<\/p>\n\n\n
L\u1ea5y 5 s\u1ed1 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t v\u00e0 5 s\u1ed1 gi\u1ea3i nh\u1ea5t ng\u00e0y h\u00f4m tr\u01b0\u1edbc \u0111\u1ec3 t\u00ednh c\u1eb7p s\u1ed1 ng\u00e0y h\u00f4m sau. VD: gi\u1ea3i DB l\u00e0 12345; gi\u1ea3i nh\u1ea5t l\u00e0 45678, gh\u00e9p l\u1ea1i ta c\u00f3 s\u1ed1 1234545678. C\u1ed9ng theo lu\u1eadt Pascal ngh\u0129a l\u00e0 2 s\u1ed1 li\u00ean ti\u1ebfp c\u1ed9ng v\u1edbi nhau (n\u1ebfu t\u1ed5ng > 10 th\u00ec ch\u1ec9 l\u1ea5y h\u00e0ng \u0111\u01a1n v\u1ecb) ta s\u1ebd \u0111\u01b0\u1ee3c d\u00e3y s\u1ed1 m\u1edbi c\u00f3 chi\u1ec1u d\u00e0i nh\u1ecf h\u01a1n 1 \u0111\u01a1n v\u1ecb. L\u1eb7p l\u1ea1i qu\u00e1 tr\u00ecnh n\u00e0y \u0111\u1ebfn khi \u0111\u01b0\u1ee3c c\u1eb7p s\u1ed1 cu\u1ed1i c\u00f9ng.<\/p>\n\n\n\n
\u1ede v\u00ed d\u1ee5 tr\u00ean ta l\u1ea5y: 1 + 2 =3; 2 + 3 = 5; 3 + 4 =7\u2026 ta s\u1ebd c\u00f3 c\u1eb7p m\u1edbi: 357999135.<\/p>\n\n\n\n
357999135<\/p>\n\n\n\n
82688048<\/p>\n\n\n\n
0846842<\/p>\n\n\n\n
82046<\/p>\n\n\n\n
0240<\/p>\n\n\n\n
264<\/p>\n\n\n\n
80<\/p>\n\n\n\n
=> 80 l\u00e0 k\u1ebft qu\u1ea3 c\u1ea7n t\u00ecm! Soi c\u1ea7u Mi\u1ec5n ph\u00ed<\/p>\n\n\n\n
C\u0103n c\u1ee9 v\u00e0o k\u1ebft qu\u1ea3 x\u1ed5 s\u1ed1 c\u1ee7a nh\u1eefng ng\u00e0y tr\u01b0\u1edbc \u1edf c\u00f9ng 1 v\u1ecb tr\u00ed c\u1ee7a gi\u1ea3i, ta ch\u1ecdn ra 1 c\u1eb7p s\u1ed1 d\u1ef1 \u0111o\u00e1n cho h\u00f4m sau, c\u1ea7u \u1ed5n \u0111\u1ecbnh khi ch\u1ea1y s\u1ebd \u0111\u00fang t\u1eeb 4 ng\u00e0y tr\u1edf l\u00ean.<\/p>\n\n\n\n
V\u00ed d\u1ee5:<\/p>\n\n\n\n
C\u0103n c\u1ee9 k\u1ebft qu\u1ea3 XSMB<\/em><\/strong> ng\u00e0y 01\/01<\/p>\n\n\n\n
46316<\/p>\n\n\n\n
21028<\/p>\n\n\n\n
D\u1ef1 \u0111o\u00e1n 68-86, k\u1ebft qu\u1ea3 ng\u00e0y 02\/01 c\u00f3 68.<\/p>\n\n\n\n
V\u1eady theo c\u1ea7u n\u00e0y ta ch\u1ecdn ti\u1ebfp \u1edf c\u00f9ng v\u1ecb tr\u00ed c\u1ee7a k\u1ebft qu\u1ea3 ng\u00e0y 02\/01 \u0111\u1ec3 d\u1ef1 \u0111o\u00e1n cho ng\u00e0y 03\/01 07012<\/p>\n\n\n\n
81525.<\/p>\n\n\n\n
D\u1ef1 \u0111o\u00e1n c\u1eb7p s\u1ed1 25-52<\/p>\n\n\n
<\/p>\n\n\n
Ta x\u00e9t k\u1ebft qu\u1ea3 trong 2 h\u00f4m \u0111\u1ec3 \u0111\u00e1nh h\u00f4m th\u1ee9 3.<\/p>\n\n\n\n
Gi\u1ea3 s\u1eed ng\u00e0y \u0110\u1ea7u ( ng\u00e0y th\u1ee9 nh\u1ea5t) c\u00f3 2 s\u1ed1 cu\u1ed1i l\u00e0 58 -> ta suy ra ng\u00e0y th\u1ee9 3 s\u1ebd c\u00f3 36 ( v\u00ec 5->3 ; 6 )<\/p>\n\n\n\n
VD c\u1ee5 th\u1ec3 h\u01a1n : h\u00f4m nay \u0110\u1ec1 v\u1ec1 32 -> ng\u00e0y kia s\u1ebd c\u00f3 l\u00f4 15 ( v\u00ec 3->1 ; 2->5)<\/p>\n\n\n\n
Tuy nhi\u00ean, c\u00e1ch t\u00ednh n\u00e0y ch\u1ec9 t\u01b0\u01a1ng \u0111\u1ed1i ch\u00ednh x\u00e1c v\u1edbi c\u00e1c gi\u1ea3i \u0111\u1eb7c bi\u1ec7t ng\u00e0y th\u1ee9 Nh\u1ea5t c\u00f3 t\u1ed5ng t\u1eeb 20 -> 30.<\/p>\n\n\n\n
