1a

2c

3b

4d

5b

6c

7a

8d

9c

10b

11b

12c

bài 1

a37.68+37.72

=37.(68+32)

=37.100

=3700

TL: 20736

\(A=100+99+98+...+3+2+1\)

\(A=1+2+3+...+98+99+100\)

\(2A=\left(100+1\right)+\left(99+2\right)+\left(98+3\right)+...+\left(3+98\right)+\left(2+99\right)+\left(1+100\right)\)

\(=101+101+101+...+101+101+101\)

\(=101.100\)

\(A=\frac{101.100}{2}=5050\)

a) \(A=2^1+2^2+2^3+...+2^{12}\)

\(=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{11}+2^{12}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{11}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{11}\right)⋮3\)

b) \(A=2^1+2^2+2^3+...+2^{12}\)

\(=\left(2+2^2+2^3+2^4\right)+...+\left(2^9+2^{10}+2^{11}+2^{12}\right)\)

\(=2\left(1+2+2^2+2^3\right)+...+2^9\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+2^9\right)⋮5\)

c) \(A=2^1+2^2+2^3+...+2^{12}\)

\(=\left(2^1+2^2+2^3\right)+...+\left(2^{10}+2^{11}+2^{12}\right)\)

\(=2\left(1+2+2^2\right)+...+2^{10}\left(1+2+2^2\right)\)

\(=7\left(2+...+2^{10}\right)⋮7\)

+A=2+22+23+...+2602+22+23+...+260

+A=(2+22)+(23+24)+...+(259+260)(2+22)+(23+24)+...+(259+260)

+A=2.(1+2)+23.(1+2)+..+259.(1+2)2.(1+2)+23.(1+2)+..+259.(1+2)

+A=2.3+23.3+..+259+32.3+23.3+..+259+3

=>A chia hết cho 3

Mấy câu sau thì nhóm 3,4 là Ok.

\(A\)chia cho \(23\)thì dư \(14\)và \(A\)chia cho \(25\)thì dư \(16\)nên \(A+9\)chia hết cho cả \(23\)và \(25\).

Mà \(\left(23,25\right)=1\)nên \(A\)chia hết cho \(23.25=575\)

Do đó \(A+9\in B\left(575\right)=\left\{0,575,1150,...\right\}\)

mà \(A\)là số có ba chữ số nên \(A+9=575\Leftrightarrow A=566\)

\(a+b+c=5+6+6=17\).

Chọn C. 

Bài 5) 

a) Ta có: \(3^{44}=3^{4.11}=\left(3^4\right)^{11}=81^{11}\)

              \(4^{33}=4^{3.11}=\left(4^3\right)^{11}=64^{11}\)

Mà 81 > 64 nên \(81^{11}>64^{11}\)từ đó \(3^{44}>4^{33}\)