[Toán 7] Chứng minh $2^9+2^{99} \vdots 100$

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0973573959thuy

Bài giải :

Ta có : A = [TEX]2^9 + 2^{99} = 2^9 + (2^{11})^9 = (2 + 2^{11})(2^8 - 2^7. 2^{11} + 2^6 . 2^{22} - ... 2 . 2^{77} + 2^{88}) = 2050 . 2A = 4100 . A \vdots 100 [/TEX]

Chúc bạn học tốt ! :)
 
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H

hthtb22

*>Ta có:
[tex]A= 2^9 +2^{99}=2^2(2^7 + 2^{97})=4(2^7 + 2^{97}) \equiv 0 [/tex](mod 4).

*> [tex]2^5 = 32 \equiv [/tex] 7 (mod 25)
\Rightarrow [tex]2^{10} \equiv 7^2[/tex] (mod 25)
\Rightarrow [tex]2^{10} \equiv -1[/tex] (mod 25)


Mặt khác:
A=[tex] 2^9 +2^{99} =2^9(1+2^{90})[/tex]
Mà [tex](1+2^{90}) = 1 + (2^{10})^9 \equiv 1 +(-1) \equiv 0[/tex] (mod 25)
\Rightarrow [tex]2^9 +2^{99} \equiv 0 [/tex](mod 25)
(4;25)=1
\Rightarrow [tex]A \equiv 0 [/tex](mod 100)
\Rightarrow A chia hết cho 100.
 
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