[Toán 6] Bài toán chia hết

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thangtru00

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harrypham

Cho S = 1-3+3^2-3^3+3^4-3^5+........+3^98-3^99
a) Chứng minh S chia hết cho 20
b) Chứng tỏ rằng 3^100:4 dư 1

b) [TEX]3 \equiv -1 \pmod{4} \Rightarrow 3^{100} \equiv (-1)^{100}=1 \pmod{4}[/TEX].
Vậy [TEX]3^{100}[/TEX] chia 4 dư 1.

a) Ta có [TEX]3S=3-3^2+3^3-3^4+...+3^{97}-3^{98}+3^{99}-3^{100}[/TEX]
[TEX]\Rightarrow 3S+S= 1-3^{100} \Rightarrow S= \frac{1-3^{100}}{4}[/TEX].

Để chứng minh S chia hết cho 20 ta chứng minh [TEX]1-3^{100}[/TEX] chia hết cho 80.

Ta có [TEX]3^2=9 \equiv -1 \pmod{5} \Rightarrow 3^{100} \equiv (-1)^{50}=1 \pmod{5} \Rightarrow 1-3^{100} \equiv 1-1=0 \pmod{5}[/TEX].
Vậy [TEX]1-3^{100} \ \vdots 5[/TEX].

Ta có [TEX]3^4 = 81 \equiv 1 \pmod{16} \Rightarrow 3^{100} \equiv 1^{25}=1 \pmod{16} \Rightarrow 1-3^{100} \equiv 1-1=0 \pmod{16}[/TEX].
Vậy [TEX]1-3^{100} \ \vdots 16[/TEX].

Do [TEX](5,16)=1 \Rightarrow 1-3^{100} \vdots 16.5=80 \Rightarrow \frac{1-3^{100}}{4} \ \vdots 20 \Rightarrow \fbox{S \vdots 20}[/TEX].
 
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toán



b) [TEX]3 \equiv -1 \pmod{4} \Rightarrow 3^{100} \equiv (-1)^{100}=1 \pmod{4}[/TEX].
Vậy [TEX]3^{100}[/TEX] chia 4 dư 1.

a) Ta có [TEX]3S=3-3^2+3^3-3^4+...+3^{97}-3^{98}+3^{99}-3^{100}[/TEX]
[TEX]\Rightarrow 3S+S= 1-3^{100} \Rightarrow S= \frac{1-3^{100}}{4}[/TEX].

Để chứng minh S chia hết cho 20 ta chứng minh [TEX]1-3^{100}[/TEX] chia hết cho 80.

Ta có [TEX]3^2=9 \equiv -1 \pmod{5} \Rightarrow 3^{100} \equiv (-1)^{50}=1 \pmod{5} \Rightarrow 1-3^{100} \equiv 1-1=0 \pmod{5}[/TEX].
Vậy [TEX]1-3^{100} \ \vdots 5[/TEX].

Ta có [TEX]3^4 = 81 \equiv 1 \pmod{16} \Rightarrow 3^{100} \equiv 1^{25}=1 \pmod{16} \Rightarrow 1-3^{100} \equiv 1-1=0 \pmod{16}[/TEX].
Vậy [TEX]1-3^{100} \ \vdots 16[/TEX].

Do [TEX](5,16)=1 \Rightarrow 1-3^{100} \vdots 16.5=80 \Rightarrow \frac{1-3^{100}}{4} \ \vdots 20 \Rightarrow \fbox{S \vdots 20}[/TEX].
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