + 2) + 1 (*)tn2 + 3n = t(t N) th (*) = t( t + 2 ) + 1 = t2 + 2t + 1 = ( t + 1 )2 = (n2 + 3n + 1)2Vn N nn n2 + 3n + 1 N Vy n(n + 1)(n + 2)(n + 3) + 1 l s chnh phng.Bi 3: Cho S = 1.2.3 + 2.3.4 + 3.4.5 + . . . + k(k+1)(k+2) Chng minh rng 4S + 1 l s chnh phng .Ta ck(k+1)(k+2) = 41 k(k+1)(k+2).4 = 41 k(k+1)(k+2).[(k+3) (k-1)] =41 k(k+1)(k+2)(k+3) - 41 k(k+1)(k+2)(k-1)S =41.1.2.3.4 -41.0.1.2.3 + 41.2.3.4.5 -41.1.2.3.4 ++41 k(k+1)(k+2)(k+3) - 41 k(k+1)(k+2)(k-1) = 41 k(k+1)(k+2)(k+3)4S + 1 = k(k+1)(k+2)(k+3) + 1Theo kt qu bi 2 k(k+1)(k+2)(k+3) + 1 l s chnh ph ng.Bi 4: Cho dy s 49; 4489; 444889; 44448889; Dy s trn c xy dng bng cch thm s 48 vo gia s ng trc n. Chng minh rng tt c cc s ca dy trn u l s chnh phng.Ta c 4448889 = 44488..8 + 1 = 444 . 10n + 8 . 111 + 1 n ch s 4n-1 ch s 8 n ch s 4 n ch s 8 n ch s 4n ch s 1

=4. 91 10 n. 10n+ 8. 91 10 n + 1 = 99 8 10 . 8 10 . 4 10 . 42+ + n n n =91 10 . 4 10 . 42+ +n n=

,_

+31 10 . 2nTa thy 2.10n +1=20001 c tng cc ch s chia ht cho 3 nn n chia ht cho 3 n-1 ch s 0

,_

+31 10 . 2n Z haycc s c dng 4448889 l s chnh phng.Bi 5: Chng minh rng cc s sau y l s chnh phng: A = 111 + 444 + 1

2n ch s 1 n ch s 4 B =111 + 111 + 666 + 8Trang 222Chuyn Bi Dng HSG Ton 6

2n ch s 1n+1 ch s 1n ch s 6

C = 444 + 222 + 888 + 7

2n ch s 4 n+1 ch s 2n ch s 8Kt qu: A =

,_

+32 10n; B =

,_

+38 10n ; C =

,_

+37 10 . 2nBi 6: Chng minh rng cc s sau l s chnh phng:

a.A = 22499910009n-2 ch s 9 n ch s 0 b.B = 1115556 n ch s 1 n-1 ch s 5a. A = 224.102n + 999.10n+2 + 10n+1 + 9= 224.102n + ( 10n-2 1 ) . 10n+2 + 10n+1 + 9 = 224.102n + 102n 10n+2 + 10n+1 + 9= 225.102n 90.10n + 9 = ( 15.10n 3 ) 2 A l s chnh phng b.B = 11115555 + 1 = 111.10n + 5.111 + 1 n ch s 1 n ch s 5 n ch s 1 n ch s 1= 91 10 n. 10n + 5. 91 10 n + 1 = 99 5 10 . 5 10 102+ + n n n=94 10 . 4 102+ +n n =

,_

+32 10n l s chnh phng ( iu phi chng minh)Bi 7:Chng minh rng tng cc bnh phng ca 5 s t nhin lin tip khng th l mt s chnh phngGi 5 s t nhin lin tip l n-2, n-1, n , n+1 , n+2(n N , n 2 ).Ta c ( n-2)2 + (n-1)2 + n2 + ( n+1)2 + ( n+2)2 = 5.( n2+2)V n2 khng th tn cng bi 3 hoc 8 do n2+2 khng th chia ht cho 5 5.( n2+2) khng l s chnh phng hay A khng l s chnh phngBi 8:Chng minh rngs c dng n6 n4 + 2n3 + 2n2 trong nN v n>1 khng phi l s chnh phngTrang 322 22Chuyn Bi Dng HSG Ton 6n6 n4 + 2n3 +2n2 = n2.( n4 n2 + 2n +2 ) = n2.[ n2(n-1)(n+1) + 2(n+1) ] = n2[ (n+1)(n3 n2 + 2) ] = n2(n+1).[ (n3+1) (n2-1) ]= n2( n+1 )2.( n22n+2)Vi nN, n >1 th n2-2n+2 = (n - 1)2+ 1 > ( n 1 )2 v n2 2n + 2 = n2 2(n - 1) < n2 Vy ( n 1)2 < n2 2n + 2 < n2 n2 2n + 2 khng phi l mt s chnh phng. Bi 9: Cho 5 s chnh phng bt k c ch s hng chc khc nhau cn ch s hng n v u l 6. Chng minh rng tng cc ch s hng chc ca 5 s chnh phng l mt s chnh phng Cch 1: Ta bit mt s chnh phng c ch s hng n v l 6 th ch s hng chc ca n l s l. V vy ch s hng chc ca 5 s chnh phng cho l 1,3,5,7,9 khi tng ca chng bng 1 + 3 + 5 + 7 + 9 = 25 = 52 l s chnh phng Cch 2: Nu mt s chnh phng M = a2 c ch s hng n v l 6 th ch s tn cng ca a l 4 hoc 6 a2 a2 4 Theo du hiu chia ht cho 4 th hai ch s tn cng ca M ch c th l 16, 36, 56, 76, 96 Ta c: 1 + 3 + 5 + 7 + 9 = 25 = 52 l s chnh phng.Bi 10: Chng minh rng tng bnh phng ca hai s l bt k khng phi l mt s chnh phng.a v b l nn a = 2k+1, b = 2m+1 (Vi k, m N) a2 + b2= (2k+1)2 + (2m+1)2 = 4k2 + 4k + 1 + 4m2 + 4m + 1 = 4(k2 + k + m2 + m) + 2=4t + 2(Vi t N)Khng c s chnh phng no c dng 4t + 2(t N) do a2 + b2khng th l s chnh phng.Bi 11: Chng minh rng nu p l tch ca n s nguyn t u tin th p-1 v p+1 khng th l cc s chnh phng.V p l tch ca n s nguyn t u tin nn p2 v p khng chia ht cho 4 (1)a. Gi s p+1 l s chnh phng . t p+1 = m2 (m N)V p chn nn p+1 l m2 l m l.t m = 2k+1(k N). Ta c m2 = 4k2 + 4k + 1 p+1 = 4k2 + 4k + 1 p = 4k2 + 4k = 4k(k+1) 4 mu thun vi (1) p+1 l s chnh phngb. p = 2.3.5 l s chia ht cho 3 p-1 c dng 3k+2.Khng c s chnh phng no c dng 3k+2 p-1 khng l s chnh phng .Vy nu p l tch n s nguyn t u tin th p-1 v p+1 khng l s chnh phngTrang 4Chuyn Bi Dng HSG Ton 6Bi 12: Gi s N = 1.3.5.72007.Chng minh rng trong 3 s nguyn lin tip 2N-1, 2N v 2N+1 khng c s no l s chnh phng.a. 2N-1 = 2.1.3.5.72007 1C 2N 3 2N-1 khng chia ht cho 3 v 2N-1 = 3k+2(k N) 2N-1 khng l s chnh phng.b. 2N = 2.1.3.5.72007V N l N khng chia ht cho 2 v 2N 2 nhng 2N khng chia ht cho 4.2N chn nn 2N khng chia cho 4 d 1 2N khng l s chnh phng.c. 2N+1 = 2.1.3.5.72007 + 12N+1 l nn 2N+1 khng chia ht cho 4 2N khng chia ht cho 4 nn 2N+1 khng chia cho 4 d 1 2N+1 khng l s chnh phng.Bi 13: Cho a = 111 ;b = 10005

2008 ch s 12007 ch s 0Chng minh 1 + abl s t nhin.Cch 1: Ta c a = 111 = 91 102008;b = 10005 = 1000 + 5 = 102008 + 5 2008 ch s 1 2007 ch s 02008 ch s 0

ab+1 =9) 5 10 )( 1 10 (2008 2008+ + 1 = 99 5 10 . 4 ) 10 (2008 2 2008+ + =

,_

+32 1020081 + ab=

,_

+32 102008 = 32 102008+Ta thy 102008 + 2 = 10002 3 nn32 102008+ N hay1 + abl s t nhin.2007 ch s 0Cch 2: b = 10005 = 1000 1 + 6 = 999 + 6 = 9a +6

