<\/span>L\u00f4 4 c\u00e0ng \u0103n bao nhi\u00eau?<\/h3>\n\u0110i\u1ec1u m\u00e0 anh em quan t\u00e2m nhi\u1ec1u nh\u1ea5t ch\u1eafc ch\u1eafn l\u00e0 ti\u1ec1n \u0103n con l\u00f4 4 c\u00e0ng l\u00e0 bao nhi\u00eau? Anh em y\u00ean t\u00e2m v\u00ec c\u00e1ch \u0111\u00e1nh n\u00e0y ch\u1eafn ch\u1eafn s\u1ebd c\u00f3 t\u1ef7 l\u1ec7 ti\u1ec1n th\u01b0\u1edfng r\u1ea5t cao v\u00e0 h\u1ea5p d\u1eabn, nh\u01b0ng \u0111\u1ed5i l\u1ea1i l\u00e0 x\u00e1c su\u1ea5t tr\u00fang c\u0169ng kh\u00f4ng h\u1ec1 d\u1ec5 d\u00e0ng.<\/p>\n
N\u1ebfu nh\u01b0 t\u1ef7 l\u1ec7 \u0103n \u0111\u1ec1 4 c\u00e0ng l\u00e0 0.01%, \u0111\u1ec1 3 c\u00e0ng l\u00e0 0.1%, th\u00ec \u0103n l\u00f4 4 s\u1ebd r\u01a1i v\u00e0o kho\u1ea3ng g\u1ea7n 0.2%<\/p>\n
\u1ede m\u1ed7i m\u1ed9t nh\u00e0 c\u00e1i\/ch\u1ee7 l\u00f4 \u0111\u1ec1 s\u1ebd quy \u0111\u1ecbnh ti\u1ec1n \u0103n l\u00f4 4 kh\u00e1c nhau. Th\u00f4ng th\u01b0\u1eddng th\u00ec t\u1ef7 l\u1ec7 n\u00e0y r\u01a1i v\u00e0o kho\u1ea3ng 1 \u0103n 8880 ho\u1eb7c 1 \u0103n 9000, v\u1edbi s\u1ed1 ti\u1ec1n v\u1ed1n b\u1ecf ra ban \u0111\u1ea7u th\u1ea5p nh\u1ea5t ch\u1ec9 l\u00e0 1.000 vn\u0111.<\/p>\n
Ti\u1ec1n v\u1ed1n = S\u1ed1 ti\u1ec1n \u0111\u1eb7t c\u01b0\u1ee3c 1 con X s\u1ed1 l\u01b0\u1ee3ng con l\u00f4 \u0111\u00e1nh.<\/strong><\/p>\nTi\u1ec1n th\u01b0\u1edfng = Ti\u1ec1n v\u1ed1n X 8880 X (S\u1ed1 con tr\u00fang + s\u1ed1 nh\u00e1y n\u1ed5 m\u1ed7i con).<\/strong><\/p>\nTi\u1ec1n l\u1eddi = Ti\u1ec1n th\u01b0\u1edfng \u2013 ti\u1ec1n v\u1ed1n.<\/strong><\/p>\nV\u00ed d\u1ee5: Anh em \u0111\u00e1nh 10.000 vn\u0111 cho con 2612 v\u00e0 1458, chi\u1ec1u h\u00f4m \u0111\u00f3 con 2612 v\u00e0 1458 \u0111\u1ec1u tr\u00fang 1 nh\u00e1y th\u00ec s\u1ed1 ti\u1ec1n th\u01b0\u1edfng c\u1ee7a anh em theo c\u00f4ng th\u1ee9c ph\u00eda tr\u00ean s\u1ebd l\u00e0 20.000 X 8880 X 2 = 355.200.000 vn\u0111.<\/p>\n
V\u1eady s\u1ed1 ti\u1ec1n m\u00e0 anh em th\u1ef1c nh\u1eadn s\u1ebd l\u00e0 355.200.000 \u2013 20.000 = 355.180.000 vn\u0111. M\u1ed9t s\u1ed1 ti\u1ec1n qu\u00e1 cao n\u1ebfu anh em tr\u00fang \u0111\u01b0\u1ee3c con l\u00f4 4.<\/p>\n
\u0110\u00e2y ch\u1ec9 l\u00e0 c\u00e1ch t\u00ednh ti\u1ec1n tr\u00fang khi t\u1ef7 l\u1ec7 l\u00e0 1 \u0103n 8880, c\u00f3 nh\u1eefng nh\u00e0 c\u00e1i v\u1edbi t\u1ef7 l\u1ec7 \u0103n l\u00e0 1 \u0103n 9000 th\u00ec s\u1ed1 ti\u1ec1n th\u1ef1c t\u1ebf nh\u1eadn n\u00e0y s\u1ebd cao h\u01a1n nhi\u1ec1u.<\/p>\n
