Download pdf - 50 Bài Toán Bồi Dưỡng Học Sinh Giỏi Lớp 5 (Có Lời Giải)

50 bi ton Bi dng HSG Lp 5 (Li gii)

1

Bi 51 : Cho hai hnh vung ABCD v MNPQ nh trong hnh v. Bit BD = 12 cm. Hy tnh din tch phn gch cho.

Bi gii : Din tch tam gic ABD l : (12 x (12 : 2))/2 = 36 (cm2)

Din tch hnh vung ABCD l : 36 x 2 = 72 (cm2)

Din tch hnh vung AEOK l :

72 : 4 = 18 (cm2)

Do : OE x OK = 18 (cm2) r x r = 18 (cm2)

Din tch hnh trn tm O l : 18 x 3,14 = 56,92 (cm2)

Din tch tam gic MON = r x r : 2 = 18 : 2 = 9 (cm2) Din tch hnh vung MNPQ l : 9 x 4 = 36 (cm2) Vy din tch phn gch cho l : 56,52 - 36 = 20,52 (cm2)

Bi 52 : Bn Ton nhn mt s vi 2002 nhng ng tr qun vit 2 ch s 0 ca s 2002 nn kt qu b gim i 3965940 n v. Ton nh nhn s no vi 2002 ?

Bi gii : V "ng tr" nn bn Ton nhn nhm s vi 22. Tha s th hai b gim i s n v l : 2002 - 22 = 1980 (n v). Do kt qu b gim i 1980 ln tha s th nht, v bng 3965940 n v. Vy tha s th nht l : 3965940 : 1980 = 2003.

Bi 53 : Ngi ta cng 5 s v chia cho 5 th c 138. Nu xp cc s theo th t ln dn th cng 3 s u tin v chia cho 3 s c 127, cng 3 s cui v chia cho 3 s c 148. Bn c bit s ng gia theo th t trn l s no khng ?


2

Bi gii : 138 l trung bnh cng ca 5 s, nn tng 5 s l : 138 x 5 = 690. Tng ca ba s u tin l : 127 x 3 = 381. Tng ca ba s cui cng l : 148 x 3 = 444. Tng ca hai s u tin l : 690 - 444 = 246. S gia l s ng th ba, nn s gia l : 381 - 246 = 135.

Bi 54 : Cho bng vung gm 10 dng v 10 ct. Hai bn Tn v Nhi t mu cc , mi mt mu trong 3 mu : xanh, , tm. Bn Tn bo : "Ln no t xong ht cc cng c 2 dng m trn 2 dng c mt mu t s dng ny bng t s dng kia". Bn Nhi bo : "T pht hin ra bao gi cng c 2 ct c t nh th". No, bn hy cho bit ai ng, ai sai ?

Bi gii : Gi s s t mu tt c cc dng u khc nhau m mi dng c 10 nn s c t mu t nht l : 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 ().

L lun tng t vi mu xanh, mu tm ta cng c kt qu nh vy. Do bng s c t nht 45 + 45 + 45 = 135 (). iu ny mu thun vi bng ch c 100 . Chng t t nht phi c 2 dng m s t bi cng mt mu l nh nhau. i vi cc ct, ta cng lp lun tng t nh trn. Do c hai bn u ni ng.

Bi 55 : Tm 4 s t nhin c tng bng 2003. Bit rng nu xa b ch s hng n v ca s th nht ta c s th hai. Nu xa b ch s hng n v ca s th hai ta c s th ba. Nu xa b ch s hng n v ca s th ba ta c s th t.

Bi gii : S th nht khng th nhiu hn 4 ch s v tng 4 s bng 2003. Nu s th nht c t hn 4 ch s th s khng tn ti s th t. Vy s th nht phi c 4 ch s. Gi s th nht l abcd (a > 0, a, b, c, d < 10). S th hai, s th ba, s th t ln lt s l : abc ; ab ; a. Theo bi ra ta c php tnh : abcd + abc + ab + a = 2003. Theo phn tch cu to s ta c : aaaa + bbb + cc + d = 2003 (*) T php tnh (*) ta c a < 2, nn a = 1. Thay a = 1 vo (*) ta c : 1111 + bbb + cc + d = 2003.

bbb + cc + d = 2003 - 1111 bbb + cc + d = 892 (**)

b > 7 v nu b nh hn hoc bng 7 th bbb + cc + d nh hn 892 ; b < 9 v nu b = 9 th bbb = 999 > 892. Suy ra b ch c th bng 8. Thay b = 8 vo (**) ta c : 888 + cc + d = 892

cc + d = 892 - 888


3

cc + d = 4 T y suy ra c ch c th bng 0 v d = 4. Vy s th nht l 1804, s th hai l 180, s th ba l 18 v s th t l 1. Th li : 1804 + 180 + 18 + 1 = 2003 (ng)

Bi 56 : Mt ngi mang ra ch 5 gi to gm hai loi. S to trong mi gi ln lt l : 20 ; 25 ; 30 ; 35 v 40. Mi gi ch ng mt loi to. Sau khi bn ht mt gi to no , ngi y thy rng : S to loi 2 cn li ng bng na s to loi 1. Hi s to loi 2 cn li l bao nhiu ?

Bi gii : S to ngi mang ra ch l : 20 + 25 + 30 + 35 + 40 = 150 (qu) V s to loi 2 cn li ng bng na s to loi 1 nn sau khi bn, s to cn li phi chia ht cho 3. V tng s to mang ra ch l 150 qu chia ht cho 3 nn s to bn phi chia ht cho 3. Trong cc s 20, 25, 30, 35, 40 ch c 30 chia ht cho 3. Do vy ngi y bn gi to ng 30 qu. Tng s to cn li l : 150 - 30 = 120 (qu) Ta c s biu din s to ca loi 1 v loi 2 cn li :

S to loi 2 cn li l : 120 : (2 + 1) = 40 (qu) Vy ngi y cn li gi ng 40 qu chnh l s to loi 2 cn li. p s : 40 qu

Bi 57 : Khng c thay i v tr ca cc ch s vit trn bng : 8 7 6 5 4 3 2 1 m ch c vit thm cc du cng (+), bn c th cho c kt qu ca dy php tnh l 90 c khng ?