\u0110\u1ec3 ti\u1ec7n cho vi\u1ec7c tr\u00ecnh b\u00e0y, t\u00f4i s\u1ebd \u0111\u01b0a ra m\u1ed9t s\u1ed1 quy \u0111\u1ecbnh soi c\u1ea7u h\u00f4m nay chu\u1ea9n:<\/p>\n\n\n\n
\u0110\u1ec1 ng\u00e0y 1 = MN<\/p>\n\n\n\n
\u0110\u1ec1 ng\u00e0y 2 = CD<\/p>\n\n\n\n
L\u00f4 ng\u00e0y 3 (d\u1ef1 \u0111o\u00e1n \u0111\u01b0\u1ee3c) = AB<\/p>\n\n\n\n
trong c\u00f9ng m\u1ed9t s\u1ed1 th\u00ec A, B l\u00e0 kh\u00e1c nhau. Nh\u01b0ng \u1edf 2 ng\u00e0y kh\u00e1c nhau th\u00ec c\u00f3 th\u1ec3 gi\u1ed1ng nhau<\/p>\n\n\n\n
V\u00ed d\u1ee5: MN = 35 ( 3#5) CD = 39 ( 3#9, C=M) , xin nh\u1edb \u1edf \u0111\u00e2y l\u00e0 CD ch\u01b0a c\u00f3 quan h\u1ec7 g\u00ec v\u1edbi nhau.<\/p>\n\n\n\n
Ng\u00e0y m\u1ed9t c\u00f3 \u0110\u1ec1 = MN -> L\u00f4 ng\u00e0y ba = AB => \u0110\u00e1nh BA<\/p>\n\n\n\n
Ng\u00e0y 1 \u0111\u1ec1 = MN -> d\u1ef1 \u0111o\u00e1n L\u00f4 ng\u00e0y 3 = AB<\/p>\n\n\n\n
l\u1ea5y 10 \u2013 B = I , gi\u1eef nguy\u00ean A => \u0110\u00e1nh AI<\/p>\n\n\n\n
Ng\u00e0y m\u1ed9t \u0111\u1ec3 = MN -> d\u1ef1 \u0111o\u00e1n L\u00f4 ng\u00e0y ba = AB => \u0110\u00e1nh AB<\/p>\n\n\n\n
t\u1ee9c l\u00e0 \u1edf \u0111\u00e2y AB=CD ( k\u1ebft qu\u1ea3 ng\u00e0y 2 = v\u1edbi l\u00f4 d\u1ef1 t\u00ednh ng\u00e0y 3)<\/p>\n\n\n\n
=> \u0110\u00e1nh BA<\/p>\n\n\n\n
=> \u0110\u00e1nh BA<\/p>\n\n\n\n
vd: \u0111\u1ec1 ng\u00e0y 1 = 35 -> l\u00f4 ng\u00e0y 3 = 13. Nh\u01b0ng t\u1ed5ng ng\u00e0y 2 c\u0169ng = 35 -> 13 => ng\u00e0y 3 \u0111\u00e1nh l\u00f4 31<\/p>\n\n\n\n
=> \u0110\u00e1nh BA<\/p>\n\n\n\n
N\u1ebfu k\u1ebft qu\u1ea3 ng\u00e0y 1 c\u00f3 CD 2 nh\u00e1y, CE 1 nh\u00e1y, v\u1edbi E l\u00e0 b\u00f3ng c\u1ee7a D => l\u00f4 ng\u00e0y th\u1ee9 3 c\u00f3 th\u1ec3 \u0111\u01b0\u1ee3c suy ra t\u1eeb CE.<\/p>\n\n\n\n
V\u00ed d\u1ee5: ng\u00e0y 1 c\u00f3 l\u00f4 35 l\u00e0 2 nh\u00e1y, l\u00f4 30 c\u00f3 m\u1ed9t ph\u00e1t -> l\u00f4 ng\u00e0y th\u1ee9 3 = 18 ( v\u00ec 3->1 ; 0->8)<\/p>\n\n\n\n
\u0110\u1eb7c bi\u1ec7t: n\u1ebfu ng\u00e0y 1 c\u00f3 2 con th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n nh\u01b0 v\u1eady -> l\u00f4 ng\u00e0y 3 d\u1ef1 \u0111o\u00e1n \u0111\u01b0\u1ee3c 2 con. X\u1ebfp con th\u1ea5p trong 2 con \u0111\u00f3 l\u00ean tr\u01b0\u1edbc ( vd HI , TU v\u00ec HI< TU) , l\u1ea5y \u0111\u1ea7u 2 con \u0111\u00f3 gh\u00e9p v\u00e0o nhau ( \u0111\u01b0\u1ee3c HT) => \u0110\u00e1nh HT<\/p>\n\n\n\n
Ch\u00fac c\u00e1c b\u1ea1n th\u00e0nh c\u00f4ng!<\/p>\n\n\n
<\/p>\r\n
C\u00e1ch soi c\u1ea7u chu\u1ea9n theo lu\u1eadt Pascal L\u1ea5y 5 s\u1ed1 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t v\u00e0 5 s\u1ed1 gi\u1ea3i nh\u1ea5t ng\u00e0y h\u00f4m tr\u01b0\u1edbc \u0111\u1ec3<\/p>\n","protected":false},"author":1,"featured_media":10833,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[468],"tags":[378,251],"yoast_head":"\n