Bạn đã học đồng dư chưa ?
Nếu rồi thì có thể tham khảo cách này :
Ta có :
\(331\text{≡}1\) ( mod 3 )
\(\Rightarrow331^{332}\text{≡}1^{332}\)( mod 3 )
\(\Rightarrow331^{332}\text{≡}1\)( mod 3 )
\(332\text{≡}2\)( mod 3 )
\(\Rightarrow332^2\text{≡}2^2\)( mod 3 )
\(\Rightarrow332^2\text{≡}4\text{≡}1\)( mod 3 )
\(\Rightarrow\left(332^2\right)^{166}\text{≡}1^{166}\)( mod 3 )
\(\Rightarrow332^{332}\text{≡}1\)( mod 3 )
\(\Rightarrow332^{333}\text{≡}1.332\text{≡}332\text{≡}2\) ( mod 3 )
\(333\text{≡}0\) ( mod 3 )
\(\Rightarrow333^{334}\text{≡}0\) ( mod 3 )
\(\Rightarrow A=331^{332}+332^{333}+333^{334}\text{≡}1+2+0\text{≡}3\text{≡}0\)( mod 3 )
Vì vậy A chia 3 dư 0 ; hay A chia hết cho 3.
Lại có :
\(A=331^{332}+332^{333}+333^{334}\)
\(=\left(...1\right)^{332}+332^{4.83}.332+333^{4.83}.333^2\)
\(=\left(...1\right)+\left(...6\right)\left(...1\right)+\left(...1\right).\left(...9\right)\)
\(=\left(...1\right)+\left(..6\right)+\left(...9\right)\)
\(=\left(...6\right)\)
A có tận cùng 6 nên A chia 5 dư 1.