2007 ch s 02008 ch s 02008 ch s 9 ab+1 = a(9a +6) + 1 = 9a2 + 6a + 1 = (3a+1)21 + ab= 2) 1 3 ( + a = 3a + 1 NB. DNG 2 : TM GI TR CA BIN BIU THC L S CHNH PHNGBi1: Tm s t nhin n sao cho cc s sau l s chnh phng:a.n2 + 2n + 12 b.n ( n+3 ) Trang 522Chuyn Bi Dng HSG Ton 6c. 13n + 3 d. n2 + n + 1589Giia. V n2 + 2n + 12 l s chnh phng nn t n2 + 2n + 12 = k2(k N)

(n2 + 2n + 1) + 11 = k2

k2 (n+1)2 = 11 (k+n+1)(k-n-1) = 11Nhn xt thy k+n+1 > k-n-1 v chng l nhng s nguyn dng, nn ta c th vit (k+n+1)(k-n-1) = 11.1 k+n+1 = 11 k = 6k n - 1 = 1n = 4b. tn(n+3) = a2(n N) n2 + 3n = a2 4n2 + 12n = 4a2 (4n2 + 12n + 9) 9 = 4a2 (2n + 3)2- 4a2 = 9(2n + 3 + 2a)(2n + 3 2a)= 9Nhn xt thy 2n + 3 + 2a > 2n + 3 2a v chng l nhng s nguyn dng, nn ta c th vit (2n + 3 + 2a)(2n + 3 2a) = 9.1 2n + 3 + 2a = 9 n = 1 2n + 3 2a = 1 a = 2c. t 13n + 3 = y2 ( y N) 13(n 1) = y2 16 13(n 1) = (y + 4)(y 4) (y + 4)(y 4) 13 m 13 l s nguyn t nn y + 4 13 hoc y 4 13 y = 13k t 4(Vi k N) 13(n 1) = (13k t 4)2 16 = 13k.(13k t 8) n = 13k2 t 8k + 1Vy n = 13k2 t 8k + 1 (Vi k N) th 13n + 3 l s chnh phng.d. t n2 + n + 1589 = m2 (m N) (4n2 + 1)2 + 6355 = 4m2

(2m + 2n +1)(2m 2n -1) = 6355Nhn xt thy 2m + 2n +1> 2m 2n -1 > 0 v chng l nhng s l, nn ta c th vit (2m + 2n +1)(2m 2n -1) = 6355.1 = 1271.5 = 205.31 = 155.41Suy ra n c th c cc gi tr sau: 1588; 316; 43; 28.Bi 2:Tm a cc s sau l nhng s chnh phng:a. a2 + a + 43 b. a2 + 81c. a2 + 31a + 1984Kt qu: a.2; 42; 13b.0; 12; 40c.12; 33; 48; 97; 176; 332; 565; 1728Bi 3:Tm s t nhin n 1 sao cho tng 1! + 2! + 3! + + n! l mt s chnh phng .Trang 6Chuyn Bi Dng HSG Ton 6Vi n = 1 th 1! = 1 = 12l s chnh phng .Vi n = 2 th 1! + 2! = 3 khng l s chnh phng Vi n = 3 th 1! + 2! + 3! = 1+1.2+1.2.3 = 9 = 32 l s chnh phng Vi n 4 ta c 1! + 2! + 3! + 4! = 1+1.2+1.2.3+1.2.3.4 = 33 cn 5!;6!; ; n! u tn cng bi 0 do 1! + 2! + 3! + + n! c tn cng bi ch s 3 nn n khng phi l s chnh phng .Vy c 2 s t nhin n tha mn bi l n = 1; n = 3.Bi 4: Tm n N cc s sau l s chnh phng: a. n2 + 2004 ( Kt qu: 500; 164)b. (23 n)(n 3)( Kt qu: 3; 5; 7; 13; 19; 21; 23)c. n2 + 4n + 97 d. 2n + 15Bi 5:C hay khng s t nhin n 2006 + n2 l s chnh phng. Gi s 2006 + n2 l s chnh phng th 2006 + n2 = m2 (m N)T suy ra m2 n2 = 2006 (m + n)(m - n) = 2006 Nh vy trong 2 s m v n phi c t nht 1 s chn (1)Mt khcm + n + m n = 2m 2 s m + n v m n cng tnh chn l (2)T (1) v (2) m + n v m n l 2 s chn

(m + n)(m - n) 4Nhng 2006 khng chia ht cho 4

iu gi s sai. Vy khng tn ti s t nhin n 2006 + n2 l s chnh phng.Bi 6: Bit x Nv x>2. Tm x sao cho x(x-1).x(x-1) = (x-2)xx(x-1) ng thc cho c vit li nh sau: x(x-1) =(x-2)xx(x-1)Do v tri l mt s chnh phng nn v phi cng l mt s chnh phng .Mt s chnh phng ch c th tn cng bi 1 trong cc ch s 0; 1; 4; 5; 6; 9 nn x ch c th tn cng bi 1 trong cc ch s 1; 2; 5; 6; 7; 0 (1)Do x l ch s nn x 9, kt hp vi iu kin bi ta c xN v 2 < x 9 (2)T (1) v (2) x ch c th nhn 1 trong cc gi tr5; 6; 7.Bng php th ta thy ch c x = 7 tha mn bi, khi 762 = 5776Bi 7: Tm s t nhin n c 2 ch s bit rng 2n+1 v 3n+1 u l cc s chnh phng.Ta c 10 n 99 nn 21 2n+1 199. Tm s chnh phng l trong khong trn ta c 25; 49; 81; 121; 169 tng ng vi s n bng 12; 24; 40; 60; 84.Trang 72Chuyn Bi Dng HSG Ton 6S 3n+1 bng 37; 73; 121; 181; 253. Ch c 121 l s chnh phng.Vy n = 40Bi 8: Chng minh rng nu n l s t nhin sao cho n+1 v 2n+1 u l cc s chnh phng th n l bi s ca 24.V n+1 v 2n+1 l cc s chnh phng nn t n+1 = k2 , 2n+1 = m2(k, m N)Ta c m l s l m = 2a+1 m2 = 4a (a+1) + 1

n = 212 m = 2) 1 ( 4 + a a = 2a(a+1) n chn n+1 l k l t k = 2b+1 (Vi b N) k2= 4b(b+1) +1

n = 4b(b+1) n 8(1)Ta c k2 + m2 = 3n + 2 2 (mod3)Mt khc k2 chia cho 3 d 0 hoc 1, m2 chia cho 3 d 0 hoc 1. Nn k2 + m2

2 (mod3) th k2 1 (mod3)m2 1 (mod3) m2 k2 3 hay (2n+1) (n+1) 3 n 3(2)M (8; 3) = 1 (3)T (1), (2), (3) n 24.Bi 9: Tm tt c cc s t nhin n sao cho s 28 + 211 + 2n l s chnh phng .Gi s 28 + 211 + 2n = a2 (a N) th 2n = a2 482 = (a+48)(a-48) 2p.2q = (a+48)(a-48) Vi p, q N ; p+q = nv p > q a+48 = 2p

2p 2q = 96 2q (2p-q -1) = 25.3 a- 48 = 2q

q = 5 v p-q = 2 p = 7 n = 5+7 = 12Th li ta c:28 + 211 + 2n = 802C.DNG 3: TM S CHNH PHNG Bi 1: Cho A l s chnh phng gm 4 ch s. Nu ta thm vo mi ch s ca A mtn v th ta c s chnh phng B. Hy tm cc s A v B.Gi A = abcd = k2. Nu thm vo mi ch s ca A mtn v th ta c s B = (a+1)(b+1)(c+1)(d+1) = m2vi k, m N v 32 < k < m < 100a, b, c, dN ; 1 a 9 ; 0 b, c, d 9 Ta c A = abcd = k2

