Aratati ca numarul \( N=8^{16}-6^{16} \) este divizibil cu 400 .
Indicatie : \( N=2^{16}(4^{16}-3^{16}) \) si \( 400=2^4\cdot 5^2 \)
Divizibilitate
Divizibilitate
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Andi Brojbeanu
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re:divizibilitate
\( 2^{16}=2^4\cdot 2^{12};
4^{16}-3^{16}=(4^{8})^2-(3^{8})^2=(4^{8}+3^{8})(4^{8}-3^{8})=(4^{8}+3^{8})[(4^{4})^2-(3^{4})^2] \)\( =(4^{8}+3^{8})(4^4+3^4)(4^4-3^4) \)\( =(4^{8}+3^{8})(4^4+3^4)[(4^2)^2-(3^2)^2]= \)
\( =(4^{8}+3^{8})(4^4+3^4)(4^2-3^2)(4^2+3^2) \)\( =(4^{8}+3^{8})(4^4+3^4)(4^2-3^2)\cdot 5^2 \)
Atunci \( N=2^4\cdot 2^{12}\cdot 5^2\cdot (4^{8}+3^{8})(4^4+3^4)(4^2-3^2)=400\cdot 2^{12}\cdot (4^{8}+3^{8})(4^4+3^4)(4^2-3^2) \), divizibil cu 400.
4^{16}-3^{16}=(4^{8})^2-(3^{8})^2=(4^{8}+3^{8})(4^{8}-3^{8})=(4^{8}+3^{8})[(4^{4})^2-(3^{4})^2] \)\( =(4^{8}+3^{8})(4^4+3^4)(4^4-3^4) \)\( =(4^{8}+3^{8})(4^4+3^4)[(4^2)^2-(3^2)^2]= \)
\( =(4^{8}+3^{8})(4^4+3^4)(4^2-3^2)(4^2+3^2) \)\( =(4^{8}+3^{8})(4^4+3^4)(4^2-3^2)\cdot 5^2 \)
Atunci \( N=2^4\cdot 2^{12}\cdot 5^2\cdot (4^{8}+3^{8})(4^4+3^4)(4^2-3^2)=400\cdot 2^{12}\cdot (4^{8}+3^{8})(4^4+3^4)(4^2-3^2) \), divizibil cu 400.