\u0110\u1ed1i v\u1edbi ph\u01b0\u01a1ng ph\u00e1p n\u00e0y b\u1ea1n s\u1ebd t\u00ednh t\u1ed5ng t\u1ea5t c\u1ea3 c\u00e1c ch\u1eef s\u1ed1 l\u1ea1i v\u1edbi nhau, sau \u0111\u00f3 l\u1ea5y k\u1ebft qu\u1ea3 \u0111\u00f3 \u0111\u1ec3 \u0111\u00e1nh \u0111\u1ec1 k\u00e9p t\u01b0\u01a1ng \u1ee9ng v\u1edbi t\u1eebng ch\u1eef s\u1ed1 trong t\u1ed5ng gi\u1ea3i \u0111\u1eb7c bi\u1ec7t.<\/p>\n
V\u00ed d\u1ee5 h\u00f4m nay XSMB ra gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 051036. Ta s\u1ebd c\u00f3 t\u1ed5ng l\u00e0 0+5+1+0+3+6= 15 n\u00ean ta s\u1ebd nu\u00f4i c\u1eb7p l\u00f4 k\u00e9p b\u1eb1ng 11 v\u00e0 55 cho 3 \u0111\u1ebfn 5 ng\u00e0y ti\u1ebfp theo.<\/p>\n
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ghi nh\u1edb: <\/span>h\u00e3y l\u01b0u l\u1ea1i v\u00e0 gi\u1edbi thi\u1ec7u t\u1edbi b\u1ea1n b\u00e8 trang n\u00e0y \u0111\u1ec3 c\u00f3 nh\u1eefng con s\u1ed1 \u0111\u1eb9p v\u00e0 \u0111\u01b0\u1ee3c soi c\u1ea7u x\u1ed5 s\u1ed1 ch\u00ednh x\u00e1c nh\u1ea5t<\/strong><\/span><\/strong><\/span><\/p>\n