B = abcd + 1111 = m2 Trang 8Chuyn Bi Dng HSG Ton 6 m2 k2 = 1111 (m-k)(m+k) = 1111(*)Nhn xt thy tch (m-k)(m+k) > 0 nn m-k v m+k l 2 s nguyn dng.Vm-k < m+k < 200 nn (*) c th vit(m-k)(m+k) = 11.101Do m k == 11 m = 56A = 2025m + k = 101n = 45 B = 3136 Bi 2: Tm 1 s chnh phng gm 4 ch s bit rng s gm 2 ch s u ln hn s gm 2 ch s sau 1 n v.t abcd = k2 ta c ab cd = 1v k N, 32 k < 100 Suy ra 101cd = k2 100 = (k-10)(k+10) k +10 101 hoc k-10 101M (k-10; 101) = 1 k +10 101V 32 k < 100 nn 42 k+10 < 110 k+10 = 101 k = 91 abcd = 912 = 8281Bi 3: Tm s chnh phng c 4 ch s bit rng 2 ch s u ging nhau, 2 ch s cui ging nhau.Gi s chnh phng phi tm l aabb = n2vi a, b N,1 a 9; 0 b 9Ta c n2 = aabb = 11.a0b = 11.(100a+b) = 11.(99a+a+b) (1)Nhn xt thy aabb 11 a + b 11M 1 a 9 ;0 b 9 nn 1 a+b 18 a+b = 11Thay a+b = 11 vo (1) c n2 = 112(9a+1) do 9a+1 l s chnh phng .Bng php th vi a = 1; 2; ; 9 ta thy ch c a = 7 tha mn b = 4S cn tm l 7744Bi 4: Tm mt s c 4 ch s va l s chnh phng va l mt lp phng.Gi s chnh phng l abcd . V abcd va l s chnh phng va l mt lp phng nn t abcd = x2 = y3 Vi x, y NV y3 = x2 nn y cng l mt s chnh phng .Ta c 1000 abcd 9999 10 y 21 v y chnh phng y = 16 abcd = 4096 Bi 5: Tm mt s chnh phng gm 4 ch s sao cho ch s cui l s nguyn t,cn bc hai ca s c tng cc ch s l mt s chnh phng.Gi s phi tm l abcdvi a, b, c, d nguyn v 1 a 9 ; 0 b,c,d 9abcd chnh phng d{ 0,1,4,5,6,9}d nguyn t d = 5Trang 9Chuyn Bi Dng HSG Ton 6t abcd = k2 < 10000 32 k < 100k l mt s c hai ch s m k2 c tn cng bng 5 k tn cng bng 5Tng cc ch s ca k l mt s chnh phng k = 45 abcd = 2025Vy s phi tm l 2025Bi 6: Tm s t nhin c hai ch s bit rng hiu cc bnh phng ca s v vit s bi hai ch s ca s nhng theo th t ngc li l mt s chnh phngGi s t nhin c hai ch s phi tm lab ( a,b N, 1 a,b 9 )S vit theo th t ngc li ba Ta cab-ba

= ( 10a + b ) 2 ( 10b + a )2 = 99 ( a2 b2 ) 11 a2 - b2 11Hay ( a-b )(a+b ) 11 V 0 < a - b 8 , 2 a+b 18 nn a+b 11 a + b = 11Khi ab -ba = 32 . 112 . (a - b)ab -ba l s chnh phng th a - b phi l s chnh phng do a-b = 1 hoc a - b = 4 Nu a-b = 1 kt hp vi a+b = 11 a = 6, b = 5, ab = 65 Khi 652 562 = 1089 = 332 Nu a - b = 4 kt hp vi a+b = 11 a = 7,5 ( loi )Vy s phi tm l 65Bi 7:Cho mt s chnh phng c 4 ch s. Nu thm 3 vo mi ch s ta cng c mt s chnh phng. Tm s chnh phng ban u ( Kt qu: 1156 )Bi 8:Tm s c 2 ch s m bnh phng ca s y bng lp phng ca tng cc ch s ca n. Gi s phi tm lab vi a,b N v 1 a 9 , 0 b 9Theo gi thit ta c : ab = ( a + b )3 (10a+b)2 = ( a + b )3 ab l mt lp phng v a+b l mt s chnh phngt ab= t3 ( t N ) , a + b = l 2 ( l N )V 10 ab 99 ab= 27 hoc ab = 64 Nu ab = 27 a + b = 9 l s chnh phng Nuab = 64 a + b = 10 khng l s chnh phng loiVy s cn tm l ab = 27Trang 102 22 22 22Chuyn Bi Dng HSG Ton 6Bi 9: Tm 3 s l lin tip m tng bnh phng l mt s c 4 ch s ging nhau.Gi 3 s l lin tip l 2n-1, 2n+1, 2n+3( n N)Ta cA= ( 2n-1 )2 + ( 2n+1)2 + ( 2n+3 )2= 12n2 + 12n + 11Theo bi ta t 12n2 + 12n + 11 = aaaa= 1111.avi a l v 1 a 9 12n( n + 1 ) = 11(101a 1 )