Bi gii : C hai cch in : 8 + 7 + 65 + 4 + 3 + 2 + 1 = 90 8 + 7 + 6 + 5 + 43 + 21 = 90 tm c hai cch in ny ta c th c nhn xt sau : Tng 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 ; 90 - 36 = 54. Nh vy mun c tng 90 th trong cc s hng phi c mt hoc hai s l s c hai ch s. Nu s c hai ch s l 87 hoc 76 m 87 > 54, 76 > 54 nn khng th c. Nu s c hai ch s l 65 ; 65 + 36 - 6 - 5 = 90, ta c th in : 8 + 7 + 65 + 4 + 3 + 2 + 1 - 90.


4

Nu s c hai ch s l 54 th cng khng th c tng l 90 c v 54 + 36 - 5 - 4 < 90. Nu s c hai ch s l 43 ; 43 < 54 nn cng khng th c. Nu trong tng c 2 s c hai ch s l 43 v 21 th ta c 43 + 21 - (4 + 3 + 2 + 1) = 54. Nh vy ta c th in: 8 + 7 + 6 + 5 + 43 + 21 = 90.

Bi 58 : Cho phn s M = (1 + 2 +... + 9)/(11 + 12 +... +19).

Hy bt mt s hng t s v mt s hng mu s sao cho gi tr phn s khng thay i.

Tm tt bi gii : M = (1 + 2 +... + 9)/(11 + 12 +... +19) = 45/135 = 1/3.

Theo tnh cht ca hai t s bng nhau th 45/135 = (45 - k)/(135 - kx3)(k l s t nhin nh hn 45). Do t s ca M bt i 4 ; 5 ; 6 th tng ng mu s phi bt i 12 ; 15 ; 18.

Bi 59 : Ch c mt chic ca ng y va mt lt Bn hy mau cho bit ong na lt th no ?

Bi gii : Ai kho tay tinh mt Nghing ca nh hnh trn S t yu cu lin Trong ca : ng na lt !

Bi 60 : in s thch hp theo mu :

Bi gii : Bi ny c hai cch in :


5

Cch 1 : Theo hnh 1, ta c 4 l trung bnh cng ca 3 v 5 (v (3 + 5) : 2 = 4). Khi hnh 2, gi A l s cn in, ta c A l trung bnh cng ca 5 v 13. Do A = (5 + 13) : 2 = 9. hnh 3, gi B l s cn in, ta c 15 l trung bnh cng ca 8 v B. Do 8 + B = 15 x 2. T tm c B = 22.

Cch 2 : Theo hnh 1, ta c

3 x 3 + 4 x 4 = 5 x 5. Khi hnh 2 ta c : 5 x 5 + A x A = 13 x 13. suy ra A x A = 144. Vy A = 12 (v 12 x 12 = 144). hnh 3 ta c : 8 x 8 + 15 x 15 = B x B. suy ra B x B = 289. Vy B = 17 (v 17 x 17 = 289).

Bi 61 : C lp 4A phi lm mt bi kim tra ton gm c 3 bi ton. Gio vin ch nhim lp bo co vi nh trng rng : c lp mi em u lm c t nht mt bi, trong lp c 20 em gii c bi ton th nht, 14 em gii c bi ton th hai, 10 em gii c bi ton th ba, 5 em gii c bi ton th hai v th ba, 2 em gii c bi ton th nht v th hai, c mi mt em c 10 im v gii c c ba bi. Hi rng lp hc c bao nhiu em tt c ?

Bi gii :

Mi hnh trn ghi s bn gii ng mt bi no . V ch c mt bn gii ng 3 bi nn in s 1 vo phn chung ca 3 hnh trn. S bn gii ng bi I v bi II l 2 nn phn chung ca hai hnh trn ny m khng chung vi hnh trn cn li s c ghi s 1 (v 2 - 1 = 1). Tng t, ta ghi c cc s vo cc phn cn li.


6

S hc sinh lp 4A chnh l tng cc s in vo cc phn : 13 + 5 + 1 + 1 + 4 + 8 + 0 = 32 (HS)

Bi 62 : Bn hy in cc s t 1 n 9 vo cc trng cc php tnh u thc hin ng (c hng dc v hng ngang).

Bi gii : Ta t tn cho cc s phi tm nh trong bng. Cc s in vo trng l cc s c 1 ch s nn tng cc s ln nht ch c th l 17.

ct 1, c A + D : H = 6, nn H ch c th ln nht l 2. Ct 5 c C + G : M = 5 nn M ch c th ln nht l 3. * Nu H = 1 th A + D = 6 = 2 + 4, do M = 3 v H + K = 2 x 3 = 6 = 1 + 5. K = 5 th B x E = 4 + 5 = 9, nh th ch c th B hoc E bng 1, iu chng t H khng th bng 1. * Nu H = 2 th M phi bng 1 hoc 3; nu M = 1 th H + K = 2, nh vy K = 0, iu ny cng khng th c. Vy M = 3 ; H + K = 6 th K = 4. H = 2 th A + D = 12 = 5 + 7 ; nh vy A = 5, D = 7 hoc D = 5, A = 7. K = 4 th B x E = 4 + 4 = 8 = 1 x 8 ; nh vy B = 1, E = 8 hoc E = 1, B = 8. M = 3 th C + G = 15 = 6 + 9 ; nh vy C = 6, G = 9 hoc G = 6, C = 9 ; G ch c th bng 9 v nu G = 6 th D + E = 10, m trong cc s 1, 5, 7, 8 khng c hai s no c tng bng 10. Vy C = 6 v A + B = 8, nh vy B ch c th bng 1, A = 7 th D = 5 v E = 8. Cc s in vo bng nh hnh sau.