<\/span>T\u1ed5ng h\u1ee3p c\u00e1ch b\u1eaft 4 c\u00e0ng hi\u1ec7u qu\u1ea3 nh\u1ea5t<\/h2>\nD\u00f9 l\u00e0 b\u1ed1n c\u00e0ng<\/strong>, nh\u01b0ng c\u00e1ch soi c\u1ea7u b\u1eaft s\u1ed1 l\u00e0 kh\u00e1c nhau. T\u00f4i xin t\u1ed5ng h\u1ee3p c\u00e1c c\u00e1ch b\u1eaft l\u00f4 \u0111\u1ec1 4 c\u00e0ng hi\u1ec7u qu\u1ea3 nh\u1ea5t b\u00ean d\u01b0\u1edbi. C\u1ee5 th\u1ec3 nh\u01b0 sau:<\/p>\n<\/span>I. C\u00e1ch \u0111\u00e1nh \u0111\u1ec1 4 c\u00e0ng<\/h3>\nN\u1ebfu nh\u01b0 t\u1ef7 l\u1ec7 \u0111o\u00e1n tr\u00fang c\u1ee7a ch\u01a1i \u0111\u1ec1 truy\u1ec1n th\u1ed1ng l\u00e0 1\/100, \u0111\u1ec1 3 c\u00e0ng kh\u00f3 h\u01a1n l\u00e0 1\/1000, th\u00ec \u0111\u1ec3 b\u1eaft \u0111\u01b0\u1ee3c con \u0111\u1ec1 4 c\u00e0ng s\u1ebd l\u00e0 1\/10000 t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi 0.01%. V\u1edbi m\u1ee9c ti\u1ec1n th\u01b0\u1edfng cao ng\u1ea5t ng\u01b0\u1ee1ng m\u00e0 c\u00e1c nh\u00e0 c\u00e1i h\u1ea5p d\u1eabn anh em th\u00ec t\u1ef7 l\u1ec7 tr\u00fang th\u1ea5p c\u0169ng l\u00e0 \u0111i\u1ec1u \u0111\u01b0\u01a1ng nhi\u00ean.<\/p>\n
V\u1eady n\u00ean, anh em c\u1ea7n ph\u1ea3i c\u00f3 cho m\u00ecnh nh\u1eefng b\u00ed quy\u1ebft ho\u1eb7c kinh nghi\u1ec7m ch\u01a1i l\u00f4 \u0111\u1ec1 l\u00e2u n\u0103m thay v\u00ec ch\u1ec9 \u0111o\u00e1n b\u1eeba theo qu\u00e1n t\u00ednh.<\/p>\n
<\/span>1. Ph\u01b0\u01a1ng ph\u00e1p b\u1eaft t\u1ed5ng \u0111\u1ec1 4 ng\u00e0y<\/h4>\nC\u00e1ch b\u1eaft t\u1ed5ng \u0111\u1ec1 4 ng\u00e0y l\u00e0 m\u1ed9t ph\u01b0\u01a1ng ph\u00e1p b\u1eaft c\u00e0ng 4 chu\u1ea9n x\u00e1c c\u1ee7a m\u1ed9t tay d\u00e2n ch\u01a1i mi\u1ec1n B\u1eafc 3 l\u1ea7n li\u00ean t\u1ee5c th\u1eafng \u0111\u1ec1 4 c\u00e0ng.<\/p>\n
D\u00e2n ch\u01a1i m\u1ecdi n\u01a1i lan truy\u1ec1n nhau v\u1ec1 c\u00e1ch \u0111\u00e1nh n\u00e0y v\u00e0 t\u1ea5t nhi\u00ean r\u1eb1ng \u0111\u00e2y l\u00e0 m\u1ed9t c\u00e1ch \u0111\u00e1nh hi\u1ec7u qu\u1ea3 th\u00ec m\u1edbi \u0111\u01b0\u1ee3c lan r\u1ed9ng \u0111\u1ebfn nh\u01b0 v\u1eady.<\/p>\n
T\u1ed5ng \u0111\u1ec1 (ng\u00e0y 1 + ng\u00e0y 2 + ng\u00e0y 3 + ng\u00e0y 4) => CON \u0110\u1ec0 4 C\u00c0NG.<\/p>\n
V\u00ed d\u1ee5: Anh em mu\u1ed1n \u0111\u00e1nh \u0111\u1ec1 4 c\u00e0ng ng\u00e0y th\u1ee9 7, anh em xem th\u1ed1ng k\u00ea \u0111\u1ec1 \u0111\u1eb7c bi\u1ec7t c\u1ee7a c\u00e1c ng\u00e0y v\u1ec1 tr\u01b0\u1edbc:<\/p>\n
