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xóa chữ số hđv thì số đó giảm 10 lần
số ban đầu là: 1188:(10-1)=132
đáp số:...
mk nghĩ bài này chỉ tìm dc số có 3 chữ số thôi, k mk nhé
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Gọi các chữ số hàng nghìn , trăm , chục , đơn vị lần lượt là abcd
Theo đề bài ta có:abcd = 1188+ab
Trong phép cộng, khi cộng 2 chữ số trong cùng một hàng thì có nhớ nhiều nhất là 1 nên ab chỉ có thể là 11 hoặc 12
Nếu ab = 11 thì abcd =1188 + 11 = 1199
Nếu ab = 12 thì abcd = 1188 + 12 = 1200
Vậy ta tìm đc 2 số thỏa mãn đề bài là : 1199 và 1200
Đs:1199 và 1200
Nhớ k mk nhe!!!
gọi số cần tìm là abcd .
ta có :
ab + 1188 = abcd
ab + 1188 = ab x 100 + cd
1188 + cd = ab x 99
suy ra ab = 12
và cd = 00
đ/s : 1200
nhé !
Gọi số phải tìm là abcd. Xoá đi chữ số hàng chục và hàng đơn vị ta được số ab
Theo đề bài ta có : abcd – ab = 4455
100 x ab + cd – ab = 4455
cd + 100 x ab – ab = 4455
cd + 99 x ab = 4455
cd = 99 x (45 – ab)
Ta nhận xét tích của 99 với 1 số tự nhiên là 1 số tự nhiên nhỏ hơn 100. Cho nên 45 – ab phải bằng 0 hoặc 1.
- Nếu 45 – ab = 0 thì ab = 45 và cd = 0.
- Nếu 45 – ab = 1 thì ab = 44 và cd = 99.
Số phải tìm là 4500 hoặc 4499
Gọi số cần tìm là \(\overline{abcd}\) Theo đề bài
\(\overline{abcd}-\overline{ab}=1188\)
\(\Rightarrow100x\overline{ab}+\overline{cd}-\overline{ab}=1188\)
\(\Rightarrow99x\overline{ab}=1188-\overline{cd}\) (1)
Ta có
\(99x\overline{ab}⋮99\Rightarrow1188-\overline{cd}⋮99\)
Mà \(1188⋮99\Rightarrow\overline{cd}⋮99\Rightarrow\overline{cd}=0\) hoặc \(\overline{cd}=99\)
+ Với \(\overline{cd}=0\) thay vào (1) \(\Rightarrow99x\overline{ab}=1188\Rightarrow\overline{ab}=1188:99=12\)
\(\Rightarrow\overline{abcd}=1200\)
+ Với \(\overline{cd}=99\) thay vào (1) \(\Rightarrow99x\overline{ab}=1188-99\Rightarrow\overline{ab}=11\)
\(\Rightarrow\overline{abcd}=1199\)
gọi số cần tìm là abcd,ta có :
ab + 1188 = abcd
ab + 1188 = ab x 100 + cd
1188 + cd = ab x 99
suy ra ab = 12
và cd = 00
Vậy số đó = 1200
Học vv!
Gọi số cần tìm là abcd (a\(\ne\)0; a;b;c;d < 10)
Theo bài ra ta có:
abcd - abc = 1188
\(\Rightarrow\)10 x abc + d - abc = 1188
\(\Rightarrow\)9abc + d = 1188
\(\Rightarrow\)abc = (1188 - d) : 9
\(\Rightarrow\)d là số dư trong phép chia của 1188 cho 9 và abc là thương của phép chia đó
Ta có: 1188 : 9 = 132 (dư 0)
Vậy d = 0; abc = 132
Vậy số cần tìm là 1320
Gọi số phải tìm là abcd ( a > 0 ; a,b,c,d < 10 )
Khi xóa đi cd ta được số mới ab .
Theo đề ra ta có : abcd = 1188 + ab
Trong phép cộng , khi cộng hai số trong cùng 1 hàng thì có nhớ nhiều nhất là 1 nên ab chỉ có thể là 11 hoặc 12
Nếu ab = 11 thì abcd = 1188 + 11 = 1199
Nếu ab là 12 thì abcd = 1188 + 12 = 1200
Vậy ta có hai số thỏa mãn đề bài là : 1199 và 1200
Đáp số : 1199 ; 1200