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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
Gọi số đó là ab (a; b là chữ số ; a khác 0)
theo bài cho ta có:
ab = 3ab
=> 10a + b = 3ab
=> 10a - 3ab + b = 0
=> a(10 - 3b) - \(\frac{1}{3}\)(10 - 3b) + \(\frac{10}{3}\)=0
=> 3a(10 - 3b) - (10 - 3b) + 10 = 0
=> (3a - 1)(10 - 3b) + 10 = 0
=> (3a - 1)(3b - 10) = 10 = 1.10 = 2.5 = 5.2 = 10.1
+) Nếu 3a - 1 = 1 thì a = 2/3(Loại vì a là chữ số)
+) Nếu 3a - 1 = 2 thì a = 1 và 3b - 10 = 5 => b = 5 (Chọn)
+) Nếu 3a - 1 = 5 thì a = 2 và 3b - 10 = 2 => b = 4 (Chọn)
+) Nếu 3a - 1 = 10 thì a = 11/3 (loại)
Vậy các số thỏa mãn yêu cầu là 15; 24
MK CHỈ LÀM BÀI ĐẦU CÒN BÀI SAU CẬU GIỰA VÀO NÓ MÀ LÀM NHA .VÌ MK HƠI BẬN :
VÌ SỐ ĐÓ GẤP 3 LẦN TỔNG NÊN :\(ab=3.\left(a+b\right)\)
\(=>10.a+b=3.a+3.b\)
\(=>7.a=2b=>3,5.a=b\)
VỚI \(a=1\)THÌ \(b=3,5\)(loại )
VỚI \(a=2\)THÌ \(b=7\)(CHỌN)\(=>ab=27\)
VỚI \(a=3\)thì \(b\)>=9(loại)
vẬY SỐ CẦN TÌM LÀ :27
- Gọi số cần tìm là ab (a khác 0 , a,b<10)
Theo bài ra ta có: (a+b) x 3 =ab
a x 3 +b x 3 = ab
a x 3 + b x 3 = a x 10 + b (áp dụng phân tích số)
b x 3 - b = a x 10 - a x 3
b x ( 3- 1 ) = a x ( 10 - 3) (áp dụng 1 số nhân 1 hiệu )
b x 2 = a x 7
Vì b x 2 = a x 7 nên a phải = 2 .
Với a = 2 thì b = 2 x 7 : 2 = 7
Ta được số :27
Thử lại : 27 : 3 = (2+7) (Đúng)
Vậy số cần tìm là 27
yêu cầu thứ 2 bạn tự làm nhé!
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