x⋮14 và x<92

xϵB(14) và <92

B(14)=0,14,28,42,56,68,80,92,104,...

x<92 nên x=0,14,28,42,56,68,80,92.

x là 14,28,42,56,70,84

+ Nếu trong 2 số a;b có ít nhất 1 số chẵn

=> ab(a+b) chẵn \(\Rightarrow ab\left(a+b\right)⋮2\)

+ Nếu a;b cùng lẻ => a+b chẵn \(\Rightarrow ab\left(a+b\right)\) chẵn \(\Rightarrow ab\left(a+b\right)⋮2\)

\(\Rightarrow ab\left(a+b\right)⋮2\forall a;b\)

Gọi \(d=ƯCLN\left(2n+2;2n\right)\)

\(\Rightarrow\left(2n+2\right)⋮d\) và \(2n⋮d\)

\(\Rightarrow\left(2n+2-2n\right)⋮d\)

\(\Rightarrow2⋮d\)

\(\Rightarrow d=2\)

2612=2000+600+10+2

=2.1000+6.100+1.10+2

=2.10^3+6.10^2+1.10^1+2.1

(-76). (-2)=152

152

a) Đặt A = \(6^5.5-3^5\)

\(=\left(2.3\right)^5.5-3^5\)

\(=2^5.3^5.5-3^5\)

\(=3^5.\left(2^5.5-1\right)\)

\(=3^5.\left(32.5-1\right)\)

\(=3^5.159\)

\(=3^5.3.53⋮53\)

Vậy \(A⋮53\)

b) Đặt \(B=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\)

\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{119}.\left(1+2\right)\)

\(=2.3+2^3.3+...+2^{119}.3\)

\(=3.\left(2+2^3+...+2^{59}\right)⋮3\)

Vậy \(B⋮3\)

\(B=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2\right)+3^4.\left(1+2+2^2\right)+...+2^{118}.\left(1+2+2^2\right)\)

\(=2.7+2^4.7+...+2^{118}.7\)

\(=7.\left(2+2^4+...+2^{118}\right)⋮7\)

Vậy \(B⋮7\)

\(B=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)

\(+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2+2^3+2^4\right)+2^6.\left(1+2+2^2+2^3+2^4\right)\)

\(+2^{116}.\left(1+2+2^2+2^3+2^4\right)\)

\(=2.31+2^6.31+...+2^{116}.31\)

\(=31.\left(2+2^6+...+2^{116}\right)⋮31\)

Vậy \(B⋮31\)

\(B=\left(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}+2^{13}+2^{14}+2^{15}+2^{16}\right)\)

\(+...+\left(2^{113}+2^{114}+2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)+2^9.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)

\(+...+2^{113}.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)

\(=2.255+2^9.255+...+2^{113}.255\)

\(=255.\left(2+2^9+...+2^{113}\right)\)

\(=17.15.\left(2+2^9+...+2^{113}\right)⋮17\)

Vậy \(B⋮17\)

c) Đặt C = \(3^{4n+1}+2^{4n+1}\)

Ta có:

\(3^{4n+1}=\left(3^4\right)^n.3\)

\(2^{4n}=\left(2^4\right)^n.2\)

\(3^4\equiv1\left(mod10\right)\)

\(\Rightarrow\left(3^4\right)^n\equiv1^n\left(mod10\right)\equiv1\left(mod10\right)\)

\(\Rightarrow3^{4n+1}\equiv\left(3^4\right)^n.3\left(mod10\right)\equiv1.3\left(mod10\right)\equiv3\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(3^{4n+1}\) là \(3\)

\(2^4\equiv6\left(mod10\right)\)

\(\Rightarrow\left(2^4\right)^n\equiv6^n\left(mod10\right)\equiv6\left(mod10\right)\)

\(\Rightarrow2^{4n+1}\equiv\left(2^4\right)^n.2\left(mod10\right)\equiv6.2\left(mod10\right)\equiv2\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(2^{4n+1}\) là \(2\)

\(\Rightarrow\) Chữ số tận cùng của C là 5

\(\Rightarrow C⋮5\)

toán lớp 6 khó thật ý.

Dĩ nhiên.

=4.28.25

=(4.25).28

=100.28

=2800

  2². 28. 5²

= ( 2² . 5² ) . 28

= ( 4 . 25 ) .28

= 100 . 28

= 2800

Ta có số: \(\overline{513xy}\) để số này chia hết cho 2 thì \(y\in\left\{0;2;4;6;8\right\}\)

Để số này chia hết cho 5 thì \(y\in\left\{0;5\right\}\)

Và vừa chia hết cho 5 vừa chia hết cho 2 thì: \(y=0\)

Mà số này lại chia hết cho 3 

\(\Rightarrow5+1+3+x+0=9+x\) ⋮ 3

TH1: \(9+x=9\text{⇒}x=0\)

TH2: \(9+x=12\text{⇒}x=3\)

TH3: \(9+x=15\text{⇒}x=6\)

TH4: \(9+x=18\text{⇒}x=9\)

Vậy các cặp (x;y) thỏa mãn là: (0;0); (3;0); (6;0); (9;0) 

x=3          y=0