Bi 63 : S = 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 c phi l s t nhin khng ? V sao ?


7

Bi gii : Cc bn gii theo 3 hng sau y : Hng 1 : Tnh S = 1 201/280 Hng 2 : Khi qui ng mu s tnh S th mu s chung l s chn. Vi mu s chung ny th 1/2 ; 1/3 ; 1/4 ; 1/5 ; 1/6 ; 1/7 s tr thnh cc phn s m t s l s chn, ch c 1/8 l tr thnh phn s m t s l s l. Vy S l mt phn s c t s l s l v mu s l s chn nn S khng phi l s t nhin. Hng 3 : Chng minh 5/4 < S < 2 Tht vy 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 > 6 x 1/8 = 3/4 nn S > 3/4 + 1/2 = 5/4

Mt khc : 1/4 + 1/5 + 1/6 + 1/7 < 4 x 1/4 = 1 nn S < 1 + 1/2 + 1/3 + 1/8 = 1 + 1/2 + 11/24


8

Bi 65: Cn phng c 4 bc tng, trn mi bc tng treo 3 l c m khong cch gia 3 l c trn mt bc tng l nh nhau. Bn c bit cn phng treo my l c khng ? Bi gii: n gin, ta s treo tt c cc l c cao ngang nhau trn c 4 bc tng. Khi cch treo c s ging nh bi ton trng cy. Ta c 5 cch trng ng vi s l c l 8, 9, 10, 11, 12 l c nh sau (coi mi l c l mt im chm trn):

Nu cc l c c treo cao khc nhau trn mi bc tng th v tr 3 l c trn mt bc tng s to thnh 3 nh ca mt hnh tam gic u. Khi ta s c cc cch treo khc ng vi s l c l 6,] 7, 8, 9, 10, 11, 12 l c. Xin nu ra 2 cch treo ng vi s l c l 6 l v 7 l nh sau:

Vy s l c trong cn phng c th t 6 n 12 l c. Bi 66: L Lem chia mt qu da (da ) thnh 9 phn cho 9 c gi. Nhng khi cc c n xong, L Lem thy c 10 ming v da. L Lem chia da kiu g y nh ? Bi gii: C nhiu cch b da, Lo Lem b da nh sau:


9

Ct ngang qu da lm 3 phn, sau li b dc qu da lm 3 phn s c 9 ming da (nh hnh v) chia cho 9 c, sau khi n xong s c 10 ming v da. V ring ming s 5 c v 2 u, nn khi n xong s c 2 ming v.

Bi 67: Bn hy in cc s t 1 n 10 vo cc vung sao cho tng cc s nt dc (1 nt) cng nh nt ngang (3 nt) u l 16.

Bi gii: Tt c cc bn u nhn ra mt phng n in s: a = 1; b = 9; c = 5; d = 4; e = 6; g = 10; h = 3; i = 1; k = 8; l = 7. T s c cc phng n khc bng cch: 1) i cc b v c. 2) i cc k v l. 3) i cc d v h. 4) i ng thi c 3 a, b, c cho 3 i, k, l. Nh vy cc bn s c 16 cch in s khc nhau. Bi 68: Trong mt cuc thi ti Ton Tui th c 51 bn tham d. Lut cho im nh sau: + Mi bi lm ng c 4 im. + Mi bi lm sai hoc khng lm s b tr 1 im. Bn chng t rng tm c 11 bn c s im bng nhau. Bi gii: Thi ti gii Ton Tui th c 5 bi. S im ca 51 bn thi c th xp theo 5 loi im sau y: + Lm ng 5 bi c: 4 x 5 = 20 (im). + Lm ng 4 bi c: 4 x 4 - 1 x 1 = 15 (im). + Lm ng 3 bi c: 4 x 3 - 1 x 2 = 10 (im). + Lm ng 2 bi c: 4 x 2 - 1 x 3 = 5 (im). + Lm ng 1 bi c:


10

4 x 1 - 1 x 4 = 0 (im). V 51 : 5 = 10 (d 1) nn phi c t nht 11 bn c s im bng nhau. Bi 69:

V Hu cng vi Lng Th Vinh Hai nh ton hc, mt nm sinh

Thc hnh, tnh ton u thng tho V vang dn tc nc non mnh

Nm sinh ca hai ng l mt s c bn ch s, tng cc ch s bng 10. Nu vit nm sinh theo th t ngc li th nm sinh khng i. Bn bit nm sinh ca hai ng cha?

Bi gii: Gi nm sinh ca hai ng l abba (a 0, a < 3, b 99% nn ngy th nht bn c nhiu cam hn ngy th ba.

Bi 71: Cu T chn 4 ch s lin tip nhau v dng 4 ch s ny vit ra 3 s gm 4 ch s khc nhau. Bit rng s th nht vit cc ch s theo th t tng dn, s th hai vit cc ch s theo th t gim dn v s th ba vit cc ch s theo th t no . Khi cng ba s va vit th c tng l 12300. Bn hy cho bit cc s m cu T vit.

Bi gii : Gi 4 s t nhin lin tip t nh n ln l a, b, c, d. S th nht cu T vit l abcd, s th hai cu T vit l dcba. Ta xt cc ch s hng nghn ca ba s c tng l 12300: a l s ln hn 1 v nu a = 1 th d = 4, khi s th ba c ch s hng nghn ln nht l 4 v tng ca ba ch s ny ln nht l: 1 + 4 + 4 = 9 < 12; nh vy tng ca ba s nh hn 12300.