101a 1 3 2a 1 3 V 1 a 9 nn 1 2a-1 17 v 2a-1 l nn 2a 1 { 3; 9; 15 } a { 2; 5; 8 } V a l a = 5 n = 21 3 s cn tm l 41; 43; 45Bi 10: Tm s c 2 ch s sao cho tch ca s vi tng cc ch s ca n bng tng lp phng cc ch s ca s .ab (a + b ) = a3 + b310a + b = a2 ab + b2 = ( a + b )2 3ab 3a( 3 + b ) = ( a + b ) ( a + b 1 )a + b v a + b 1 nguyn t cng nhau do a + b = 3a hoca + b 1 = 3aa+ b 1 = 3 + ba + b = 3 + b a = 4 , b = 8hoca = 3 , b = 7Vy ab= 48hoc ab= 37.... Ht .Trang 11Chuyn Bi Dng HSG Ton 6S nguyn tI. Kin thc cn nh:1. Dnh ngha: * S nguyn t l s t nhin ln hn 1, ch c hai c l 1 v chnh n.* Hp s l s t nhin ln hn 1, c nhiu hn hai c.2. Tnh cht:* Nu s nguyn t p chia ht cho s nguyn t q th p = q.* Nu tch abc chia ht cho s nguyn t p th t nht mt tha s ca tch abc chia ht cho s nguyn t p.* Nu a v b khng chia ht cho s nguyn t p th tch ab khng chia ht cho s nguyn t p .3. Cch nhn bit mt s nguyn t:a) Chia s ln lt cho cc s nguyn t bit t nh n ln.- Nu c mt php chia ht th s khng phi l s nguyn t.- Nu chia cho n lc s thng nh hn s chia m cc php chia vn cn s d th ss l s nguyn t.b) Mt s c 2 c s ln hn 1 th s khng phi l s nguyn t.4. Phn tch mt s ra tha s nguyn t:* Phn tch mt s t nhin ln hn 1 ra tha s nguyn t l vit s di dng mt tch cc tha s nguyn t.- Dng phn tch ra tha s nguyn t ca mi s nguyn t l chnh s .- Mi hp s u phn tch c ra tha s nguyn t.. ..... i, , nhng s nguyn t.,, ...,N v,, ...,1A a b cV a b c l 5. S cc c s v tng cc c s ca mt s:+1 1 1 s. ..... i, , nhng s nguyn t.,, ...,N v,, ...,11. S cc c s ca A l: ( +1)( +1)...( +1).a 1 1 12. Tng cc c s ca A l:. ...1 1 1Gi A a b cV a b c lb ca b c + + 6. S nguyn t cng nhau: * Hai s nguyn t cng nhau l hai s c CLN bng 1.Trang 12Chuyn Bi Dng HSG Ton 6Hai s a v b nguyn t cng nhau CLN(a, b) = 1.Cc s a, b, c nguyn t cng nhau CLN(a, b, c) = 1.Cc s a, b, c i mt nguyn t cng nhau CLN(a, b) = CLN(b, c) = CLN(c, a) =1.II. Cc v d:VD1: Ta bit rng c 25 s nguyn t nh hn 100. Tng ca 25 s nguyn t l s chn hay s l.HD:Trong 25 s nguyn t nh hn 100 c cha mt s nguyn t chn duy nht l 2, cn 24 s nguyn t cn li l s l. Do tng ca 25 s nguyn t l s chn.VD2: Tng ca 3 s nguyn t bng 1012. Tm s nguyn t nh nht trong ba s nguyn t .HD:V tng ca 3 s nguyn t bng 1012, nn trong 3 s nguyn t tn ti t nht mt s nguyn t chn. M s nguyn t chn duy nht l 2 v l s nguyn t nh nht. Vy s nguyn t nh nht trong 3 s nguyn t l 2.VD3: Tng ca 2 s nguyn t c th bng 2003 hay khng? V sao?HD:V tng ca 2 s nguyn t bng 2003, nn trong 2 s nguyn t tn ti 1 s nguyn t chn. M s nguyn t chn duy nht l 2. Do s nguyn t cn li l 2001. Do 2001 chia ht cho 3 v 2001 > 3. Suy ra 2001 khng phi l s nguyn t.VD4: Tm s nguyn t p, sao cho p + 2 v p + 4 cng l cc s nguyn t.HD:Gi s p l s nguyn t.- Nu p = 2 th p + 2 = 4 v p + 4 = 6 u khng phi l s nguyn t.- Nu p 3 th s nguyn t p c 1 trong 3 dng: 3k, 3k + 1, 3k + 2 vi k N*.+) Nu p = 3k p = 3 p + 2 = 5 v p + 4 = 7 u l cc s nguyn t.+) Nu p = 3k +1 th p + 2 = 3k + 3 = 3(k + 1) p + 2 M3 v p + 2 > 3. Do p + 2 l hp s.+) Nu p = 3k + 2 th p + 4 = 3k + 6 = 3(k + 2) p + 4 M3 v p + 4 > 3. Do p + 4 l hp s.Vy vi p = 3 th p + 2 v p + 4 cng l cc s nguyn t.Trang 13Chuyn Bi Dng HSG Ton 6VD5: Cho p v p + 4 l cc s nguyn t (p > 3). Chng minh rng p + 8 l hp s.HD:V p l s nguyn t v p > 3, nn s nguyn t p c 1 trong 2 dng: 3k + 1, 3k + 2 vi k N*.- Nu p = 3k + 2 th p + 4 = 3k + 6 = 3(k + 2) p + 4 M3 v p + 4 > 3. Do p + 4 l hp s ( Tri vi bi p + 4 l s nguyn t).- Nu p = 3k + 1 th p + 8 = 3k + 9 = 3(k + 3) p + 8 M3 v p + 8 > 3. Do p + 8 l hp s.Vy s nguyn t p c dng: p = 3k + 1 th p + 8 l hp s.VD6: Chng minh rng mi s nguyn t ln hn 2 u c dng 4n + 1 hoc 4n 1.HD:Mi s t nhin n khi chia cho 4 c th c 1 trong cc s d: 0; 1; 2; 3. Do mi s t nhin n u c th vit c di 1 trong 4 dng:4k, 4k + 1, 4k + 2, 4k + 3 vi k N*.- Nu n = 4k nM4 n l hp s.- Nu n = 4k + 2 nM2 n l hp s.Vy mi s nguyn t ln hn 2 u c dng 4k + 1 hoc 4k 1. Hay mi s nguyn t ln hn 2 u c dng 4n + 1 hoc 4n 1 vi n N*.VD7: Tm ss nguyn t, bit rng s bng tng ca hai s nguyn t v bng hiu ca hai s nguyn t.HD: s a, b, c, d, e l cc s nguyn t v d >e.Theo bi ra: a =b +c =d - e(*).T (*) a >2 a l s nguyn t l . b +c v d - e l s l .Do b, d l cc s nguyn t b, d l s lc, eGi l s chn. c =e =2 (do c, e l cc s nguyn t). a =b +2 =d - 2d =b +4.Vy ta cn tm s nguyn t b sao cho b +2 v b +4 cng l cc s nguyn t. VD8: Tm tt c cc s nguyn t x, y sao cho: x2 6y2 = 1.HD:Trang 14Chuyn Bi Dng HSG Ton 62 2 2 2 222 22: x 6 1 1 6 ( 1)( 1) 66 2 ( 1)( 1) 2 x - 1 +x +1 =2x x - 1 v x +1 c c ng tnh chn l .x - 1 v x +1 l hai s chn lin tip( 1)( 1) 8 6 8 3 42 2 2 5Ta c y x y x x yDoy x xMx x y yy y y x + + + M MM M MM MVD9: Cho p v p + 2 l cc s nguyn t (p > 3). Chng minh rng p + 1M6.HD:V p l s nguyn t v p > 3, nn s nguyn t p c 1 trong 2 dng: 3k + 1, 3k + 2 vi k N*.- Nu p = 3k + 1 th p + 2 = 3k + 3 = 3(k + 1) p + 2 M3 v p + 2 > 3. Do p + 2 l hp s ( Tri vi bi p + 2 l s nguyn t).- Nu p = 3k + 2 th p + 1 = 3k + 3 = 3(k + 1)(1). Do p l s nguyn t v p > 3 p l k l k + 1 chn k + 1M2 (2)T (1) v (2) p + 1M6.II. Bi tp vn dng:Bi 1: Tm s nguyn t p sao cho cc s sau cng l s nguyn t:a) p + 2 v p + 10.b) p + 10 v p + 20.c) p + 10 v p + 14.d) p + 14 v p + 20.e) p + 2v p + 8.f) p + 2 v p + 14.g) p + 4 v p + 10.h) p + 8 v p + 10.Bi 2: Tm s nguyn t p sao cho cc s sau cng l s nguyn t:a) p + 2, p + 8, p + 12, p + 14.b) p + 2, p + 6, p + 8, p + 14.c) p + 6, p + 8, p + 12, p + 14.d) p + 2, p + 6, p + 8, p + 12, p + 14.e) p + 6, p + 12, p + 18, p + 24.f) p + 18, p + 24, p + 26, p + 32.g) p + 4, p + 6, p + 10, p + 12, p+16.Bi 3:a) Cho p v p + 4 l cc s nguyn t (p > 3). Chng minh rng: p + 8 l hp s.Trang 15Chuyn Bi Dng HSG Ton 6b) Cho p v 2p + 1 l cc s nguyn t (p > 3). Chng minh rng: 4p + 1 l hp s.c) Cho p v 10p + 1 l cc s nguyn t (p > 3). Chng minh rng: 5p + 1 l hp s.d) Cho p v p + 8 l cc s nguyn t (p > 3). Chng minh rng: p + 4 l hp s.e) Cho p v 4p + 1 l cc s nguyn t (p > 3). Chng minh rng: 2p + 1 l hp s.f) Cho p v 5p + 1 l cc s nguyn t (p > 3). Chng minh rng: 10p + 1 l hp s.g) Cho p v 8p + 1 l cc s nguyn t (p > 3). Chng minh rng: 8p - 1 l hp s.h) Cho p v 8p - 1 l cc s nguyn t (p > 3). Chng minh rng: 8p + 1 l hp s.i) Cho p v 8p2 - 1 l cc s nguyn t (p > 3). Chng minh rng: 8p2 + 1 l hp s.j) Cho p v 8p2 + 1 l cc s nguyn t (p > 3). Chng minh rng: 8p2 - 1 l hp s.Bi 4: Chng minh rng:a) Nu p v q l hai s nguyn t ln hn 3 th p2 q2 M24.b) Nu a, a + k, a + 2k (a, k N*) l cc s nguyn t ln hn 3 th k M6.Bi 5: a) Mt s nguyn t chia cho 42 c s d r l hp s. Tm s d r.b) Mt s nguyn t chia cho 30 c s d r. Tm s d r bit rng r khng l s nguyn t.Bi 6: Hai s nguyn t gi l sinh i nu chng l hai s nguyn t l lin tip. Chng minh rng mt s t nhin ln hn 3 nm gia hai s nguyn t sinh i th chia ht cho 6.Bi 7: Cho 3 s nguyn t ln hn 3, trong s sau ln hn s trc l d n v. Chng minh rng d chia ht cho 6.Bi 8: Tm s nguyn t c ba ch s, bit rng nu vit s theo th t ngc li th ta c mt s l lp phng ca mt s t nhin.Bi 9: Tm s t nhin c 4 ch s, ch s hng nghn bng ch