AMC8 2011

AMC8 2011 · Q22

AMC8 2011 · Q22. It mainly tests Remainders & modular arithmetic, Powers & residues.

What is the tens digit of $7^{2011}$?

$7^{2011}$的十位数字是多少？

(A) 0 0

(B) 1 1

(D) 4 4

(E) 7 7

Answer

Correct choice: (D)

正确答案：（D）

Solution

Since we want the tens digit, we can find the last two digits of $7^{2011}$. We can do this by using modular arithmetic. \[7^1\equiv 07 \pmod{100}.\] \[7^2\equiv 49 \pmod{100}.\] \[7^3\equiv 43 \pmod{100}.\] \[7^4\equiv 01 \pmod{100}.\] We can write $7^{2011}$ as $(7^4)^{502}\times 7^3$. Using this, we can say: \[7^{2011}\equiv (7^4)^{502}\times 7^3\equiv 7^3\equiv 343\equiv 43\pmod{100}.\] From the above, we can conclude that the last two digits of $7^{2011}$ are 43. Since they have asked us to find the tens digit, our answer is $\boxed{\textbf{(D)}\ 4}$.

由于要找十位数字，我们可以找到$7^{2011}$的最后两位数字，使用模运算。 \[7^1\equiv 07 \pmod{100}.\] \[7^2\equiv 49 \pmod{100}.\] \[7^3\equiv 43 \pmod{100}.\] \[7^4\equiv 01 \pmod{100}.\] 我们可以写成$7^{2011} = (7^4)^{502}\times 7^3$。因此： \[7^{2011}\equiv (7^4)^{502}\times 7^3\equiv 7^3\equiv 343\equiv 43\pmod{100}.\] 最后两位是43，所以十位数字是$\boxed{\textbf{(D)}\ 4}$。

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