11

a l s nh hn 5 v nu a = 5 th d = 8 v a + d = 13 > 12; nh vy tng ca ba s ln hn 12300. a ch c th nhn 3 gi tr l 2, 3, 4. - Nu a = 2 th s th nht l 2345, s th hai l 5432. S th ba l: 12300 - (2345 + 5432) = 4523 (ng, v s ny c cc ch s l 2, 3, 4, 5). - Nu a = 3 th s th nht l 3456, s th hai l 6543. S th ba l : 12300 - (3456 + 6543) = 2301 (loi, v s ny c cc ch s khc vi 3, 4, 5, 6). - Nu a = 4 th s th nht l 4567, s th hai l 7654. S th ba l: 12300 - (4567 + 7654) = 79 (loi). Vy cc s m cu T vit l : 2345, 5432, 4523.

Bi 72: Vi 4 ch s 2 v cc du php tnh bn c th vit c mt biu thc c kt qu l 9 c khng? Ti c gng vit mt biu thc c kt qu l 7 nhng cha c. Cn bn? Bn th sc xem no! Bi gii: Vi bn ch s 2 ta vit c biu thc c gi tr bng 9 l: 22 : 2 - 2 = 9. Khng th dng bn ch s 2 vit c biu thc c kt qu l 7. Bi 73: Vi 36 que dim c xp nh hnh di.

1) Bn m c bao nhiu hnh vung? 2) Bn hy nhc ra 4 que dim ch cn 4 hnh vung c khng? Bi gii : 1) Nhn vo hnh v, ta thy c 2 loi hnh vung, hnh vung c cnh l 1 que dim v hnh vung c cnh l 2 que dim. Hnh vung c cnh l 1 que dim gm c 13 hnh, hnh vung c cnh l 2 que dim gm c 4 hnh. Vy c tt c l 17 hnh vung. 2) Mi que dim c th nm trn cnh ca nhiu nht l 3 hnh vung, nu nht ra 4 que dim th ta bt i nhiu nht l : 4 x 3 = 12 (hnh vung), cn li

17 - 12 = 5 (hnh vung). Nh vy khng th nht ra 4 que dim cn li 4 hnh vung c. Bi 74: C 7 thng ng y du, 7 thng ch cn na thng du v 7 v thng. Lm sao c th chia cho 3 ngi mi ngi u c lng du nh nhau v s thng nh nhau ?


12

Bi gii: Gi thng y du l A, thng c na thng du l B, thng khng c du l C. Cch 1: Khng phi du t thng ny sang thng kia. Ngi th nht nhn: 3A, 1B, 3C. Ngi th hai nhn: 2A, 3B, 2C. Ngi th ba nhn: 2A, 3B, 2C. Cch 2: Khng phi du t thng ny sang thng kia. Ngi th nht nhn: 3A, 1B, 3C. Ngi th hai nhn: 3A, 1B, 3C. Ngi th ba nhn: 1A, 5B, 1C. Cch 3: du t thng ny sang thng kia. Ly 4 thng cha na thng du (4B) y sang 2 thng khng (2C) c 2 thng y du (2A). Khi c 9A, 3B, 9C v mi ngi s nhn c nh nhau l 3A, 1B, 3C.

Bi 75: Hy v 4 on thng i qua 9 im hnh bn m khng c nhc bt hay t li. Bi gii: Ci kh bi ton ny l ch c v 4 on thng v ch c v bng mt nt nn cn phi to thm hai im bn ngoi 9 im th mi thc hin c yu cu ca bi. Xin nu ra mt cch v vi hai ng i khc nhau (bt u t im 1 v kt thc im 2 vi ng i theo chiu mi tn) nh sau:

Khi xoay hoc lt hai hnh trn ta s c cc cch v khc. Bi 76:

Chic bnh trung thu Nhn trn gia

Hy ct 4 ln

Thnh 12 ming Nhng nh iu kin Cc ming bng nhau

V ln ct no Cng qua gia bnh


13

Bi gii: C nhiu cch ct c cc bn xut. Xin gii thiu 3 cch. Cch 1: Nht th nht chia i theo b dy ca chic bnh v nguyn v tr ny ct thm 3 nht (nh hnh v).

Lu l AM = BN = DQ = CP = 1/6 AB v IA = ID = KB = KC = 1/2 AB. Cc bn c th d dng chng minh c 12 ming bnh l bng nhau v c 3 nht ct u i qua ng ... tm bnh. Cch 2: Ct 2 nht theo 2 ng cho c 4 ming ri chng 4 ming ny ln nhau ct 2 nht chia mi ming thnh 3 phn bng nhau (lu : BM = MN = NC).

Cch 3: Nht th nht ct nh cch 1 v nguyn v tr ny ct thm 3 nht

nh hnh v. Lu : AN = AM = CQ = CP = 1/2 AB.

Bi 77: Mi nh ca mt tm ba hnh tam gic c nh s ln lt l 1; 2; 3. Ngi ta chng cc tam gic ny ln nhau sao cho khng c ch s no b che lp. Mt bn cng tt c cc ch s nhn thy th c kt qu l 2002. Liu bn c tnh nhm khng?

Bi gii: Tng cc s trn ba nh ca mi hnh tam gic l 1 + 2 + 3 = 6. Tng ny l mt s chia ht cho 6. Khi chng cc hnh tam gic ny ln nhau sao cho khng c ch s no b che lp, ri tnh tng tt c cc ch s nhn thy c phi c kt qu l s chia ht cho 6. V s 2002 khng chia ht cho 6 nn bn tnh sai.


