Tìm hai số tự nhiên a và b, biết rằng BCNN(a,b) = 300; ƯCLN(a,b) = 15.
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Do ƯCLN(a,b)=15 => a = 15 x m; b = 15 x n (m,n)=1
=> BCNN(a,b) = 15 x m x n = 300
=> m x n = 300 : 15 = 20
Giả sử a > b => m > n do (m,n)=1 => m = 20; n = 1 hoặc m = 5; n = 4
+ Với m = 20; n = 1 thì a = 15 x 20 = 300; b = 15 x 1 = 15
+ Với m = 5; n = 4 thì a = 15 x 5 = 75; b = 15 x 4 = 60
Vậy các cặp giá trị (m;n) thỏa mãn đề bài là: (300;15) ; (75;60) ; (15;300) ; (60;75)
Do ƯCLN(a,b)=15 => a = 15 x m; b = 15 x n (m,n)=1
=> BCNN(a,b) = 15 x m x n = 300
=> m x n = 300 : 15 = 20
Giả sử a > b => m > n do (m,n)=1 => m = 20; n = 1 hoặc m = 5; n = 4
+ Với m = 20; n = 1 thì a = 15 x 20 = 300; b = 15 x 1 = 15
+ Với m = 5; n = 4 thì a = 15 x 5 = 75; b = 15 x 4 = 60
Vậy các cặp giá trị (m;n) thỏa mãn đề bài là: (300;15) ; (75;60) ; (15;300) ; (60;75)
a) goi hai so la a ; b va a >b
vi UCLN(a,b)=18=>a=18k ; b=18q (trong do UCLN (k,q)=1 va k>q)
=>a+b=162
18k+18q =162
18(k+q)=162
k+q=9
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Vì BCNN (a,b) = 300 và ƯCLN (a,b)=15
Suy ra: a.b = 300.15 = 4500
Vì ƯCLN (a,b) =15 nên: a= 15m và b= 15n (với ƯCLN (m,n) = 1).
Vì a+15 =b,=>15m+15 =15n, =>15(m+1) =15n, => m+1= n.
Mà a.b =4500 nên ta có: 15m.15n =4500=>15.15.m.n =4500=> m.n = 20
Suy ra: m=1 và n=20 hoặc m=4 và n=5