s hng n v, ch s hng trm bng ch s hng chc v s vit c di dng tch ca 3 s nguyn t lin tip.Bi 10: Tm 3 s nguyn t l lin tip u l cc s nguyn t.Bi 11: Tm 3 s nguyn t lin tip p, q, r sao cho p2 + q2 + r2 cng l s nguyn t.Bi 12: Tm tt c cc b ba s nguyn t a, b, c sao cho a.b.c < a.b + b.c + c.a.Trang 16Chuyn Bi Dng HSG Ton 6Bi 13: Tm 3 s nguyn t p, q, r sao cho pq + qp = r.Bi 14: Tm cc s nguyn t x, y, z tho mn xy + 1 = z.Bi 15: Tm s nguyn t 2, cc s nguyn t v b . abcd sao cho ab ac l cdb c + Bi 16: Cho cc s p = bc + a, q = ab + c, r = ca + b (a, b, c N*) l cc s nguyn t. Chng minh rng 3 s p, q, r c t nht hai s bng nhau.Bi 17: Tm tt c cc s nguyn t x, y sao cho: a) x2 12y2 = 1.b) 3x2 + 1 = 19y2.c) 5x2 11y2 = 1.d) 7x2 3y2 = 1.e) 13x2 y2 = 3.f) x2 = 8y + 1.Bi 18: Tm 3 s nguyn t sao cho tch ca chng gp 5 ln tng ca chng.Bi 19: Chng minh rng iu kin cn v p v 8p2 + 1 l cc s nguyn t l p = 3.Bi 20: Chng minh rng: Nu a2 b2 l mt s nguyn t th a2 b2 = a + b.Bi 21: Chng minh rng mi s nguyn t ln hn 3 u c dng 6n + 1 hoc 6n 1.Bi 22: Chng minh rng tng bnh phng ca 3 s nguyn t ln hn 3 khng th l mt s nguyn t.Bi 23: Cho s t nhin n2. Gi p1, p2, ..., pn l nhng s nguyn t sao cho pn n + 1. t A = p1.p2 ...pn. Chng minh rng trong dy s cc s t nhin lin tip: A + 2, A + 3, ..., A + (n + 1). Khng cha mt s nguyn t no.Bi 24: Chng minh rng: Nu p l s nguyn t th2.3.4...(p 3)(p 2) - 1Mp.Bi 25: Chng minh rng: Nu p l s nguyn t th2.3.4...(p 2)(p 1) + 1Mp.Trang 17Chuyn Bi Dng HSG Ton 6Chuyn tm ch s tn cngI. Tm mt ch s tn cngTnh cht 1: a) Cc s c ch s tn cng l 0, 1, 5, 6 khi nng ln ly tha bc bt kth ch s tn cng vn khng thay i. b) Cc s c ch s tn cng l 4, 9 khi nng ln ly tha bc l th ch s tn cng vn khng thay i. c) Cc s c ch s tn cng l 3, 7, 9 khi nng ln ly tha bc 4n (n thuc N) th ch s tn cng l 1. d) Cc s c ch s tn cng l 2, 4, 8 khi nng ln ly tha bc 4n (n thuc N) th ch s tn cng l 6. e) Tch ca mt s t nhin c ch s tn cng l 5 vi bt k s t nhin l no cng cho ta s c ch s tn cng l 5.Tnh cht 2: Mt s t nhin bt k, khi nng ln ly tha bc 4n + 1 (n thuc N) th ch s tn cng vn khng thay i. Tnh cht 3: a) S c ch s tn cng l 3 khi nng ln ly tha bc 4n + 3 s c ch s tn cng l 7 ; s c ch s tn cng l 7 khi nng ln ly tha bc 4n + 3 s c ch stn cng l 3. b) S c ch s tn cng l 2 khi nng ln ly tha bc 4n + 3 s c ch s tn cng l 8 ; s c ch s tn cng l 8 khi nng ln ly tha bc 4n + 3 s c ch s tn cng l 2. c) Cc s c ch s tn cng l 0, 1, 4, 5, 6, 9, khi nng ln ly tha bc 4n + 3 skhng thay i ch s tn cng. Bi 1: Tm ch s tn cng ca cc s: a) 799b) 141414c) 6754 Gii: a) Trc ht, ta tm s d ca php chia 99 cho 4: 99 1 = (9 1)(98 + 97 + + 9 + 1) chia ht cho 4 99 = 4k + 1 (k N) 799 = 74k + 1 = 74k.7 Do 74k c ch s tn cng l 1 799 c ch s tn cng l 7.b) D thy 1414 = 4k (k N) 141414 = 144k c ch s tn cng l 6.c) Ta c 567 1 M4 567 = 4k + 1 (k N) 4567 = 44k + 1 = 44k.4 44k c ch s tn cng l 6 nn 4567 c ch s tn cng l 4. Bi 2: Tm ch s tn cng ca cc s:a) 71993b) 21000c) 31993d) 4161e) 432g) 999h) 1945819i) 193023Bi 3: Chng minh rng: a) 8102 2102 M10 b) 175 + 244 1321 M10c) 4343 1717 M10Bi 4: Tm cc s t nhin n n10 + 1 10Trang 18Chuyn Bi Dng HSG Ton 6Bi 5: C tn ti hay khng s t nhin n n2 + n + 2 chia ht cho 5?Bi 6: Tm ch s tn cng ca C = 1.3.5.7..99Ch s tn cng ca mt tng cc ly tha c xc nh bng cch tnh tng cc ch s tn cng ca tng ly tha trong tng. Bi 2: Tm ch s tn cng ca tng S = 21 + 35 + 49 + + 20048009. Gii: Trc ht ta c nhn xt: Mi ly tha trong S u c s m khi chia cho 4 th d 1 (cc ly tha u c dng n4(n 2) + 1, n {2, 3, , 2004}). Theo tnh cht 2, mi ly tha trong S v cc c s tng ng u c ch s tn cng ging nhau, bng ch s tn cng ca tng: (2 + 3 + + 9) + 199.(1 + 2 + + 9) + 1 + 2 + 3 + 4 = 200(1 + 2 + + 9) + 9 = 9009. Vy ch s tn cng ca tng S l 9. Bi 3: Tm ch s tn cng ca tng T = 23 + 37 + 411 + + 20048011. Gii: Trc ht ta c nhn xt: Mi ly tha trong T u c s m khi chia cho 4 th d 3 (cc ly tha u c dng n4(n 2) + 3, n thuc {2, 3, , 2004}). Theo tnh cht 3 th 23 c ch s tn cng l 8 ; 37 c ch s tn cng l 7 ; 411 c ch s tn cng l 4 ; Nh vy, tng T c ch s tn cng bng ch s tn cng ca tng:(8 + 7 + 4 + 5 + 6 + 3 + 2 + 9) + 199.(1 + 8 + 7 + 4 + 5 + 6 + 3 + 2 + 9) + 1 + 8 + 7 + 4 = 200(1 + 8 + 7 + 4 + 5 + 6 + 3 + 2 + 9) + 8 + 7 + 4 = 9019. Vy: ch s tn cng ca tng T l 9. Bi 4: Tn ti hay khng s t nhin n sao cho n2 + n + 1 chia ht cho 19952000. Gii: 19952000 tn cng bi ch s 5 nn chia ht cho 5. V vy, ta t vn l liu n2 + n + 1 c chia ht cho 5 khng? Ta c n2 + n = n(n + 1), l tch ca hai s t nhin lin tip nn ch s tn cng ca n2 + n ch c th l 0; 2; 6 n2 + n + 1 ch c th tn cng l 1; 3; 7 n2 + n + 1 khng chia ht cho 5. Vy: khng tn ti s t nhin n sao cho n2 + n + 1 chia ht cho 19952000. S dng tnh cht Mt s chnh phng ch c th tn cng bi cc ch s 0 ; 1 ; 4 ; 5 ; 6 ; 9, ta c th gii c Bi sau: Bi 5: Chng minh rng cc tng sau khng th l s chnh phng: a) M = 19k + 5k + 1995k + 1996k (vi k chn) b) N = 20042004k + 2003 S dng tnh cht mt s nguyn t ln hn 5 ch c th tn cng bi cc ch s 1 ; 3 ; 7 ; 9Bi 6: Cho p l s nguyn t ln hn 5. Chng minh rng: p8n +3.p4n 4 chia ht cho 5. Bi 7: Tm s d ca cc php chia: a) 21 + 35 + 49 + + 20038005 cho 5 b) 23 + 37 + 411 + + 20038007 cho 5 Bi 8: Tm ch s tn cng ca X, Y: X = 22 + 36 + 410 + + 20048010 Y = 28 + 312 + 416 + + 20048016 Bi 9: Chng minh rng ch s tn cng ca hai tng sau ging nhau: Trang 19Chuyn Bi Dng HSG Ton 6U = 21 + 35 + 49 + + 20058013 V = 23 + 37 + 411 + + 20058015 Bi 10: Chng minh rng khng tn ti cc s t nhin x, y, z tha mn: 19x + 5y + 1980z = 1975430 + 2004. II. Tm hai ch s tn cng Nhn xt: Nu x N v x = 100k + y, trong k; y N th hai ch s tn cng ca x cng chnh l hai ch s tn cng ca y. Hin nhin l y x. Nh vy, n gin vic tm hai ch s tn cng ca s t nhin x th thay vo ta i tm hai ch s tn cng ca s t nhin y (nh hn). R rng s y cng nh th vic tm cc ch s tn cng ca y cng n gin hn. T nhn xt trn, ta xut phng php tm hai ch s tn cng ca s t nhin x = am nh sau: Trng hp 1: Nu a chn th x = am M2m. Gi n l s t nhin sao cho an 1 M25. Vit m = pn + q (p ; q N), trong q l s nh nht aq M4 ta c:x = am = aq(apn 1) + aq. V an 1 M25 apn 1 M25. Mt khc, do (4, 25) = 1 nn aq(apn 1) M100. Vy hai ch s tn cng ca am cng chnh l hai ch s tn cng ca aq. Tip theo, ta tm hai ch s tn cng ca aq. Trng hp 2: Nu a l , gi n l s t nhin sao cho an 1 M100. Vit m = un + v (u ; v N, 0 v < n) ta c: x = am = av(aun 1) + av V an 1M100 aun 1M100. Vy hai ch s tn cng ca am cng chnh l hai ch s tn cng ca av. Tip theo, ta tm hai ch s tn cng ca av. Trong c hai trng hp trn, cha kha gii c Bi l chng ta phi tm c s t nhin n. Nu n cng nh th q v v cng nh nn s d dng tm hai ch s tn cng ca aq v av. Bi 11: Tm hai ch s tn cng ca cc s: a) a2003 b)799 Gii: a) Do 22003 l s chn, theo trng hp 1, ta tm s t nhin n nh nht sao cho 2n 1 M25. Ta c 210 = 1024 210 + 1 = 1025 M25 220 1 = (210 + 1)(210 1) M25 23(220 1) M100. Mt khc: 22003 = 23(22000 1) + 23 = 23((220)100 1) + 23 = 100k + 8 (k N). Vy hai ch s tn cng ca 22003 l 08. b) Do 799 l s l, theo trng hp 2, ta tm s t