1990 AMC 12 真题（P版）

1 答案 E

方程化简为 \(\frac{x}{8}=\frac8x\)。由于 \(x\ne0\)，两边乘以 \(8x\)，得 \(x^2=64\)，所以 \(x=\pm8\)。

所以正确答案是 E。

English

The equation simplifies to \(\frac{x}{8}=\frac8x.\) Since \(x\ne0,\) multiplying by \(8x\) gives \(x^2=64,\) so \(x=\pm8.\)

Thus the correct answer is E.

2 答案 E

取倒数后得到 \(4^{\frac{1}{4}}=(2^2)^{\frac{1}{4}}=2^{\frac{1}{2}}=\sqrt2\)。

所以正确答案是 E。

English

Taking the reciprocal gives \(4^{\frac{1}{4}}=(2^2)^{\frac{1}{4}}=2^{\frac{1}{2}}=\sqrt2.\)

Thus the correct answer is E.

3 答案 C

四个角的和为 \(360^\circ\)，因此 \[ \begin{aligned} &75+(75+d)\\ &\quad +(75+2d)\\ &\quad +(75+3d)=360 \end{aligned}\text{。}\]所以 \(d=10\)，最大角为 \(75+3(10)=105^\circ\)。

所以正确答案是 C。

English

The four angles sum to \(360^\circ,\) so \[ \begin{aligned} &75+(75+d)\\ &\quad +(75+2d)\\ &\quad +(75+3d)=360. \end{aligned} \] Thus \(d=10,\) and the largest angle is \(75+3(10)=105^\circ.\)

Thus the correct answer is C.

4 答案 B

由于 \(AB\parallel ED\)，三角形 \(ABF\) 与 \(DEF\) 相似。因此 \[ \frac{AF}{FD}=\frac{AB}{DE}=\frac{16}{4}=4\text{。}\]又因 \(AF+FD=AD=BC=10\)，所以 \(5FD=10\)，从而 \(FD=2\)。

所以正确答案是 B。

English

Because \(AB\parallel ED,\) triangles \(ABF\) and \(DEF\) are similar. Hence \[ \frac{AF}{FD}=\frac{AB}{DE}=\frac{16}{4}=4. \] Since \(AF+FD=AD=BC=10,\) we get \(5FD=10,\) so \(FD=2.\)

Thus the correct answer is B.

5 答案 B

对五个正数分别取六次方不改变它们的大小顺序，依次得到 \[ \begin{aligned} 30,\qquad &6^3\cdot5=1080,\\ 5^3\cdot6&=750,\\ 5^2\cdot6&=150,\qquad 6^2\cdot5=180 \end{aligned}\text{。}\]其中第二个最大。

所以正确答案是 B。

English

Raising the five positive choices to the sixth power preserves their order and gives, respectively, \[ \begin{aligned} 30,\qquad &6^3\cdot5=1080,\\ 5^3\cdot6&=750,\\ 5^2\cdot6&=150,\qquad 6^2\cdot5=180. \end{aligned} \] The largest is the second.

Thus the correct answer is B.

6 答案 D

符合条件的直线，是以 \(A\) 为圆心、半径为 \(2\) 的圆与以 \(B\) 为圆心、半径为 \(3\) 的圆的公切线。两圆的圆心距等于半径之和，所以两圆外切。它们有两条外公切线，并在切点处另有一条公切线，共有 \(3\) 条。

所以正确答案是 D。

English

A qualifying line is a common tangent to the circle centered at \(A\) with radius \(2\) and the circle centered at \(B\) with radius \(3.\) Their center distance equals the sum of their radii, so they are externally tangent. They have two external common tangents and one common tangent at their point of contact, for a total of \(3.\)

Thus the correct answer is D.

7 答案 A

和为 \(8\) 且满足三角不等式的无序正整数边长只有 \(2,3,3\)。半周长为 \(4\)，因此海伦公式给出 \[ \begin{aligned} K&=\sqrt{4(4-2)(4-3)(4-3)}\\ &=2\sqrt2 \end{aligned}\text{。}\]

所以正确答案是 A。

English

The only unordered positive integral side lengths summing to \(8\) and satisfying the triangle inequality are \(2,3,3.\) Their semiperimeter is \(4,\) so Heron’s formula gives \[ \begin{aligned} K&=\sqrt{4(4-2)(4-3)(4-3)}\\ &=2\sqrt2. \end{aligned} \]

Thus the correct answer is A.

