Giải:

1ab +36 = ab1(a khác 0;a,b<10)

100+ ab +36 = ab  x 10 +1

136 +ab = ab x 10 +1

 135 +ab  = ab x 10

 ab x 9 = 135

 ab = 15

Vậy ab=15

\(\overline{1ab}+36=\overline{ab1}\)

\(100+10a+b+36=100a+10b+1\)

\(136+10a+b=100a+10b+1\)

\(135=90a+9b\)

\(135=9\left(10a+b\right)\)

\(135:9=\overline{ab}\)

\(\Rightarrow\overline{ab}=15\)

1AB + 36 = AB1

100 + 10A + B + 36 = 100A + 10B + 1

136 +10A + B = 100A + 10B + 1

135 = 90A + 9B

135 = 9(10A + B)

135 : 9 = AB

135 : 9 = AB. tick đg cho mil vs

ab=15 dung10000000

000000000000000000

00000000%

Đáp án:

$13520$ hoặc $63504.$

Giải thích các bước giải:

$\overline{abcde}=2\overline{ab}.\overline{cde}\Rightarrow 1000\overline{ab}+\overline{cde}=2\overline{ab}.\overline{cde}\Rightarrow 1000\overline{ab}=-\overline{cde}+2\overline{ab}.\overline{cde}\Rightarrow 1000\overline{ab}=(2\overline{ab}-1)\overline{cde}(\ast )\Rightarrow 1000\overline{ab}⋮2\overline{ab}-1$

Do $(\overline{ab};2\overline{ab}-1)=1$

$\Rightarrow 1000⋮2\overline{ab}-1$

$2\overline{ab}-1\ge 19(\overline{ab}$ nhỏ nhất là $10)$

Ước dương của $1000$

$Ư\left(1000\right)=\{1;2;4;5;8;10;20;25;40;50;100;125;200;250;500;1000\}$

Do $2\overline{ab}-1$ lẻ và $2\overline{ab}-1\ge 19$

$\Rightarrow (2\overline{ab}-1)\in \{25;125\}⊛2\overline{ab}-1=25\Rightarrow 2\overline{ab}=26\Rightarrow \overline{ab}=13(\ast )\Rightarrow 1000.13=(2.13-1)\overline{cde}\Rightarrow 13000=25\overline{cde}\Rightarrow \overline{cde}=520⊛2\overline{ab}-1=125\Rightarrow 2\overline{ab}=126\Rightarrow \overline{ab}=63(\ast )\Rightarrow 1000.63=(2.63-1)\overline{cde}\Rightarrow 63000=125\overline{cde}\Rightarrow \overline{cde}=504$

Vậy số thoả mãn là $13520$ hoặc $63504.$

\(\left(10a+b\right)b=100+10a+b\)

\(\Leftrightarrow10ab+b^2=100+10a+b\)

\(\Leftrightarrow10a\left(b-1\right)+b\left(b-1\right)=100\)

\(\Leftrightarrow\left(10a+b\right)\left(b-1\right)=100\)

Do \(10a+b>b-1\) ; \(10\le10a+b< 100\), các trường hợp có thể xảy ra:

TH1: \(\left\{{}\begin{matrix}b-1=2\\10a+b=50\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=3\\10a=47\left(l\right)\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}b-1=4\\10a+b=25\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=5\\a=2\end{matrix}\right.\)

TH3: \(\left\{{}\begin{matrix}b-1=5\\10a+b=20\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=6\\10a=14\left(l\right)\end{matrix}\right.\)

Vậy chữ số cần tìm là 25

\(\overline{ab}\cdot1,01=\overline{2b,a3}\)

=>1,01(10a+b)=20+b+0,1a+0,03

=>10,1a+1,01b-20,03-b-0,1a=0

=>10a+0,01b-20,03=0

=>10a+0,01b=20,03

=>a=2 và b=3

Vậy: Số cần tìm là 23