26.Demonstrati ca numarul A=63^n +7^n+1 *3^2n+1 -21^n *3^n+2,n∈ N este divizibil cu 13. b)Demonstrati ca numarul B=35^n +7^n *5^n+2 +3*7^n+1 *5^n , n∈ N este divizibil cu 47. c)Aratati ca numarul A=7*12^n *3^n+1 +6*4^n+1 *9^n+2 +18^n+1 *2^n+1 este divizibil cu 2001,oricare ar fi n∈ N *.

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Răspunsuri

2014-08-31T17:40:58+03:00
A=63^n+7^(n+1)*3^(2n+1)-21^n*3^(n+2)
A=63^n+7*7^n*3*9^n-21^n*3^2*3^n=63^n+21*63^n-9*63^n=63^n(1+21-9)=13*63^n
⇒ A divizibil cu 13
S-au folosit formulele a^(m+n)=a^m*a^n , a^n*b^n=(a*b)^n

b)B=35^n+5^2*35^n+3*7*7^n*5^n=35^n+25*35^n+21*35^n=35^n(1+25+21)=47*35^n
⇒ B divizibil cu 47
c)C=7*12^n*3^n*3+6*4*4^n*9^n*9^2+18*18^n*2^n*2=
21*36^n+24*81*36^n+18*2*36^n=36^n(21+1944+36)=2001*36^n ⇒ C divizibil cu 2001
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