8 答案 E

对每个 \(x\in[2,3]\)，\[ \begin{aligned} &|x-2|+|x-3|\\ &\quad=(x-2)+(3-x)\\ &=1 \end{aligned}\text{。}\]因此整个区间 \([2,3]\) 中的数都是解，所以解的个数多于 \(3\)。

所以正确答案是 E。

English

For every \(x\in[2,3],\) \[ \begin{aligned} &|x-2|+|x-3|\\ &\quad=(x-2)+(3-x)\\ &=1. \end{aligned} \] Thus the entire interval \([2,3]\) consists of solutions, so there are more than \(3.\)

Thus the correct answer is E.

9 答案 B

每条黑棱只能覆盖与它相邻的两个面，所以要覆盖全部 \(6\) 个面，至少需要 \(3\) 条黑棱。选择分别属于“顶面与前面”“底面与左面”“后面与右面”这三对面的棱。这三条棱覆盖全部六个面，所以最小值为 \(3\)。

所以正确答案是 B。

English

Each black edge can cover only its two incident faces, so covering all \(6\) faces requires at least \(3\) black edges. Choose edges incident to the face-pairs top/front, bottom/left, and back/right. These three edges cover all six faces, so the minimum is \(3.\)

Thus the correct answer is B.

10 答案 D

从适当的观察点可以看到三个两两相邻的面。它们共包含 \[ \begin{aligned} &3(11^2)-3(11)+1\\ &=363-33+1\\ &=331 \end{aligned} \] 个不同的单位立方体：减去三条重复计算的公共棱，再补回顶角处的单位立方体。

所以正确答案是 D。

English

From a suitable point one can see three mutually adjacent faces. They contain \[ \begin{aligned} &3(11^2)-3(11)+1\\ &=363-33+1\\ &=331 \end{aligned} \] distinct unit cubes: subtract the three shared edges and restore the corner cube.

Thus the correct answer is D.

11 答案 C

一个正整数的因数个数为奇数，当且仅当它是完全平方数。小于 \(50\) 的完全平方数为 \[ 1,4,9,16,25,36,49\text{，}\]所以共有 \(7\) 个。

所以正确答案是 C。

English

A positive integer has an odd number of divisors exactly when it is a perfect square. The squares below \(50\) are \[ 1,4,9,16,25,36,49, \] so there are \(7.\)

Thus the correct answer is C.

12 答案 D

由于 \(a\gt0\)，方程 \(f(y)=ay^2-\sqrt2=-\sqrt2\) 迫使 \(y=0\)。因此 \[ f(\sqrt2)=2a-\sqrt2=0\text{，}\]所以 \(a=\frac{\sqrt2}{2}\)。

所以正确答案是 D。

English

Because \(a\gt0,\) the equation \(f(y)=ay^2-\sqrt2=-\sqrt2\) forces \(y=0.\) Hence \[ f(\sqrt2)=2a-\sqrt2=0, \] so \(a=\frac{\sqrt2}{2}.\)

Thus the correct answer is D.

13 答案 E

执行 \(k\) 轮后，\(X=2k+3\)，且 \[ \begin{aligned} S&=5+7+\cdots+(2k+3)\\ &=k(k+4) \end{aligned}\text{。}\]由于 \(98(102)=9996\lt10000\)，而 \(99(103)=10197\ge10000\)，循环在 \(k=99\) 时停止，此时 \(X=2(99)+3=201\)。

所以正确答案是 E。

English

After \(k\) passes, \(X=2k+3,\) and \[ \begin{aligned} S&=5+7+\cdots+(2k+3)\\ &=k(k+4). \end{aligned} \] Now \(98(102)=9996\lt10000,\) while \(99(103)=10197\ge10000.\) Thus the loop stops at \(k=99,\) when \(X=2(99)+3=201.\)

Thus the correct answer is E.