14

Bi 78: Bn hy in 12 s t 1 n 12, mi s vo mt vung sao cho tng 4 s cng nm trn mt ct hay mt hng u nh nhau. Bi gii: Tng cc s t 1 n 12 l: (12+1) x 12 : 2 = 78 V tng 4 s cng nm trn mt ct hay mt hng u nh nhau nn tng s ca 4 hng v ct phi l mt s chia ht cho 4. t cc ch ci A, B, C, D vo cc vung gia (hnh v).

Khi tnh tng s ca 4 hng v ct th cc s cc A, B, C, D c tnh hai ln. Do tng 4 hng, ct chia ht cho 4 th tng 4 s ca 4 A, B, C, D phi chia cho 4 d 2 (v 78 chia cho 4 d 2). Ta thy tng ca 4 s c th l: 10, 14, 18, 22, 26, 30, 34, 38, 42.

Ta xt mt vi trng hp: 1) Tng ca 4 s b nht l 10. Khi 4 s s l 1, 2, 3, 4. Do tng ca mi hng (hay mi ct) l: (78 + 10) : 4 = 22. Xin nu ra mt cch in nh hnh di:

2) Tng ca 4 s l 14. Ta c: 14 = 1 + 2 + 3 + 8 = 1 + 2 + 4 + 7 = 1 + 3 + 4 + 6 = 2 + 3 + 4 + 5.

Do tng ca mi hng (hay mi ct) l: (78 + 14) : 4 = 23. Xin nu ra mt cch in nh hnh sau:

Cc trng hp cn li s cho ta kt qu mi hng (hay mi ct) ln lt l 24, 25, 26, 27, 28, 29, 30. C rt nhiu cch in y! Cc bn th tm tip xem sao? Bi 79:


15

Mt i tuyn tham d k thi hc sinh gii 3 mn Vn, Ton, Ngoi ng do thnh ph t chc t c 15 gii. Hi i tuyn hc sinh gii c bao nhiu hc sinh? Bit rng: Hc sinh no cng c gii. Bt k mn no cng c t nht 1 hc sinh ch t 1 gii. Bt k hai mn no cng c t nht 1 hc sinh t gii c hai mn. C t nht 1 hc sinh t gii c 3 mn. Tng s hc sinh t 3 gii, 2 gii, 1 gii tng dn.

Bi gii: Gi s hc sinh t gii c 3 mn l a (hc sinh) Gi s hc sinh t gii c 2 mn l b (hc sinh) Gi s hc sinh ch t gii 1 mn l c (hc sinh) Tng s gii t c l: 3 x a + 2 x b + c = 15 (gii). V tng s hc sinh t 3 gii, 2 gii, 1 gii tng dn nn a < b < c. V bt k 2 mn no cng c t nht 1 hc sinh t gii c 2 mn nn: - C t nht 1 hc sinh t gii c 2 mn Vn v Ton. - C t nht 1 hc sinh t gii c 2 mn Ton v Ngoi Ng. - C t nht 1 hc sinh t gii c 2 mn Vn v Ngoi Ng. Do vy b= 3. Gi s a = 2 th b b nht l 3, c b nht l 4; do tng s gii b nht l: 3 x 2 + 2 x 3 + 4 = 16 > 15 (loi). Do a < 2, nn a = 1. Ta c: 3 x 1 + 2 x b + c = 15 suy ra: 2 x b + c = 12. Nu b = 3 th c = 12 - 2 x 3 = 6 (ng). Nu b = 4 th c = 12 - 2 x 4 = 4 (loi v tri vi iu kin b < c) Vy c 1 bn t 3 gii, 3 bn t 2 gii, 6 bn t 1 gii. i tuyn c s hc sinh l: 1 + 3 + 6 = 10 (bn).

Bi 80: in s S dng cc s 3, 5, 8, 10 v cc du +, - , x in vo mi cn trng bng sau: ( Ch c in mt du hoc mt s vo mi hng hoc mi ct. in t tri sang phi, t trn xung di)


16

Bi gii: Bn c c th xt cc tng theo tng hng, tng ct v khng kh khn lm s c kt qu sau:

Bi 81: 20 Gi da hu Tr v Dng gip b m xp 65 qu da hu mi qu nng 1kg, 35 qu da hu mi qu nng 2kg v 15 qu da hu mi qu nng 3kg vo trong 20 gi. Mi ngi cng ang lm vic, Tr chy n bn hc ly giy bt ra ghi... ghi v Tr la ln: C xp th no i chng na, chng ta lun tm c 2 gi trong 20 gi ny c khi lng bng nhau. Cc bn hy chng t l Tr ni ng.

Bi gii: Tng khi lng da l: 1 x 65 + 2 x 35 + 3 x 15 = 180 (kg). Gi s khi lng da mi gi khc nhau th tng khi lng da 20 gi b nht l: 1 + 2 + 3 + ... + 19 + 20 = 210 (kg). V 210 kg > 180 kg nn chc chn phi c t nht 2 gi trong 20 gi c khi lng bng nhau. Vy Tr ni ng.

Bi 82: Hong mua 6 quyn v, Hng mua 3 quyn v. Hai bn gp s v ca mnh vi s v ca bn Sn, ri chia u cho nhau. Sn tnh rng mnh phi tr cc bn ng 800 ng. Tnh gi tin 1 quyn v, bit rng c ba bn u mua cng mt loi v.

