aaaa = 1000a+100a+10a+a=1111a

mà 1111 chia hết cho 37

nên aaaa chia hết cho 37

Ta có :

aaaa=1111 . a

Mà \(1111⋮37\)

\(\Rightarrow1111.a⋮37\)

\(\Leftrightarrow\)aaaa \(⋮37\)

\(\RightarrowĐPCM\)

\(1458=2\times3^6\)

1458 = 2 x 3 ⁶

6*7 chia hết cho 3

=> 6 + * + 7 chia hết cho 3

=> 13 + * chia hết cho 3

=> * = {2;5;8}

* thuộc {2;5;8}

Ta có :

\(a:2=18\)

\(\Rightarrow a=18.2=36\)

Vậy \(a=36\)

ta có

a:2=18

=>a=18x2

=>a=36

......

\(5.\left[\left(85-35:7\right):8+90\right]\)

\(=5.\left[\left(85-5\right):8+90\right]\)

\(=5.\left(80:8+90\right)\)

\(=5.\left(10+90\right)\)

\(=5.100\)

\(=500\)

\(5\left[\left(85-35:7\right):8+90\right]\)

\(=5\left[\left(85-5\right):8=90\right]-50\)

\(=5\left[80:8+90\right]-50\)

\(=5\left[10+90\right]-50\)

\(=5.100-50\)

\(=450\)

A=4+2^2+2^3+.......+2^20 .Hỏi A có chia hết cho 128 không?
A = 4 + 4 + 2^3+.......+2^20
A = 2^3 + 2^3+ 2^4 + 2^5 +.......+2^20
A = 2.2^3 + 2^4 + 2^5 +2^6 + 2^7 + .......+2^20
A = 2^4 + 2^4 + 2^5 + 2^6 + 2^7 + .......+2^20
A= 2.2^4 + 2^5 + 2^6 + 2^7 + .......+2^20
A= 2^5 + 2^5 + 2^6 + 2^7 + .......+2^20
A = 2.2^5 + 2^6 + 2^7 + .......+2^20
A = 2^6 + 2^6 + 2^7 + .......+2^20
A= 2.2^6 + 2^7 + 2^8 + .......+2^20
A = 2^7 + 2^7 + 2^8 +.......+2^20
A = 2^7( 1 + 1+ 2 + ....+ 2^13)
A = 128 (1 + 1+ 2 + ....+ 2^13)
Vậy A chia hết cho 128