14 答案 A

两个底角均为 \(\frac{\pi-x}{2}\)。小弧 \(BC\) 对应的圆心角为 \(2x\)，所以两条切线所成的角为 \(\pi-2x\)。由已知关系可得 \[ \frac{\pi-x}{2}=2(\pi-2x)\text{。}\]因此 \(7x=3\pi\)，即 \(x=\frac{3\pi}{7}\)。

所以正确答案是 A。

English

The two base angles are each \(\frac{\pi-x}{2}.\) The minor arc \(BC\) has central angle \(2x,\) so the angle between the tangents is \(\pi-2x.\) The given relation yields \[ \frac{\pi-x}{2}=2(\pi-2x). \] Therefore \(7x=3\pi,\) or \(x=\frac{3\pi}{7}.\)

Thus the correct answer is A.

15 答案 C

设四个数的和为 \(T\)。将四个三数之和相加，得 \[ \begin{aligned} 3T&=180+197+208+222\\ &=807 \end{aligned}\text{，}\]所以 \(T=269\)。被省略的四个数分别为 \(269-180\)、\(269-197\)、\(269-208\) 和 \(269-222\)，其中最大的是 \(89\)。

所以正确答案是 C。

English

If \(T\) is the sum of the four numbers, adding the four triple-sums gives \[ \begin{aligned} 3T&=180+197+208+222\\ &=807, \end{aligned} \] so \(T=269.\) The omitted numbers are \(269-180,\) \(269-197,\) \(269-208,\) and \(269-222,\) whose largest is \(89.\)

Thus the correct answer is C.

16 答案 C

所有可能的两人组合共有 \(\binom{26}{2}=325\) 个。去掉 \(13\) 对夫妻和 \(\binom{13}{2}=78\) 个由两位女士组成的组合。握手次数为 \[ 325-13-78=234\text{。}\]

所以正确答案是 C。

English

There are \(\binom{26}{2}=325\) possible pairs. Exclude the \(13\) married pairs and the \(\binom{13}{2}=78\) pairs of women. The number of handshakes is \[ 325-13-78=234. \]

Thus the correct answer is C.

17 答案 C

递增的三位数不能使用 \(0\)，所以递增数共有 \(\binom93=84\) 个。从 \(0\) 到 \(9\) 中任取三个数字，都能组成一个有效的递减三位数，因为最大的数字位于首位；这样的数有 \(\binom{10}{3}=120\) 个。总数为 \(84+120=204\)。

所以正确答案是 C。

English

An increasing three-digit number cannot use \(0,\) so there are \(\binom93=84\) increasing numbers. Any three digits chosen from \(0\) through \(9\) form a valid decreasing three-digit number, because the largest digit comes first; this gives \(\binom{10}{3}=120.\) The total is \(84+120=204.\)

Thus the correct answer is C.

18 答案 C

\(3^a\) 在 \(a\equiv1,2,3,0\pmod4\) 时的个位数字依次为 \(3,9,7,1\)；而 \(7^b\) 的个位数字依次为 \(7,9,3,1\)。和的个位数字为 \(8\) 时，余数对为 \[ \begin{aligned} (a,b)\equiv{}&(2,2),(3,0),\\ &(0,1)\pmod4 \end{aligned}\text{。}\]每种余数都出现 \(25\) 次（在 \(1,\ldots,100\) 中），所以所求概率为 \(\frac{3}{16}\)。

所以正确答案是 C。

English

The units digits of \(3^a\) for \(a\equiv1,2,3,0\pmod4\) are \(3,9,7,1,\) while those of \(7^b\) are \(7,9,3,1.\) A sum ending in \(8\) occurs for the residue pairs \[ \begin{aligned} (a,b)\equiv{}&(2,2),(3,0),\\ &(0,1)\pmod4. \end{aligned} \] Each residue occurs \(25\) times among \(1,\ldots,100,\) so the probability is \(\frac{3}{16}.\)

Thus the correct answer is C.

19 答案 B

对 \(N+4\) 取模，有 \(N\equiv-4\)，所以 \[ N^2+7\equiv16+7=23\text{。}\]因此，该分数可约分，当且仅当 \(N+4\) 能被 \(23\) 整除，即 \(N\equiv19\pmod{23}\)。这些值为 \(19,42,\ldots,1974\)，一共有 \(86\) 个。

所以正确答案是 B。

English

Modulo \(N+4,\) we have \(N\equiv-4,\) so \[ N^2+7\equiv16+7=23. \] Thus the fraction is reducible exactly when \(N+4\) is divisible by \(23,\) or \(N\equiv19\pmod{23}.\) The values are \(19,42,\ldots,1974,\) a total of \(86.\)

Thus the correct answer is B.

