Câu hỏi của tan suong Nguyen

Question

Chữ số tận cùng của 3²⁰¹⁵.7²⁰¹⁶.9²⁰¹⁷.19²⁰¹⁵ là mấy ?

giang ho dai ca · Accepted Answer

$3^{2015}=3^{4.503+3}=\left(3^4\right)^{503}.27=\left(...1\right).27=\left(...7\right)$

$7^{2016}=\left(7^4\right)^{504}=\left(...1\right)^{504}=\left(...1\right)$

$9^{2017}=\left(9^2\right)^{1008}.9=\left(...1\right).9=\left(...9\right)$

$19^{2015}=\left(19^2\right)^{1007}.19=\left(...1\right)^{1007}.19=\left(...1\right).19=\left(...9\right)$

=> 3²⁰¹⁵.7²⁰¹⁶.9²⁰¹⁷.19^{2015 = $\left(...7\right).\left(...1\right).\left(...9\right).\left(...9\right)=\left(...7\right)$}

Câu trả lời:

$3^{2015}=3^{4.503+3}=\left(3^4\right)^{503}.27=\left(...1\right).27=\left(...7\right)$

$7^{2016}=\left(7^4\right)^{504}=\left(...1\right)^{504}=\left(...1\right)$

$9^{2017}=\left(9^2\right)^{1008}.9=\left(...1\right).9=\left(...9\right)$

$19^{2015}=\left(19^2\right)^{1007}.19=\left(...1\right)^{1007}.19=\left(...1\right).19=\left(...9\right)$

=> 3²⁰¹⁵.7²⁰¹⁶.9²⁰¹⁷.19^{2015 = $\left(...7\right).\left(...1\right).\left(...9\right).\left(...9\right)=\left(...7\right)$}