\n- Ng\u00e0y th\u1ee9 3: X\u1ed5 s\u1ed1 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t Mi\u1ec1n B\u1eafc l\u00e0 41512 \u2013 12: T\u1ed5ng \u0111\u1ec1 1 + 2 = 3.<\/li>\n
- Ng\u00e0y th\u1ee9 4: X\u1ed5 s\u1ed1 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 12771 \u2013 71: T\u1ed5ng \u0111\u1ec1 7 + 1 = 8.<\/li>\n
- Ng\u00e0y th\u1ee9 5: X\u1ed5 s\u1ed1 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 54193 \u2013 93: T\u1ed5ng \u0111\u1ec1 9 + 3 = 2.<\/li>\n
- Ng\u00e0y th\u1ee9 6: X\u1ed5 s\u1ed1 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 47156 \u2013 56: T\u1ed5ng \u0111\u1ec1 5 + 6 = 1.<\/li>\n<\/ul>\n
Theo ph\u01b0\u01a1ng ph\u00e1p ta c\u00f3 con 4 c\u00e0ng \u0111\u00e1nh ng\u00e0y th\u1ee9 7 l\u00e0: 3 8 2 1.<\/strong><\/p>\nL\u01b0u \u00fd: Ph\u01b0\u01a1ng ph\u00e1p n\u00e0y th\u01b0\u1eddng ch\u1ec9 c\u00f3 hi\u1ec7u qu\u1ea3 v\u00e0o th\u1ee9 7 v\u00e0 ch\u1ee7 nh\u1eadt, anh em l\u01b0u \u00fd.<\/p>\n
>>> <\/strong>Xem ph\u00e2n t\u00edch chuy\u00ean s\u00e2u v\u1ec1 T\u1ed5ng l\u00f4 \u0111\u1ec1<\/strong> t\u1ea1i \u0111\u00e2y.<\/p>\n<\/span>2. Ph\u01b0\u01a1ng ph\u00e1p b\u1eaft \u0111\u1ec1 theo t\u1ed5ng ng\u00e0y\/ th\u00e1ng v\u00e0 con b\u1ea1ch th\u1ee7<\/h4>\nPh\u01b0\u01a1ng ph\u00e1p n\u00e0y l\u00e0 kinh nghi\u1ec7m ch\u01a1i l\u00f4 \u0111\u1ec1 c\u1ee7a h\u01a1n 30 anh em l\u00f4 th\u1ee7 trong su\u1ed1t h\u01a1n 5 n\u0103m qua v\u1edbi t\u1ef7 l\u1ec7 tr\u00fang h\u01a1n 30%. \u0110\u01b0\u01a1ng nhi\u00ean l\u00e0 t\u1eeb 0.01% l\u00ean \u0111\u01b0\u1ee3c 30% th\u00ec c\u0169ng l\u00e0 m\u1ed9t v\u1ea5n \u0111\u1ec1 kh\u00f4ng h\u1ec1 \u0111\u01a1n gi\u1ea3n.<\/p>\n
V\u00ed d\u1ee5: H\u00f4m nay l\u00e0 ng\u00e0y 22\/11 => T\u1ed5ng ng\u00e0y + th\u00e1ng = 22 + 11 = 33.<\/p>\n
Sau \u0111\u00f3 anh em l\u1ea5y 3 + 3 = 6. V\u1edbi 6 l\u00e0 t\u1ed5ng c\u1ee7a c\u00e1c con s\u1ed1 nh\u01b0 06 \u2013 60 \u2013 15 \u2013 51 \u2013 24 \u2013 42 \u2013 33.<\/p>\n
Gi\u1ea3 s\u1eed con b\u1ea1ch th\u1ee7 \u0111\u1ec1 c\u1ee7a anh em ng\u00e0y h\u00f4m nay l\u00e0 51 th\u00ec anh em c\u00f3 th\u1ec3 \u0111\u00e1nh nh\u01b0 sau: 0651 \u2013 6051 \u2013 1551 \u2013 5151 \u2013 2451 \u2013 4251 \u2013 3351.<\/p>\n
