L\u00f4 \u0111\u1ec1 l\u00e0 m\u1ed9t trong nh\u1eefng tr\u00f2 ch\u01a1i c\u00f3 t\u00ednh may r\u1ee7i cao. Tuy nhi\u00ean, n\u1ebfu bi\u1ebft t\u00ednh to\u00e1n h\u1ee3p l\u00fd, b\u1ea1n ho\u00e0n to\u00e0n c\u00f3 th\u1ec3 mang v\u1ec1 s\u1ed1 ti\u1ec1n kh\u00f4ng h\u1ec1 nh\u1ecf th\u00f4ng qua tr\u00f2 gi\u1ea3i tr\u00ed n\u00e0y. Hi\u1ec7n nay c\u00f3 r\u1ea5t nhi\u1ec1u ph\u01b0\u01a1ng ph\u00e1p <\/span>soi c\u1ea7u l\u00f4 h\u00f4m nay mi\u1ec1n b\u1eafc<\/b> b\u1ea1n c\u00f3 th\u1ec3 \u00e1p d\u1ee5ng. Tham kh\u1ea3o b\u00e0i vi\u1ebft sau \u0111\u1ec3 hi\u1ec3u r\u00f5 h\u01a1n v\u1ec1 nh\u1eefng ph\u01b0\u01a1ng ph\u00e1p n\u00e0y nh\u00e9.<\/span><\/p>\n Ph\u01b0\u01a1ng ph\u00e1p n\u00e0y t\u00ednh qua c\u00e1ch \u00e1p d\u1ee5ng \u0111\u1ea7u \u0111u\u00f4i c\u1ee7a gi\u1ea3i \u0111\u1eb7c bi\u1ec7t 2 ng\u00e0y \u0111\u1ea7u tu\u1ea7n. Sau \u0111\u00f3, d\u00f9ng \u0111\u1ec3 t\u00ednh t\u1ed5ng \u0111\u1ec1 nu\u00f4i cho 5 ng\u00e0y cu\u1ed1i tu\u1ea7n. C\u1ee5 th\u1ec3, ta l\u1ea5y s\u1ed1 \u0111\u1ea7u ti\u00ean c\u1ee7a gi\u1ea3i \u0111\u1eb7c bi\u1ec7t ng\u00e0y th\u1ee9 2 v\u00e0 3 r\u1ed3i c\u1ed9ng l\u1ea1i l\u00e0m t\u1ed5ng \u0111\u1ec1 \u0111\u00e1nh cho n\u0103m ng\u00e0y cu\u1ed1i tu\u1ea7n. Tr\u01b0\u1eddng h\u1ee3p t\u1ed5ng l\u1edbn h\u01a1n 10, l\u1ea5y 2 s\u1ed1 c\u1ed9ng ti\u1ebfp l\u00e0m c\u1ea7u ch\u1ea1m \u0111\u00e1nh 5 ng\u00e0y cu\u1ed1i tu\u1ea7n.<\/span><\/p>\n Gi\u1ea3 d\u1ee5 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t ng\u00e0y th\u1ee9 2 l\u00e0 54863, ng\u00e0y th\u1ee9 3 l\u00e0 89573. Ta c\u00f3 2 \u0111\u1ea7u gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 5 v\u00e0 8. K\u1ebft qu\u1ea3 t\u00ednh l\u00e0 5+8=13, con s\u1ed1 13 l\u1edbn h\u01a1n 10 n\u00ean ta c\u1ed9ng ti\u1ebfp 1+3 = 4. Nh\u1eefng ng\u00e0y c\u00f2n l\u1ea1i c\u1ee7a tu\u1ea7n 5 ng\u00e0y, b\u1ea1n ch\u01a1i \u0111\u1ec1 ch\u1ea1m 4.<\/span><\/p>\nSoi c\u1ea7u l\u00f4 h\u00f4m nay mi\u1ec1n B\u1eafc 5 ng\u00e0y cu\u1ed1i tu\u1ea7n qua gi\u1ea3i \u0111\u1eb7c bi\u1ec7t <\/b><\/span><\/h2>\n
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