Tìm hai số tự nhiên a và b (14 < a < b) có BCNN bằng 140 và ƯCLN bằng 14
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TC
752
NM
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LT
25 tháng 2 2020
Vi ƯCLN(a,b) . BCNN(a,b) = 770. 14 = a.b
nên a.b=10780
Ta có UCLN (a,b) =14 nên tồn tại hai số tự nhiên m,n (m<n) sao cho a = 14m, b=14n và UCLN(m,n) =1
suy ra 14m . 14n = 10780
m.n=55
vì m<n và UCLN(m,n) =1
TH1: m=1 suy ra n=55 suy ra a=14, b = 770 (loại)
TH2: m=5 suy ra n = 11 suy ra a=70, b = 154 (loại)
Vậy không tìm được a,b thỏa mãn
LP
0
LP
0
KC
0
8 tháng 12 2015
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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29 tháng 10 2016
21453
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
2 tháng 7 2023
Theo bài ra ta có:
\(\left\{{}\begin{matrix}a=13.k\\b=13.d\end{matrix}\right.\) (k;d)=1;k<d
13.k.13.d = 715,13 =9295
k.d = 9295:13:13 = 55 = 5. 11
⇒k = 5; d = 11
a = 13.5 = 65
b = 13.11 = 143
Kết luận: a = 65; b = 143
a.b=ƯCLN(a,b).BCNN(a,b)=140.14=1960
Vì ƯCLN(a,b)=14=> a=14m; b=14n (ƯCLN (m,n)=1)
Mà a.b=1960 => 14.m.14.n=1960
m.n=10
mà 14<a<b nên 1<m<n
Ta có bảng:
Vậy a = 28; b=70