Ta có số ngày cập bến ít nhất của 3 thuyền là BCNN ( 5;10;8 )

5 = 5

8 = 2^3

10 = 2.5

=> BCNN ( 5;10;8 ) = 40

Vậy sau ít nhất 40 ngày thì cả 3 thuyền cùng cập bến.

40 ngày

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Ta có số ngày cập bến gần nhất của ba thuyền là BCNN( 5; 10; 8 )

\(5=5^1\)

10 = 2 . 5

8 = 2\(^3\)

\(\Rightarrow\) BCNN( 5; 10; 8 ) = \(5^1.2^3\) = 40

Vậy sau ít nhất 40 ngày thì cả ba thuyền cùng cập bến

ta có: 

BCNN(5,8,10)

5=5

8=2.2.2

10=2.5

=>BCNN(5,8,10)=2.2.2.5=40

Vậy sau ít nhất 40 ngày cả 3 thuyền cùng cập bến

ta có: 

BCNN(5,8,10)

5=5

8=2.2.2

10=2.5

=>BCNN(5,8,10)=2.2.2.5=40

Vậy sau ít nhất 40 ngày cả 3 thuyền cùng cập bến

Ta có số ngày cập bến ít nhất của ba thuyền là BCNN_{( 5;10;8)}

8 = 2³ 

10 = 2.5

Suy ra: BCNN_(5;10;8)= 40

Vậy sau ít nhất 40 ngày thì cả ba thuyền cùng cập bến

4.Vì a chia cho 4 dư 3, chia 5 dư 4 ,chia 6 dư 5

\(\Rightarrow\)a-1 chia hết cho 4;5;6

\(\Rightarrow\)a-1 \(\in\) BCNN(4,5,6)

4=4

5=5

6=6

BCNN (4;5;6)=4.5.6=120

a-1\(\in\) B(120)=BC(4,5,6)={0;120;240;360;..............}

a-1\(\Rightarrow\){-1;119;239;.............}

Mà 200<a<300

Vậy a=239

Số cần tìm là 239

Kết quả của bạn mà trừ 1 là sai, 239 - 1 = 238 không chia hết cho 4.

Tính như bạn thì thiếu 299 rồi =w=

Ta có: (a + 1) chia hết cho 4, 5, 6 => (a + 1) thuộc BC(4, 5, 6).

Mà BCNN(4, 5, 6) = 60 => (a + 1) thuộc B(60) => a + 1 thuộc {0; 60; 120; 180; 240; 300; 360;...} => a thuộc {59, 119; 179; 239; 299; 359}

Vì a trong khoảng 200 đến 300 => a = 239; 299.

a) BSCNN của 6 và 5 là 30,

Vậy thuyền thứ nhất sau 30 ngày thì cùng cập bến với thuyền thứ 2

b) BSCNN của 6 và 9 là 18

Sau 18 ngày thì thuyền thứ nhất cùng cập bến với thuyền thứ ba

c) BSCNN của 5,6 và 9 là 90

Sau 90 ngày thì cả 3 thuyền cùng cập bến 1 lúc

126 ngày thì cả 3 thuyền gặp nhau