Bi gii:


17

V Hong v Hng gp s v ca mnh vi s v ca Sn, ri chia u cho nhau, nn tng s v ca ba bn l mt s chia ht cho 3. S v ca Hong v Hng u chia ht cho 3 nn s v ca Sn cng l s chia ht cho 3. S v ca Sn phi t hn 6 v nu s v ca Sn bng hoc nhiu hn s v ca Hong (6 quyn) th sau khi gp v li chia u Sn s khng phi tr thm 800 ng. S v ca Sn khc 0 (Sn phi c v ca mnh th mi gp chung vi cc bn c ch!), nh hn 6 v chia ht cho 3 nn Sn c 3 quyn v. S v ca mi bn sau khi chia u l: (6 + 3 + 3) : 3 = 4 (quyn) Nh vy Sn c cc bn a thm: 4 - 3 = 1 (quyn) Gi tin mt quyn v l 800 ng. Bi 83: Hy in cc s t 1 n 9 vo cc trng c cc php tnh ng

Bi gii: t cc ch ci vo cc trng:

Theo u bi ta c cc ch ci khc nhau biu th cc s khc nhau. Do : a 1; c 1; d 1; b > 1; e > 1. V 9 = 1 x 9 = 3 x 3 nn b 9 v e 9; v 7 = 1 x 7 nn b 7 v e 7. Do : b = 6 v e = 8 hoc b = 8 v e = 6. V 6 = 2 x 3 v 8 = 2 x 4 nn a = b : c = e : d = 2.

Trong cc trng a, b, c, d, e c cc s 2, 3, 4, 6, 8; do ch cn cc s 1, 5, 7, 9 in vo cc trng g, h, i, k. * Nu e = 6 th g = 7 v h = 1. Do a = i - k = 9 - 5 = 42 (loi).


18

* Nu e = 8 th g = 9 v h = 1. Do a = i - k = 7 - 5 = 2 (ng). Khi : b = 6 v c = 3.

Kt qu:

Bi 84: C 13 tm ba, mi tm ba c ghi mt ch s v xp theo th t sau:

Khng thay i th t cc tm ba, hy t gia chng du cc php tnh + , - , x v du ngoc nu cn, sao cho kt qu l 2002. Bi gii: Bi ton c rt nhiu cch t du php tnh v du ngoc. Xin nu mt s cch: Cch 1: (123 + 4 x 5) x (6 + 7 - 8 + 9 + 1 - 2 - 3 + 4) = 2002

Cch 2: (1 x 2 + 3 x 4) x (5 + 6) x [(7 + 8 + 9) - (1 + 2 x 3 + 4)] = 2002 Cch 3: (1 + 2 + 3 + 4 x 5) x (6 x 7 + 8 + 9 - 1 + 23 - 4) = 2002

Bi 85: Hai bn Huy v Nam i mua 18 gi bnh v 12 gi ko n lp lin hoan. Huy a cho c bn hng 2 t 100000 ng v c tr li 72000 ng. Nam ni: C tnh sai ri. Bn hy cho bit Nam ni ng hay sai? Gii thch ti sao? Bi gii: V s 18 v s 12 u chia ht cho 3, nn tng s tin mua 18 gi bnh v 12 gi ko phi l s chia ht cho 3. V Huy a cho c bn hng 2 t 100000 ng v c tr li 72000 ng, nn s tin mua 18 gi bnh v 12 gi ko l: 100000 x 2 - 72000 = 128000 (ng). V s 128000 khng chia ht cho 3, nn bn Nam ni C tnh sai ri l ng. Bi 86: C hai ci ng h ct 4 pht v 7 pht. C th dng hai ci ng h ny o thi gian 9 pht c khng?

Bi gii:


19

C nhiu cch o c 9 pht: Bn c th cho c 2 ci ng h ct cng chy mt lc v chy ht ct 3 ln. Khi ng h 4 pht chy ht ct 3 ln (4 x 3 = 12(pht)) th bn bt u tnh thi gian, t lc n khi ng h 7 pht chy ht ct 3 ln th va ng c 9 pht (7 x 3 - 12 = 9(pht)); hoc cho c hai ng h cng chy mt lc, ng h 7 pht chy ht ct mt ln (7 pht), ng h 4 pht chy ht ct 4 ln (16 pht). Khi ng h 7 pht chy ht ct ta bt u tnh thi gian, t lc n lc ng h 4 pht chy ht ct 4 ln l va ng 9 pht (16 - 7 = 9 (pht)); ...

Bi 87:

Vui xun mi, cc bn cng lm php ton sau, nh rng cc ch ci khc nhau cn thay bng cc ch s khc nhau, cc ch ci ging nhau thay bng cc ch s ging nhau.

NHAM + NGO = 2002

Bi gii: - V AG m ch s hng chc ca tng l 0 nn php cng c nh 1 sang hng trm nn hng trm: H + N + 1 (nh) = 10; nh 1 sang hng nghn. Do H + N = 10 - 1 = 9.

- Php cng hng nghn: N + 1 (nh) = 2 nn N = 2 - 1 = 1. Thay N = 1 ta c: H + 1 = 9 nn H = 9 - 1 = 8

- Php cng hng n v: C 2 trng hp xy ra: * Trng hp 1: Php cng hng n v khng nh sang hng chc. Khi : M + O = 0 v A + G = 10. Ta c bng: (Lu 4 ch M, O, A, G phi khc nhau v khc 1; 8)

* Trng hp 2: Php cng hng n v c nh 1 sang hng chc. Khi : M + O = 12 v A + G = 9. Ta c bng:

Vy bi ton c 24 p s nh trn.


20

Bi 88: Hy xp 8 qun min vo mt hnh vung 4x4 sao cho tng s chm trn cc hng ngang, dc, cho ca hnh vung u bng 11.

Li gii: C ba cch gii c bn sau:

T ba cch gii c bn ny c th to nn nhiu phng n khc, chng hn:

Bi 89: S dng cc con s trong mi bin s xe t 39A 0452, 38B 0088, 52N 8233 cng cc du +, -, x, : v du ngoc ( ), [ ] lm thnh mt php tnh ng.

