phep chia het

H

harrypham

a) Ta có [TEX]A_{1991}= 1991 \cdot 100^{995}+1991 \cdot 100^{994} \cdot 10+ 1991 \cdot 100^{994}+ \cdots + 1991 \cdot 100^2 \cdot 10+1991 \cdot 100^2+ 1991[/TEX].
[TEX]=1991 \cdot (100^{995}+ \cdots + 100^2+1)+ 1991 \cdot 10 \cdot \left( 100^{994}+ \cdots + 100^2 \right)[/TEX]

Ta có [TEX]100^{2k} \equiv 1 \pmod{101}[/TEX] và [TEX]100^{2k+1} \equiv -1 \pmod{101}[/TEX] với mọi [TEX]k \in \mathbb{N}[/TEX].
Do đó [TEX]A \equiv 1991 \cdot [ (-1)+1+(-1)+1+ \cdots + 1+1]+1991 \cdot 10 \cdot (1-1+1- \cdots +1) \equiv 1991+1991 \cdot 10 \equiv \fbox{85} \pmod{101}[/TEX].
 
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