nhin n b nht sao cho 7n 1 M100. Ta c 74 = 2401 => 74 1 M100. Mt khc: 99 1 M4 => 99 = 4k + 1 (k N) Vy 799 = 74k + 1 = 7(74k 1) + 7 = 100q + 7 (q N) tn cng bi hai ch s 07.Bi 12: Tm s d ca php chia 3517 cho 25. Gii: Trc ht ta tm hai ch s tn cng ca 3517. Do s ny l nn theo trng hp 2, ta phi tm s t nhin n nh nht sao cho 3n 1 M100. Ta c 310 = 95 = 59049 310 + 1 M50 320 1 = (310 + 1) (310 1) M100. Trang 20Chuyn Bi Dng HSG Ton 6Mt khc: 516 1 M4 5(516 1) M20 517 = 5(516 1) + 5 = 20k + 5 3517 = 320k + 5 = 35(320k 1) + 35 = 35(320k 1) + 243, c hai ch s tn cng l 43. Vy s d ca php chia 3517 cho 25 l 18. Trong trng hp s cho chia ht cho 4 th ta c th tm theo cch gin tip. Trc tin, ta tm s d ca php chia s cho 25, t suy ra cc kh nng ca hai ch s tn cng. Cui cng, da vo gi thit chia ht cho 4 chn gi tr ng. Cc th d trn cho thy rng, nu a = 2 hoc a = 3 th n = 20 ; nu a = 7 th n = 4. Mt cu hi t ra l: Nu a bt k th n nh nht l bao nhiu ? Ta c tnh cht sau y:Tnh cht 4: Nu a MN v (a, 5) = 1 th a20 1 M25. Bi 13: Tm hai ch s tn cng ca cc tng: a) S1 = 12002 + 22002 + 32002 + ... + 20042002 b) S2 = 12003 + 22003 + 32003 + ... + 20042003 Gii: a) D thy, nu a chn th a2 chia ht cho 4 ; nu a l th a100 1 chia ht cho 4 ; nu a chia ht cho 5 th a2 chia ht cho 25. Mt khc, t tnh cht 4 ta suy ra vi mi a N v (a, 5) = 1 ta c a100 1 M25. Vy vi mi a N ta c a2(a100 1) M100. Do S1 = 12002 + 22(22000 1) + ... + 20042(20042000 1) + 22 + 32 + ... + 20042. V th hai ch s tn cng ca tng S1 cng chnh l hai ch s tn cng ca tng 12 + 22 + 32 + ... + 20042. p dng cng thc: 12 + 22 + 32 + ... + n2 = n(n + 1)(2n + 1)/6 12 + 22 + ... + 20042 = 2005 4009 334 = 2684707030, tn cng l 30. Vy hai ch s tn cng ca tng S1 l 30. b) Hon ton tng t nh cu a, S2 = 12003 + 23(22000 1) + ... + 20043(20042000 1) + 23 + 33 + 20043. V th, hai ch s tn cng ca tng S2 cng chnh l hai ch s tn cng ca 13+ 23+ 33+ ... + 20043. p dng cng thc: 23 3 3 3 2n(n 1)1 2 3 ... n (1 2 ... n)2+1+ + + + + + + 1 ]13 + 23 + ... + 20043 = (2005 1002)2 = 4036121180100, tn cng l 00. Vy hai ch s tn cng ca tng S2 l 00. Tnh cht 5: S t nhin A khng phi l s chnh phng nu: + A c ch s tn cng l 2, 3, 7, 8 ; + A c ch s tn cng l 6 m ch s hng chc l ch s chn ; + A c ch s hng n v khc 6 m ch s hng chc l l ; + A c ch s hng n v l 5 m ch s hng chc khc 2 ; + A c hai ch s tn cng l l. Bi 14: Cho n N v n 1 khng chia ht cho 4. CMR: 7n + 2 khng th l s chnh phng. Gii: Do n 1 khng chia ht cho 4 nn n = 4k + r (r {0, 2, 3}). Ta c 74 1 = 2400 M100. Ta vit 7n + 2 = 74k + r + 2 = 7r(74k 1) + 7r + 2. Vy hai ch s tn cng ca 7n + 2 cng chnh l hai ch s tn cng ca 7r+ 2 (r = 0, 2, 3) nn ch c th l 03, 51, 45. Trang 21Chuyn Bi Dng HSG Ton 6Theo tnh cht 5 th r rng 7n+ 2 khng th l s chnh phng khi n khng chia ht cho 4.III. Tm ba ch s tn cng Nhn xt:Tng t nh trng hp tm hai ch s tn cng, vic tm ba ch s tn cng ca s t nhin x chnh l vic tm s d ca php chia x cho 1000. Nu x = 1000k + y, trong k ; y N th ba ch s tn cng ca x cng chnh l ba ch s tn cng ca y (y x). Do 1000 = 8 x 125 m (8, 125) = 1 nn ta xut phng php tm ba ch s tn cng ca s t nhin x = am nh sau: Trng hp 1: Nu a chn th x = am chia ht cho 2m. Gi n l s t nhin sao cho an 1 chia ht cho 125. Vit m = pn + q (p ; q N), trong q l s nh nht aq chia ht cho 8 ta c: x = am = aq(apn 1) + aq. V an 1 chia ht cho 125 => apn 1 chia ht cho 125. Mt khc, do (8, 125) = 1 nn aq(apn 1) chia ht cho 1000. Vy ba ch s tn cng ca am cng chnh l ba ch s tn cng ca aq. Tip theo, ta tm ba ch s tn cng ca aq. Trng hp 2: Nu a l , gi n l s t nhin sao cho an 1 chia ht cho 1000. Vit m = un + v (u ; v N, 0 v < n) ta c: x = am = av(aun 1) + av. V an 1 chia ht cho 1000 => aun 1 chia ht cho 1000. Vy ba ch s tn cng ca am cng chnh l ba ch s tn cng ca av. Tip theo, ta tm ba ch s tn cng ca av. Tnh cht sau c suy ra t tnh cht 4. Tnh cht 6: Nu a N v (a, 5) = 1 th a100 1 chia ht cho 125. Chng minh: Do a20 1 M25 nn a20, a40, a60, a80 khi chia cho 25 c cng s d l 1 a20 + a40 + a60 + a80 + 1 M5. Vy a100 1 = (a20 1)( a80 + a60 + a40 + a20 + 1) M125. Bi 15: Tm ba ch s tn cng ca 123101. Gii: Theo tnh cht 6, do (123, 5) = 1 123100 1 M125 (1). Mt khc: 123100 1 = (12325 1)(12325 + 1)(12350 + 1) 123100 1 M8 (2). V (8, 125) = 1, t (1) v (2) suy ra: 123100 1 M1000 123101 = 123(123100 1) + 123 = 1000k + 123 (k N). Vy 123101 c ba ch s tn cng l 123. Bi 12: Tm ba ch s tn cng ca 3399...98. Gii: Theo tnh cht 6, do (9, 5) = 1 => 9100 1 chi ht cho 125 (1). Tng t bi 11, ta c 9100 1 chia ht cho 8 (2). V (8, 125) = 1, t (1) v (2) suy ra: 9100 1 chia ht cho 1000 3399...98 = 9199...9 = 9100p + 99 = 999(9100p 1) + 999 = 1000q + 999 (p, q N). Vy ba ch s tn cng ca 3399...98 cng chnh l ba ch s tn cng ca 999. Li v 9100 1 chia ht cho 1000 ba ch s tn cng ca 9100 l 001 m 999 = 9100: 9 ba ch Trang 22Chuyn Bi Dng HSG Ton 6s tn cng ca 999 l 889 (d kim tra ch s tn cng ca 999 l 9, sau da vo php nhn ???9 9 ...001 xc nh??9 889 ). Vy ba ch s tn cng ca 3399...98 l 889. Nu s cho chia ht cho 8 th ta cng c th tm ba ch s tn cng mt cch gin tip theo cc bc: Tm d ca php chia s cho 125, t suy ra cc kh nng ca ba ch s tn cng, cui cng kim tra iu kin chia ht cho 8 chn gi tr ng. Bi 16: Tm ba ch s tn cng ca 2004200. Gii: do (2004, 5) = 1 (tnh cht 6) 2004100 chia cho 125 d 1 2004200 = (2004100)2 chia cho 125 d 1 2004200 ch c th tn cng l 126, 251, 376, 501, 626, 751, 876. Do 2004200 M8 nn ch c th tn cng l 376.Bi tp vn dng: Bi 17: Chng minh 1n + 2n + 3n + 4n chia ht cho 5 khi v ch khi n khng chia ht cho 4. Bi 18: Chng minh 920002003, 720002003 c ch s tn cng ging nhau. Bi 19: Tm hai ch s tn cng ca: a) 3999b) 111213 Bi 20: Tm hai ch s tn cng ca: S = 23 + 223 + ... + 240023 Bi 21: Tm ba ch s tn cng ca: S = 12004 + 22004 + ... + 20032004 Bi 22: Cho (a, 10) = 1. Chng minh rng ba ch s tn cng ca a101 cng bng ba ch s tn cng ca a. Bi 23: Cho A l mt s chn khng chia ht cho 10. Hy tm ba ch s tn cng ca A200. Bi 24: Tm ba ch s tn cng ca s: 199319941995 ...2000 Bi 25: Tm su ch s tn cng ca 521.Trang 23Chuyn Bi Dng HSG Ton 6Dy s c qui lut I > Phng php d on v quy np : Trong mt s trng hp khi gp bi ton tnh tng hu hn Sn = a1 + a2 + .... an(1) Bngcchnotabit ckt qu(don, hocbi ton chng minh khi cho bit kt qu). Th ta nn s dng phng php ny v hu nh th no cng chng minh c . V d 1 : Tnh tngSn =1+3+5 +... + (2n -1 )Th trc tip ta thy : S1 = 1 S2 = 1 + 3 =22 S3 = 1+ 3+ 5 = 9 = 32 ...... ...Ta d on Sn = n2 Vi n = 1;2;3 ta thy kt qu ng gi s vi n= k ( k 1) ta c Sk = k 2(2)ta cn phi chng minh Sk + 1 = ( k +1 ) 2 ( 3) Tht vy cng 2 v ca ( 2) vi 2k +1ta c 1+3+5 +... + (2k 1) + ( 2k +1) = k2 + (2k +1) v k2 + ( 2k +1) = ( k +1) 2 nn ta c (3) tc l Sk+1= ( k +1) 2 theo nguyn l quy np bi ton c chng minh vy Sn = 1+3=5 + ... + ( 2n -1) = n2 Tng t ta c th chng minh cc kt qu sau y bng phng php quy np ton hc .1, 1 + 2+3 + .... + n = 2) 1 ( + n nTrang 24Chuyn Bi Dng HSG Ton 62, 12 + 2 2 + ..... + n 2 = 6) 1 2 )( 1 ( + + n n n3, 13+23 + ..... + n3 = 22) 1 (1]1