Li gii: * Bin s 39A 0452. Xin nu ra mt s cch: (4 x 2 - 5 + 0) x 3 = 9 5 x 2 - 4 + 3 + 0 = 9

45 : 9 - 3 - 2 = 0 (9 + 2 - 3) x 5 = 40 (4 + 5) : 9 + 2 + 0 = 3

9 : 3 - ( 5 - 4 + 2) = 0 3 - 9 : (4 + 5) - 0 = 2

9 : (4 + 5) + 2 + 0 = 3 (9 + 5) : 2 - 4 + 0 = 3

9 + 3 : (5 - 2) + 0 = 4 5 + 2 - 9 : 3 - 0 = 4

(9 : 3 + 0) + 4 - 2 = 5 (9 + 3) : 4 + 0 + 2 = 5 . . . .

* Bin s 38B 0088. C nhiu li gii da vo tnh cht nhn mt s vi s 0 38 x 88 x 0 = 0

hoc tnh cht chia s 0 cho mt s khc 0

0 : (38 + 88) = 0


21

Mt vi cch khc: (9 - 8) + 0 - 8 : 8 = 0

8 : 8 + 8 + 0 + 0 = 9 . . . . * Bin s 52N 8233. Xin nu ra mt s cch: 5 x 2 - 8 + 3 - 3 = 2 8 : (5 x 2 - 3 - 3) = 2

[(23 - 3) : 5] x 2 = 8 (5 + 2 + 2) - (3 : 3) = 8

(8 : 2 - 3) x (3 + 2) = 5 [(8 + 2) x 3 : 3] : 2 = 5

(5 x 2 + 3 + 3) : 2 = 8 3 x 3 - 5 + 2 + 2 = 8 . . . .

Bi 90: Mt chic ng h ang hot ng bnh thng, hin ti kim gi v kim pht ang khng trng nhau. Hi sau ng 24 gi (tc 1 ngy m), hai kim trng nhau bao nhiu ln? Hy lp lun lm ng sng t kt qu . Li gii: Vi mt chic ng h ang hot ng bnh thng, c mi gi tri qua th kim pht quay c mt vng, cn kim gi quay c 1/12 vng. Hiu vn tc ca kim pht v kim gi l: 1 - 1/12 = 11/12 (vng/gi) Thi gian hai kim trng nhau mt ln l: 1 : 11/12 = 12/11 (gi) Vy sau 24 gi hai kim s trng nhau s ln l : 24 : 12/11 = 22 (ln). Bi 91: C ba ngi dng chung mt kt tin. Hi phi lm cho ci kt t nht bao nhiu kho v bao nhiu cha kt ch m c nu c mt t nht hai ngi?

Li gii: V kt ch m c nu c mt t nht hai ngi, nn s kho phi ln hn hoc bng 2. a) Lm 2 kho. + Nu lm 3 cha th s c hai ngi c cng mt loi cha; hai ngi ny khng m c kt. + Nu lm nhiu hn 3 cha th t nht c mt ngi cm 2 cha khc loi; ch cn mt ngi ny m c kt. Vy khng th lm 2 kho. b) Lm 3 kho + Nu lm 3 cha th cn phi c ba ngi mi m c kt. + Nu lm 4 cha hoc 5 cha th t nht c hai ngi khng m c kt. + Nu lm 6 cha (mi kho 2 cha) th mi ngi cm hai cha khc nhau th ch cn hai ngi bt k l m c kt.

Vy t nht phi lm 3 kho v mi kho lm 2 cha.


22

Bi 92 : C 4 tm g di v 4 tm g hnh cung trn. Nu sp xp nh hnh bn th c 4 chung nht 4 ch th, nhng 1 ch li cha c chung. Bn hy xp li cc tm g c 5 chung cho mi ch th c mt chung ring.

Bi gii : Bi ton c nhiu cch xp. Xin nu ra ba cch xp nh sau:

Bi 93: Mt phn xng c 25 ngi. Hi rng trong phn xng c th c 20 ngi t hn 30 tui v 15 ngi nhiu hn 20 tui c khng?

Bi gii: V ch c 25 ngi, m trong c 20 t hn 30 tui v 15 ngi nhiu hn 25 tui, nn s ngi c im 2 ln l: (20 + 15) - 25 = 10 (ngi) y chnh l s ngi c tui t hn 30 tui v nhiu hn 20 tui (t 21 tui n 29 tui).

S ngi t 30 tui tr ln l: 25 - 20 = 5 (ngi) S ngi t 20 tui tr xung l: 25 - 15 = 10 (ngi) S ngi t hn 30 tui l: 10 + 10 = 20 (ngi) S ngi nhiu hn 20 tui l: 10 + 5 = 15 (ngi)


23

Vy c th c 20 ngi di 30 tui v 15 ngi trn 20 tui; trong t 21 n 29 tui t nht c hai ngi cng tui. Bi 94: Tm 4 s t nhin lin tip c tch l 3024 Bi gii: Gi s c 4 s u l 10 th tch l 10 x 10 x 10 x 10 = 10000 m 10000 > 3024 nn c 4 s t nhin lin tip phi b hn 10. V 3024 c tn cng l 4 nn c 4 s phi tm khng th c tn cng l 5. Do c 4 s phi hoc cng b hn 5, hoc cng ln hn 5. Nu 4 s phi tm l 1; 2; 3; 4 th: 1 x 2 x 3 x 4 = 24 < 3024 (loi) Nu 4 s phi tm l 6; 7; 8; 9 th: 6 x 7 x 8 x 9 = 3024 (ng) Vy 4 s phi tm l 6; 7; 8; 9.

Bi 95: C 3 loi que vi s lng v cc di nh sau: - 16 que c di 1 cm - 20 que c di 2 cm - 25 que c di 3 cm Hi c th xp tt c cc que thnh mt hnh ch nht c khng?

Bi gii: Mt hnh ch nht c chiu di (a) v chiu rng (b) u l s t nhin (cng mt n v o) th chu vi (P) ca hnh phi l s chn: P = (a + b) x 2

Tng di ca tt c cc que l: 1 x 16 + 2 x 20 + 3 x 25 = 131 (cm) V 131 l s l nn khng th xp tt c cc que thnh mt hnh ch nht c. Bi 96: Hy pht hin ra mi lin h gia cc s ri s dng mi lin h in s hp l vo (?)

