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NT
1
TH
7 tháng 11 2015
Gọi số cần tìm là abc (a > 0); a, b, c < 10)
Theo đề bài ta có: abc +a+b+c = 555
Nhìn vào biểu thức trên, ta thấy phép cộng ko có nhớ sang hàng trăm. Vậy a=5
Khi đó ta có: 5bc +5+b+c=555
500 +bc+5+b+c=555
505 +bb+c+c=555
bb +c x2=555-505
bb + c x2=50
Nếu c đạt giá trị lớn nhất là chín thì bb đạt giá trị nhỏ nhất là:
50-9x2=32, do đó b>2
Vì bb +cx2=50 nên bb <50 nên b<5
Vì 2<b<5 nên b=3 hoặc 4
Vì cx2 và 50 đều là số chẵn nên b phải là số chẵn. Do đó b=4
Khi đó ta có:
44+cx2=50
cx2=50-44
cx2=6
c=6:2=3
Vậy abc là 543.
Mỏi tay quá!
UK
1
BH
1
8 tháng 7 2016
Gọi số cần tìm là ab (a khác 0; a,b là các chữ số)
Ta có:
ab + a + b = 84
=> 10a + b + a + b = 84
=> 11a + 2b = 84
=> 11a < 84
=> a < 8 (1)
Do b là chữ số => b < 10 => 2b < 20
=> 11a > 64 => a > 5 (2)
Từ (1) và 92), do 2b chẵn, 84 chẵn => 11a chẵn => a = 6
Ta có: 11 x 6 + 2 x b = 84
=> 66 + 2 x b = 84
=> 2 x b = 84 - 66
=> 2 x b = 18
=> b = 18 : 2 = 9
Vậy số cần tìm là 69
Ủng hộ mk nha ^_-
VN
1
NH
3
4 tháng 9 2018
Gọi số đó là \(\overline{abcd}\)
Theo đề ta có :
\(\overline{abcd}+a+b+c+d=2005\)
\(1001a+101b+11c+2d=2005\)
tới đây là hết nha bạn vì đề thiếu nhiều nên tớ thua
4 tháng 9 2018
Sorry tớ xin lỗi tui ra rồi
Bài làm : tiếp theo nha :
\(1001a+101b+11c+2c=2005\)
\(\Rightarrow a=1\)
\(101b+11c+2c=1004\)
\(\Rightarrow b=9\)
\(\Rightarrow11c+2d=95\)
\(\Rightarrow c=8\)
\(\Rightarrow2d=7\Rightarrow d=\frac{7}{2}\)
Mà d là số tự nhiên => Không có nào thõa mãn đề bài
1 tháng 10 2018
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địt lồn mẹ mày bú buồi đầu buồi liếm lông chim
NT
2
7 tháng 7 2016
Số đó là : 263
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