+ n n4, 15 + 25 + .... + n5 = 121.n2 (n + 1) 2 ( 2n2 + 2n 1 ) II > Ph ng php kh lin tip : Gistacntnhtng(1) mtacthbiudinai, i = 1,2,3...,n, qua hiu hai s hng lin tip ca 1 dy s khc , chnh xc hn , gi s : a1 =b1-b2 a2 =b2 - b3 .... .... .....an = bn bn+ 1 khi ta c ngay :Sn = ( b1 b2 ) + ( b2 b3 ) + ......+ ( bn bn + 1 ) =b1 bn + 1 V d 2 : tnh tng :S = 100 . 991.......13 . 12112 . 11111 . 101+ + + +Ta c :11110111 . 101 , 12111112 . 111 , 1001991100 . 991 Do : S = 100910011011001991.......121111111101 + + + Dng tng qut Sn = ) 1 (1......3 . 212 . 11++ + +n n( n >1 ) = 1- 1 11++ nnnV d 3 : tnh tng Trang 25Chuyn Bi Dng HSG Ton 6 Sn = ) 2 )( 1 (1......5 . 4 . 314 . 3 . 213 . 2 . 11+ ++ + + +n n nTa c Sn =

,_

+ +++ + ,_

+ ,_

) 2 )( 1 (1) 1 (121........4 . 313 . 21213 . 212 . 1121n n n nSn =

,_

+ +++ + + ) 2 )( 1 (1) 1 (1......4 . 313 . 213 . 212 . 1121n n n nSn = ) 2 )( 1 ( 4) 3 () 2 )( 1 (12 . 1121+ ++

,_

+ +n nn nn nV d 4 : tnh tng Sn = 1! +2.2 ! + 3.3 ! + ...... + n .n!( n! = 1.2.3 ....n ) Ta c : 1! = 2! -1! 2.2! = 3 ! -2! 3.3! = 4! -3! ..... ..... ..... n.n! = (n + 1) n! Vy Sn = 2! - 1! +3! 2 ! + 4! - 3! +...... + ( n+1) ! n! = ( n+1) ! - 1! = ( n+ 1) ! - 1V d 5 : tnh tng Sn = [ ]2 2 2) 1 (1 2.......) 3 . 2 (5) 2 . 1 (3+++ + +n nnTa c : [ ];) 1 (1 1) 1 (1 22 2 2+ ++i i i iii = 1 ; 2 ; 3; ....; nDo Sn = ( 1-

