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\(A=17^{18}-17^{16}\\ =17^{16}\cdot\left(17^2-1\right)\\ =17^{16}\cdot\left(289-1\right)\\ =17^{16}\cdot288\\ =17^{16}\cdot18\cdot16⋮18\)
Vậy \(A⋮18\)
\(B=1+3+3^2+...+3^{11}\)
Ta có: \(52=4\cdot13\)
\(B=1+3+3^2+...+3^{11}\\ =\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{10}+3^{11}\right)\\ =1\cdot\left(1+3\right)+3^2\cdot\left(1+3\right)+...+3^{10}\cdot\left(1+3\right)\\ =\left(1+3\right)\cdot\left(1+3^2+...+3^{10}\right)\\ =4\cdot\left(1+3^2+...+3^{10}\right)⋮4\)
Vậy \(B⋮4\)
\(B=1+3+3^2+...+3^{11}\\ =\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^9+3^{10}+3^{11}\right)\\ =1\cdot\left(1+3+3^2\right)+3^3\cdot\left(1+3+3^2\right)+...+3^9\cdot\left(1+3+3^2\right)\\ =\left(1+3+3^2\right)\cdot\left(1+3^3+...+3^9\right)\\ =13\cdot\left(1+3^3+...+3^9\right)⋮13\)
Vậy \(B⋮13\)
Vì \(4\) và \(13\) là hai số nguyên tố cùng nhau nên tao có \(B⋮4\cdot13\Leftrightarrow B⋮52\)
Vậy \(B⋮52\)
\(C=3+3^3+3^5+...3^{31}\)
\(C=3+3^3+3^5+...+3^{31}\\ =\left(3+3^3\right)+\left(3^5+3^7\right)+...+\left(3^{29}+3^{31}\right)\\ =1\cdot\left(3+3^3\right)+3^4\cdot\left(3+3^3\right)+...+3^{28}\cdot\left(3+3^3\right)\\ =\left(3+3^3\right)\cdot\left(1+3^4+...+3^{28}\right)\\ =30\cdot\left(1+3^4+...+3^{28}\right)⋮15\left(\text{vì }30⋮15\right)\)
Vậy \(C⋮15\)
\(D=2+2^2+2^3+...+2^{60}\)
Tao có: \(21=3\cdot7;15=3\cdot5\)
\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\\ =2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\\ =\left(1+2\right)\cdot\left(2+2^3+...+2^{59}\right)\\ =3\cdot\left(2+2^3+...+2^{59}\right)⋮3\)
Vậy \(D⋮3\)
\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^3\right)+\left(2^5+2^7\right)+...+\left(2^{57}+2^{59}\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\\ =2\cdot\left(1+2^2\right)+2^5\cdot\left(1+2^2\right)+...+2^{57}\cdot\left(1+2^2\right)+2^2\cdot\left(1+2^2\right)+...+2^{58}\cdot\left(1+2^2\right)\\ =\left(1+2^2\right)\cdot\left(2+2^5+...+2^{57}+2^2+...+2^{59}\right)\\ =5\cdot\left(2+2^5+...+2^{57}+2^2+...+2^{59}\right)⋮5\)
Vậy \(D⋮5\)
\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\\ =2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+...+2^{58}\cdot\left(1+2+2^2\right)\\ =\left(1+2+2^2\right)\cdot\left(2+2^4+...+2^{58}\right)\\ =7\cdot\left(2+2^4+...+2^{58}\right)⋮7\)
Ta có:
\(D⋮3;D⋮5\Rightarrow D⋮3\cdot5\Leftrightarrow D⋮15\)
\(D⋮3;D⋮7\Rightarrow D⋮3\cdot7\Leftrightarrow D⋮21\)
Vậy \(D⋮15;D⋮21\)
Mình chỉ làm mẫu 1 câu thui nha:
\(A=17^{18}-17^{16}\)
\(A=17^{16}.17^2-17^{16}.1\)
\(A=17^{16}\left(17^2-1\right)\)
\(A=17^{16}.288\)
\(A=17^{16}.16.18\)
\(A⋮18\left(đpcm\right)\)
a) P=2+22+23+24+...+260 \(⋮\) 21 và 15
\(\Rightarrow\)P = 22+23+24+25+...+261
\(\Rightarrow\) (2P - P) = 261 - 2
\(\Rightarrow\) P = 261 - 2 = 2.(260 - 1)
Để P \(⋮\) 21 và 15 thì (260 - 1) \(⋮\)21 và 15
tức là (260 - 1) \(⋮\)3; 5; 7
*Ta có 260 - 1 = (24)15 = 1615 - 1
= (16 - 1).(1+16+162+163+...+1614)
= 15.(1+16+162+163+...+1614) \(⋮\) 15
Vậy P \(⋮\) 15 (1)
* Ta có 260 - 1 = (26)10 - 1 = 6410 - 1
= (64 - 1).(1+64+642+643+...+649 )
= 63 \(⋮\) (1+64+642+643+...+649 )
= 21.3.(1+64+642+643+...+649 ) \(⋮\) 21
P \(⋮\)21 (2)
Từ (1) và (2) \(\Rightarrow\) P \(⋮\)15 và 21