Bi gii: cho gn, ta k hiu cc s trn nhng trn theo bng sau:


24

Ly A chia cho K: 72 : 9 = Ly G chia cho C: 8 : 1 = Ly B chia cho H: 16 : 2 = Ly E chia cho D: 24 : 3 = u cho cng mt kt qu . Vy (?) l 8.

Bi 97: C gio yu cu: Cc con ly 6 im trn mt ng trn, ni cc im bi cc on thng t bi mc xanh hoc mc . Bn lp trng tp hp cc hnh v li v xem, bn tht ln: Bn no cng v c 1 tam gic m 3 cnh cng mu mc! Bn hy th lm li xem. Ai c th lp lun lm r tnh cht ny?

Bi gii: C nhiu cch gii, y l mt trong cc cch gii bi ny: Ta gi 6 im nm trn ng trn l A1, A2, A3, A4, A5, A6. Bng bt xanh v ta ni A1 vi 5 im cn li ta c 5 on thng c hai mu xanh hoc .

Theo nguyn l irchl c t nht 3 on thng cng mu. Khng lm mt tnh tng qut, ta ni 3 on A1A2, A1A3, A1A4 bng bt mu . Ta ni tip A2A4 v A2A3. tam gic A1A2A3 v tam gic A1A2A4 c 3 cnh khng cng mu th A2A4 v A2A3 phi t mu xanh. By gi ta tip tc ni A3A4, ta thy A3A4 c t bng bt k mu xanh hoc th ta cng c t nht mt tam gic c 3 cnh cng mu (hoc A1A3A4 c 3 cnh hoc A2A3A4 c 3 cnh mu xanh). Bi 98: Thi bn sng Hm nay Dng i thi bn sng. Dng bn gii lm, Dng bn hn 11 vin, vin no cng trng bia v u trng cc vng 8;9;10 im. Kt thc cuc thi, Dng c 100 im. Dng vui lm. Cn cc bn c bit Dng bn bao nhiu vin v kt qu bn vo cc vng ra sao khng?

Bi gii: S vin n Dng bn phi t hn 13 vin (v nu Dng bn 13 vin th Dng c s im t nht l: 8 x 11 + 9 x 1 + 10 x 1 = 107 (im) > 100 im, iu ny v l). Theo bi Dng bn hn 11 vin nn s vin n Dng bn l 12 vin. Mt khc 12 vin u trng vo cc vng 8, 9, 10 im nn t nht c 10 vin vo vng 8 im, 1 vin vo vng 9 im, 1 vin vo vng 10 im. Do s im Dng bn c t nht l: 8 x 10 + 9 x 1 + 10 x 1 = 99 (im) S im ht i so vi thc t l:


25

100 - 99 = 1 (im) Nh vy s c 1 vin khng bn vo vng 8 im m bn vo vng 9 im; hoc c 1 vin khng bn vo vng 9 im m bn vo vng 10 im. Nu c 1 vin Dng khng bn vo vng 9 im m bn vo vng 10 im th tng cng s c 10 vin vo vng 8 im v 2 vin vo vng 10 im (loi v khng c vin no bn vo vng 9 im). Vy s c 1 vin khng bn vo vng 8 im m bn vo vng 9 im, tc l c 9 vin vo vng 8 im, 2 vin vo vng 9 im v 1 vin vo vng 10 im.

Bi 99: Ai xem ca nhc? Mt gia nh c nm ngi: b ni, b, m v hai bn Chi, Bo. Mt hm gia nh c tng 2 v mi xem ca nhc. Nm kin ca nm ngi nh sau: a) B ni v m i b) B v m i c) B v b ni i d) B ni v Chi i e) B v Bo i Sau cng, mi ngi theo kin ca b ni v nh vy trong kin ca mi ngi khc u c mt phn ng. B ni ni cu no? Bi gii: Mt bi ton lgc c bn v kh, sau y l li gii. Ta k hiu theo th t i xem ca nhc: n (B ni), m (m), b (B), C (Chi) v B (Bo) v nm ngi trn khi h khng i l n, m, b, C v B. Nh vy theo kin ca nm ngi l: a) n v m b) b v m

c) b v n d) n v C

e) b v B. C l cn phi nhn mnh rng: Mi trong nm trn u c mt phn ng v mt phn sai (tr ca b!). Cu m b ni ni l ng vi c nm trn. - Nu chn cu a) th khng c e tc b v B. - Nu chn cu b) th khng c d tc n v C. - Nu chn cu c) th cc kin khc c mt phn ng. B ni ni cu c) Nu hc sinh thch th lgc Ton th cn tm thm c nhiu cch gii khc. Bi 100: Chi bc dim Trn mt bn c 18 que dim. Hai ngi tham gia cuc chi: Mi ngi ln lt n phin mnh ly ra mt s que dim. Mi ln, mi ngi ly ra khng qu 4 que. Ngi no ly c s que cui cng th ngi thng. Nu bn c bc trc, bn c chc chn thng c khng?


26

Bi gii: Gi s rng A v B tham gia cuc chi m A ly dim trc. chc thng th trc ln cui cng A phi li 5 que dim, trc A phi li 10 que dim v ln bc u tin A li 15 que dim, khi d B c bc bao nhiu que th vn cn li s que A ch cn bc mt ln l ht.Mun vy th ln trc A phi li 10 que dim , khi d B bc bao nhiu que vn cn li s que m A c th bc cn li 5 que . Tng t nh th th ln bc u tin A phi li 15 que dim . Vi " chin lc" ny bao gi A cng l ngi thng cuc.