,_

+ + + ,_

+2 2 2 2 2) 1 (1 1.....3121)21n n = 1- 2 2) 1 () 2 () 1 (1+++ nn nnIII > Ph ng php gii ph ng trnh vi n l tng cn tnh: Trang 26Chuyn Bi Dng HSG Ton 6V d 6 : Tnh tng S = 1+2+22 +....... + 2100 ( 4) ta vit li S nh sau :S = 1+2 (1+2+22 +....... + 299 ) S= 1+2 ( 1 +2+22+ ...... + 299 + 2 100- 2100 ) => S= 1+2 ( S -2 100 ) ( 5) T (5) suy ra S = 1+ 2S -2101 S = 2101-1V d 7 : tnh tng Sn= 1+ p + p 2 + p3 + ..... + pn ( p1) Ta vit li Sn di dng sau : Sn = 1+p ( 1+p+p2 +.... + pn-1 )Sn = 1 + p ( 1+p +p2 +..... + p n-1 + p n p n ) Sn = 1+p ( Sn pn ) Sn = 1 +p.Sn p n+1 Sn ( p -1 ) = pn+1 -1 Sn = 111+pPn V d 8 : Tnh tng Sn = 1+ 2p +3p 2 + .... + ( n+1 ) pn , ( p 1) Ta c : p.Sn = p + 2p 2 + 3p3 + ..... + ( n+ 1) p n +1 = 2p p +3p 2 p2 + 4p3p3 + ...... + (n+1) pn - pn + (n+1)pn pn + ( n+1) pn+1= ( 2p + 3p2 +4p3 + ...... +(n+1) pn ) ( p +p + p + .... pn ) + ( n+1) pn+1= ( 1+ 2p+ 3p2+4p3+ ....... + ( n+1) pn ) ( 1 + p+ p2 + .... + p n) + ( n +1 ) pn+1Trang 27Chuyn Bi Dng HSG Ton 6p.Sn=Sn- 11) 1 (11+++ +nnP nPP ( theo VD 7 ) Li c (p-1)Sn = (n+1)pn+1 - 111+Ppn Sn = 21 1) 1 (11) 1 (++ +PppP nn nIV > Ph ng php tnh qua cc tng bit Cc k hiu :nniia a a a a + + + + ......3 2 11 Cc tnh cht : 1, + +nininii i i ib a b a1 1 1) ( 2, niiniia a a a1 1.V d 9 : Tnh tng :Sn= 1.2 + 2.3 + 3.4 + ......... + n( n+1) Ta c :Sn = + + +ninininii i i i i i1 1 12 21) ( ) 1 ( V :

6) 1 2 )( 1 (2) 1 (.... 3 2 1121+ ++ + + + + n n nin nn inini(Theo I )cho nn : Sn = 3) 2 )( 1 (6) 1 2 )( 1 (2) 1 ( + ++ +++ n n n n n n n nV d 10 : Tnh tng :Sn =1.2+2.5+3.8+.......+n(3n-1)ta c : Sn = ninii i i i1 12) 3 ( ) 1 3 ( = ninii i1 123Trang 28Chuyn Bi Dng HSG Ton 6Theo (I) ta c :Sn = ) 1 (2) 1 (6) 1 2 )( 1 ( 32+ ++ +n nn n n n nV d 11 . Tnh tng Sn = 13+ +23 +53 +... + (2n +1 )3 ta c : Sn = [( 13 +2 3 +33 +43 +....+(2n+1)3 ] [23+43 +63 +....+(2n)3] = [13+23 +33 +43 + ..... + (2n +1 )3] -8 (13 +23 +33 +43 +......+ n3 ) Sn = 4) 1 ( 84) 2 2 ( ) 1 2 (2 2 2 2++ + n n n n( theo (I) 3 )=( n+1) 2(2n+1) 2 2n2 (n+1)2 = (n +1 )2 (2n2 +4n +1) V/ Vn dng trc tip cng thc tnh tng cc s hng ca dy s cch u ( Hc sinh lp 6 ) C s l thuyt : + m s hng ca 1 dy s m 2 s hng lin tip ca dy cch nhau cng 1 s n v , ta dng cng thc: S s hng = ( s cui s u 0 : ( khong cch ) + 1 + tnh tng cc s hng ca mt dy s m 2 s hng lin tip cch nhau cng 1 s n v , ta dng cng thc:Tng = ( s u s cui ) .( s s hng ) :2 V d 12 : Tnh tng A = 19 +20 +21 +.... + 132 S s hng ca A l : ( 132 19 ) : 1 +1 = 114 ( s hng )m A = 114 ( 132 +19 ) : 2 = 8607 V d 13 : Tnh tng B = 1 +5 +9 +.......+ 2005 +2009 Trang 29Chuyn Bi Dng HSG Ton 6 s s hng ca B l ( 2009 1 ) : 4 + 1 = 503 B = ( 2009 +1 ) .503 :2 = 505515 VI / Vn dng 1 s cng thc chng minh c vo lm ton V d14 : Chng minh rng : k ( k+1) (k+20 -9k-1)k(k+1) = 3k ( k +1 ) T tnh tng S = 1..2+2.3 + 3.4 +...... + n (n + 1) Chng minh : cch 1 : VT = k(k+1)(k+2) (k-1) k(k+1) = k( k+1) [ ] ) 1 ( ) 2 ( + k k= k (k+1) .3 = 3k(k+1) Cch 2 : Ta c k ( k +1) = k(k+1).3) 1 ( ) 2 ( + k k = 3) 1 )( 1 (3) 2 )( 1 ( ++ + k k k k k k * 3k ( k-1) = k (k+1)(k+2) (k-1) k(k+1) =>1.2 = 1.2.3 0.1.23 3 2.3.4 1.2.32.33 3...................................( 1)( 2) ( 1) ( 1)( 1)3 3n n n n n nn n + + ++ S = 1.2.0 ( 2) ( 1) ( 1) ( 2)3 3 3n n n n n n + + + ++ V d 15 : Chng minh rng : k (k+1) (k+2) (k+3) (k-1) k(k+1) (k+2) =4k (k+1) (k+2) t tnh tng S = 1.2 .3 + 2.3 .4 +3.4.5 +.... + n(n+1) (n+2) Chng minh : VT = k( k+1) (k+2) [ ] ) 1 ( ) 3 ( + k k= k( k+1) ( k +2 ) .4Rt ra : k(k+1) (k+2) = 4) 2 )( 1 ( ) 1 (4) 3 )( 2 )( 1 ( + + + + + k k k k k k k kp dng : 1.2.3 = 43 . 2 . 1 . 044 . 3 . 2 . 1Trang 30Chuyn Bi Dng HSG Ton 62.3.4 = 44 . 3 . 2 . 145 . 4 . 3 . 2..........................................................n(n+1) (n+2) = 4) 2 )( 1 ( ) 1 (4) 3 )( 2 )( 1 ( + + + + + n n n n n n n nCng v vi v ta c S = 4) 3 n )( 2 n )( 1 n ( n + + +* Bi tp ngh :Tnh cc tng sau 1, B = 2+ 6 +10 +14 + ..... + 202 2, a, A = 1+2 +22 +23 +.....+ 26.2 + 2 6 3 b, S = 5 + 52 + 53 + ..... + 5 99 + 5100

c, C = 7 + 10 + 13 + .... + 76 3, D = 49 +64 + 81+ .... + 169 4, S = 1.4 + 2 .5 + 3.6 + 4.7 +.... + n( n +3 ) , n = 1,2,3 ,.... 5, S = 100 . 991........4 . 313 . 212 . 11+ + + +6, S = 61 . 594....9 . 747 . 54+ + +7, A = 66 . 615......26 . 21521 . 16516 . 115+ + + +8, M = 2005 2 1 031.....313131+ + + +9, Sn = ) 2 )( 1 (1.....4 . 3 . 21. 3 . 2 . 11+ ++ + +n n n10, Sn = 100 . 99 . 982.....4 . 3 . 223 . 2 . 12+ + +11, Sn = ) 3 )( 2 )( 1 (1......5 . 4 . 3 . 214 . 3 . 2 . 11+ + ++ + +n n n n12, M = 9 + 99 + 999+...... + 99..... .....9 50 ch s 913, Cho:S1 = 1+2S3 = 6+7+8+9Trang 31Chuyn Bi Dng HSG Ton 6S2 = 3+4+5 S4 = 10 +11 +12 +13 + 14 Tnh S100=? Trong qu trnh bi dng hc sinh gii , ti kt hp cc dng ton c lin quan n dng tnh tng rn luyn cho cc em , chng hn dng ton tm x :14,a, (x+1) + (x+2) + (x+3) +...... + ( x+100 ) = 5070 b, 1 + 2 + 3 + 4 +.............+ x =820 c, 1 + 199119891) 1 (2......1016131++ + + +x xHay cc bi ton chng minh s chia ht lin quan 15, Chng minh : a, A = 4+ 22 +23 +24 +..... + 220 l lu tha ca 2 b, B =2 + 22 + 2 3 + ...... + 2 60 3 ; 7; 15 c,C = 3 + 33 +35 + ....+ 31991 13 ;41d,D = 119 + 118 +117 +......+ 11 +15 Trang 32