AMC8 2012
AMC8 2012 · Q12
AMC8 2012 · Q12. It mainly tests Digit properties (sum of digits, divisibility tests), Sequences in number theory (remainders patterns).
What is the units digit of $13^{2012}$?
$13^{2012}$ 的个位数是多少?
(A)
1
1
(B)
3
3
(C)
5
5
(D)
7
7
(E)
9
9
Answer
Correct choice: (A)
正确答案:(A)
Solution
The problem wants us to find the units digit of $13^{2012}$, therefore, we can eliminate the tens digit of $13$, because the tens digit will not affect the final result. So our new expression is $3^{2012}$. Now we need to look for a pattern in the units digit.
$3^1 \implies 3$
$3^2 \implies 9$
$3^3 \implies 7$
$3^4 \implies 1$
$3^5 \implies 3$
We observe that there is a pattern for the units digit which recurs every four powers of three. Using this pattern, we can subtract 1 from 2012 and divide by 4. The remainder is the power of three that we are looking for, plus one. $2011$ divided by $4$ leaves a remainder of $3$, so the answer is the units digit of $3^{3+1}$, or $3^4$. Thus, we find that the units digit of $13^{2012}$ is
$\boxed{{\textbf{(A)}\ 1}}$.
问题要求求 $13^{2012}$ 的个位数,因此可以忽略 13 的十位数,因为十位数不会影响结果。新表达式为 $3^{2012}$。现在寻找个位数的规律。
$3^1 \implies 3$
$3^2 \implies 9$
$3^3 \implies 7$
$3^4 \implies 1$
$3^5 \implies 3$
观察到个位数每四次幂重复。将 2012 减 1 除以 4,余数为所需幂减 1。2011 除以 4 余 3,所以是 $3^{3+1}$ 或 $3^4$ 的个位数。因此 $13^{2012}$ 的个位数是 $\boxed{{\textbf{(A)}\ 1}}$。
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