>>> <\/strong>T\u00f4i \u0111\u00e3 ph\u00e2n t\u00edch v\u1ec1 c\u00e1ch t\u00ednh l\u00f4 theo ng\u00e0y<\/strong> nh\u01b0 th\u1ea7n c\u01a1 di\u1ec7u to\u00e1n v\u00f4 c\u00f9ng hi\u1ec7u qu\u1ea3 \u1edf nh\u1eefng b\u00e0i vi\u1ebft tr\u01b0\u1edbc, m\u1eddi anh em tham kh\u1ea3o.<\/p>\nC\u00e1ch t\u00ecm con b\u1ea1ch th\u1ee7 l\u00f4 th\u00ec c\u0169ng t\u00f9y v\u00e0o t\u1eebng anh em. Nhi\u1ec1u anh em nu\u00f4i s\u1ed1 ho\u1eb7c ch\u1ecdn ngay con \u0111\u1ec1 ra ng\u00e0y h\u00f4m tr\u01b0\u1edbc \u0111\u1ec3 \u0111\u00e1nh cho ng\u00e0y h\u00f4m nay.<\/p>\n
<\/span>3. Ph\u01b0\u01a1ng ph\u00e1p soi ng\u0169 h\u00e0nh b\u00f3ng l\u00f4 \u0111\u1ec1<\/h4>\nV\u00ed d\u1ee5: Con b\u1ea1ch th\u1ee7 c\u1ee7a anh em t\u00ednh to\u00e1n \u0111\u01b0\u1ee3c trong ng\u00e0y h\u00f4m nay l\u00e0 27, th\u00ec anh em s\u1ebd c\u00f3 c\u00f4ng th\u1ee9c nh\u01b0 sau:<\/p>\n
B\u00f3ng \u00e2m \u0111\u1ea7u con b\u1ea1ch th\u1ee7 + b\u00f3ng d\u01b0\u01a1ng \u0111u\u00f4i con b\u1ea1ch th\u1ee7 + b\u1ea1ch th\u1ee7 \u0111\u1ec1.<\/p>\n
\u1ede \u0111\u00e2y con 27 g\u1ed3m 2 con s\u1ed1 2 v\u00e0 7. Ch\u00fang ta ph\u00e2n t\u00edch \u0111\u01b0\u1ee3c: B\u00f3ng \u00e2m c\u1ee7a 2 l\u00e0 9, b\u00f3ng d\u01b0\u01a1ng c\u1ee7a 7 l\u00e0 2. V\u1eady ta gh\u00e9p l\u1ea1i \u0111\u01b0\u1ee3c con 92.<\/p>\n
V\u1eady con \u0111\u1ec1 4 c\u00e0ng c\u1ee7a anh em s\u1ebd l\u00e0: 9 2 2 7<\/strong><\/p>\nL\u01b0u \u00fd: Ph\u01b0\u01a1ng ph\u00e1p n\u00e0y ch\u1ec9 c\u00f3 th\u1ec3 \u00e1p d\u1ee5ng 1 tu\u1ea7n 2 ng\u00e0y, c\u00f3 th\u1ec3 r\u01a1i v\u00e0o 2 ng\u00e0y \u0111\u1ea7u tu\u1ea7n ho\u1eb7c 2 ng\u00e0y cu\u1ed1i tu\u1ea7n.<\/p>\n
>>> <\/strong>Xem th\u00eam: C\u00e1ch b\u1eaft b\u00f3ng \u00e2m b\u00f3ng d\u01b0\u01a1ng<\/strong> chu\u1ea9n x\u00e1c nh\u1ea5t.<\/p>\n<\/span>II. B\u00ed quy\u1ebft b\u1eaft con l\u00f4 4 c\u00e0ng hi\u1ec7u qu\u1ea3<\/h3>\nT\u1ef7 l\u1ec7 tr\u00fang th\u1ea5p \u0111\u1ed3ng ngh\u0129a v\u1edbi vi\u1ec7c anh em kh\u00f4ng th\u1ec3 ch\u1ec9 d\u1ef1a v\u00e0o may m\u1eafn. \u0110\u1ec3 th\u1eafng, anh em c\u1ea7n ph\u1ea3i c\u00f3 ph\u01b0\u01a1ng ph\u00e1p \u0111\u00e1nh v\u00e0 c\u00e1ch soi c\u1ea7u l\u00f4 4 c\u00e0ng<\/strong> hi\u1ec7u qu\u1ea3.<\/p>\nM\u1eddi anh em c\u00f9ng t\u00ecm hi\u1ec3u qua nh\u1eefng c\u00e1ch b\u00ean d\u01b0\u1edbi.<\/p>\n