Tìm tất cả các số có 3 chữ số, sao cho tích 3 chữ số của nó bằng tổng 3 chữ số của nó.
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địt lồn mẹ mày bú buồi đầu buồi liếm lông chim
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
a) 102; 111; 120; 210; 201; 300
b) 4000 3100 3010 3001 2002 2020 2200 2011 2101 2110 1111 1003 1030 1300 1210 1201 1120 1102 1012 1021 v.v..
Nếu thiểu bổ sung hộ mk.
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!
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