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Bài 1
a0b = ab x 7
a x 100 + b = ( a x 10 + b ) x7
a x 100 + b = a x 10 x 7 + b x 7
Cùng bớt đi b
a x 100 = a x 70 + b x 6
Cùng bớt đi a x 70
a x 30 = b x 6
Cùng chia cho 6
a x 5 = b x 1
=>a = 1 ; b = 5
Vậy số đó là 15
2 bài kia bạn tự giải nha , mk lười lắm :)))))
cau hoi nay la cau hoi co 3 chu so chu khong hai la 2chu so
Gọi số cần tìm là abcd. Xóa đi chữ số hàng chục và đơn vị ta được số ab
Ta có :
abcd - ab = 3663
ab x100 + cd -ab = 3663
cd + ab x100 - ab = 3663
cd + ab x 99 = 3663
cd = 3663 - ab x99
cd = 37 x 99 - ab x 99
cd = (37- ab) x 99
Nhận xét : Tích của 99 với 1 số tự nhiên là một số bé hơn 100 nên 37 - ab phải = 0 hoặc1
Nếu 37 - ab = 0 thì ab = 37 và cd = 00
Nếu 37 - ab = 1 thì ab = 36 và cd = 99
Vậy số cần tìm là 3700 hoặc 3699
Gọi số có 4 chữ số cần tìm là \(\overline{abcd}\)
Theo đề ra, ta có: Khi xóa đi chữ số hàng chục và hàng đơn vị ta được số \(\overline{ab}\)
\(\overline{abcd}-\overline{ab}=3663\)
\(\overline{ab}\times100+\overline{cd}-\overline{ab}=3663\)
\(\overline{ab}\times99+\overline{cd}=3663\)
\(\overline{cd}=99\times\left(37-\overline{ab}\right)\)
Mà tích của 99 là một số tự nhiên bé hơn 100 nên có hai trường hợp là \(37-\overline{ab}=0\) hoặc \(37-\overline{ab}=1\)
Trường hợp 1: \(37-\overline{ab}=0\Rightarrow\overline{ab}=37\)
\(\Rightarrow\overline{cd}=0\)
Trường hợp 2: \(37-\overline{ab}=1\Rightarrow\overline{ab}=36\)
\(\Rightarrow\overline{cd}=99\)
Vậy số tự nhiên có 4 chữ số cần tìm là \(3700;3699\).
Gọi số cần tìm là \(\overline{abcd}\)
Xóa đi chữ số hàng chục và hàng đơn vị ta được số ab
Theo bài ra ta có:
\(\overline{abcd}-\overline{ab}=3663\)
\(100.ab+cd-ab=3663\)
\(cd+100.ab-ab=3663\)
\(cd+99.ab=3663\)
\(\Rightarrow cd=99.\left(37-ab\right)\)
\(\Rightarrow37-ab=0\) hoặc \(37-ab=1\)
Nếu 37 - ab = 0 thì ab = 37 và cd = 00
Nếu 37 - ab = 1 thì ab=36 và cd = 99
Vậy số cần tìm là: 3700 và 3699
nếu ta xóa đi chữ số hàng chục và hàng đơn vị của số đó đi thì số đó giảm đi 100 lần và giá trị của số hàng chục và hàng đơn vị ghép lại
Vậy giá trị đó là : 1983 : (100 - 1 ) = 20 dư 3
Vậy giá trị của số hàng chục và hàng đơn vị ghép lại là 03
Số đó là : (1983 - 3 ) : 99 x 100 + 3 =2003
Đáp số : 2003
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Ta gọi số cần tìm là abcd
Theo đề bài ta có'
abc+2016=abcd
abcx1+2016= abcx10+b
2016=abcx9+b
abc= 2016:9 +b
abc=224+b
Từ trên => abcd=224+2016= 2240
Gọi số cần tìm là abcd
Xóa đi chữ số hàng chục và hàng đơn vị ta được số ab
Theo bài cho , ta có :
abcd - ab = 4455
=> 100 . ab + cd - ab = 4455
=> ( 100 - 1 ) . ab + cd = 4455
=> 99 . ab + cd = 4455
=> cd = 99 . ( 45 - ab )
Ta nhận xét tích của 99 với 1 số tự nhiên là 1 số tự nhiên nhỏ hơn 100 nên ( 45 - ab ) phải bằng 0 hoặc 1 .
+ Nếu 45 - ab = 0 => ab = 45 và cd = 0
+ Nếu 45 - ab = 1 => ab = 44 và cd = 99
Vậy số phải tìm là 4500 hoặc 4499 .
Gọi số có 4 chữ số là : abcd
Nếu xóa chữ số hàng chục và đơn vị thì ta được số mới giảm đi 4455 đơn vị
=> abcd - ab =4455
=> ab x 100 + cd -ab =4455
=>99×ab+cd=45×99
=>cd=45×99−99×ab
=>cd=99×(45−ab)
Ta có: cd<100 ab)<100⇒45−ab=1 hoặc 45−ab>
+Nếu 45−ab=0thì ⇒ab=45;cd=99×0=0⇒cd=00
+Nếu 45−ab=1⇒ab=44;cd=99×1=99
Vậy số cần tìm là 4500 hoặc 4499.
gọi số ban đầu có dạng : \(\overline{abcd}\)
ta có : \(\overline{abcd}-\overline{ab}=3663\Leftrightarrow99\overline{ab}+\overline{cd}=3663\)'mà :
\(99\overline{ab}\le99\overline{ab}+\overline{cd}\le99\overline{ab}+99\)
Vậy : \(99\overline{ab}\le3663\le99\overline{ab}+99\) hay : \(36\le\text{}\overline{ab}\le37\)
vậy \(\orbr{\begin{cases}ab=36\Rightarrow3699\\ab=37\Rightarrow3700\end{cases}}\)
Vậy ta có hai số thỏa mãn đề bài là 3699 và 3700