Tìm một số có 4 chữ số sao cho số đó cộng với tổng các chữ số của nó ra 2005
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Mình biết đáp số rồi nhưng ko biết cách giải như thế nào ?
Một số tự nhiên có 4 chữ số biết rằng số đó cộng với tổng các chữ số của nó bằng 1993.Số đó là:1973
Gọi số đó là abcd
abcd + a + b + c + d = 1993
a x 1001 + b x 101 + c x 11 + d x 2 = 1993
a phải = 1 (vì nếu a = 2. 2 x 1001 = 2002, quá 1993)
1 x 1001 + b x 101 +c x 11 + d x 2 = 1993
b x 101 + c x 11 + d x 2 = 1993 - 1001 = 992
b phải = 9
9 x 101 + c x 11 + d x 2 = 992
c x 11 + d x 2 = 992 - 909 = 83
c = 7
7 x 11 + d x 2 = 83
d x 2 = 83 - 77 = 6
d = 6 : 2 = 3
Vậy số cần tìm là 1973
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địt lồn mẹ mày bú buồi đầu buồi liếm lông chim
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
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