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địt lồn mẹ mày bú buồi đầu buồi liếm lông chim
Gọi số đó là abc ta có abc=a+b+c. Giả sử \(9\ge a\ge b\ge c\ge0\) thì \(abc=a+b+c\le3a\) suy ra \(bc\le3\)
abc khác 0 (vì \(a\ge1\) nên a+b+c\(\ge1\))
bc=3 thì b=3, c=1, suy ra 3+1+a=3a suy ra a=2 (loại, vì a bé hơn b)
bc=2 => b=2, c=1 suy ra 3+a=2a suy ra a=3 (Chọn)
bc=1 => b=1, c=1 suy ra 2+a=a (loại)
Vậy Các số cần tìm là 123, 132, 213, 231, 312, 321
=>29b+38c chia hết cho 61
Nếu cả 2 số cùng chia hết cho 61. Vì (29;61)=1 và (38;61)=1 nên cả b,c đều chia hết cho 61. Vì b,c đều là chữ số nên chúng bằng 0. Từ đó dẫn đến a=0 ( loại) (1)
Nếu tổng số dư của 2 số chia hết cho 61
Giả sử b=61k+r , c=61q+n thì 29b+38c=29.(61k+r)+38.(61q+n)=29.61k+29r+38.61q+38n=(29.61k+38.61q)+(29r+38n)
Do đó tổng số dư của 2 só là 29r+38n=61
Vì 38n<61 nên n<\(1\frac{23}{38}\)
Nếu n=0 thì 29r=61 ( loại vì số dư ko là số tự nhiên)
Nếu n=1 thì 29r+38=61 =>29r=23 ( vô lí)
Do đó ko có STN thỏa mãn
bạn học lớp 6b k
T bắt quả tang con quỳnh lên mạng hỏi bài nhé