20 答案 C

令 \(A=(0,0)\)、\(E=(3,0)\)、\(C=(10,0)\)、\(D=(3,-5)\)、\(B=(f,h)\)，其中 \(h=BF\)。因为 \(AB\perp AD\)，\[ (f,h)\cdot(3,-5)=0\text{，}\]所以 \(3f=5h\)。因为 \(BC\perp DC\)，\[ (f-10,h)\cdot(-7,-5)=0\text{，}\]所以 \(7f+5h=70\)。代入可得 \(f=7\) 和 \(h=\frac{21}{5}=4.2\)。

所以正确答案是 C。

English

Set \(A=(0,0),\) \(E=(3,0),\) \(C=(10,0),\) \(D=(3,-5),\) and \(B=(f,h),\) where \(h=BF.\) Since \(AB\perp AD,\) \[ (f,h)\cdot(3,-5)=0, \] so \(3f=5h.\) Since \(BC\perp DC,\) \[ (f-10,h)\cdot(-7,-5)=0, \] so \(7f+5h=70.\) Substitution gives \(f=7\) and \(h=\frac{21}{5}=4.2.\)

Thus the correct answer is C.

21 答案 E

令 \(R=PA=PB\)。在等腰三角形 \(APB\) 中，\[ 1=AB=2R\sin\theta\text{，}\]所以 \(R=\frac{1}{2\sin\theta}\)。若 \(h\) 为棱锥的高，则正方形中心到 \(A\) 的水平距离为 \(\frac{1}{\sqrt2}\)，因此 \[ \begin{aligned} h^2 &=R^2-\frac12\\ &=\frac{\cos2\theta}{4\sin^2\theta} \end{aligned}\text{。}\]所以 \(h=\frac{\sqrt{\cos2\theta}}{2\sin\theta}\)，体积为 \(\frac13(1)h=\frac{\sqrt{\cos2\theta}}{6\sin\theta}\)。

所以正确答案是 E。

English

Let \(R=PA=PB.\) In isosceles triangle \(APB,\) \[ 1=AB=2R\sin\theta, \] so \(R=\frac{1}{2\sin\theta}.\) If \(h\) is the pyramid height, the horizontal distance from the square’s center to \(A\) is \(\frac{1}{\sqrt2},\) hence \[ \begin{aligned} h^2 &=R^2-\frac12\\ &=\frac{\cos2\theta}{4\sin^2\theta}. \end{aligned} \] Thus \(h=\frac{\sqrt{\cos2\theta}}{2\sin\theta},\) and the volume is \(\frac13(1)h=\frac{\sqrt{\cos2\theta}}{6\sin\theta}.\)

Thus the correct answer is E.

22 答案 D

这些根的模为 \(2\)，辐角为 \[ \frac\pi6+\frac{k\pi}{3},\qquad k=0,1,\ldots,5\text{。}\]实部为正的两个根的辐角分别为 \(\frac{\pi}{6}\) 和 \(\frac{11\pi}{6}\)。它们互为共轭，模均为 \(2\)，所以乘积为 \(2^2=4\)。

所以正确答案是 D。

English

The roots have modulus \(2\) and arguments \[ \frac\pi6+\frac{k\pi}{3},\qquad k=0,1,\ldots,5. \] The roots with positive real part have arguments \(\frac{\pi}{6}\) and \(\frac{11\pi}{6}.\) They are conjugates of modulus \(2,\) so their product is \(2^2=4.\)

Thus the correct answer is D.

23 答案 B

令 \(t=\log_yx\)。则 \[ t+\frac1t=\frac{10}{3}\text{，}\]所以 \(3t^2-10t+3=0\)，从而 \(t=3\) 或 \(\frac{1}{3}\)。因此，\(x,y\) 中一个是另一个的立方。设较小者为 \(u\)。则 \(u^4=xy=144\)，所以 \(u=2\sqrt3\)，且 \(u^3=24\sqrt3\)。因此 \[ \frac{x+y}{2}=\frac{2\sqrt3+24\sqrt3}{2}=13\sqrt3\text{。}\]

所以正确答案是 B。

English

Let \(t=\log_yx.\) Then \[ t+\frac1t=\frac{10}{3}, \] so \(3t^2-10t+3=0,\) giving \(t=3\) or \(\frac{1}{3}.\) Thus one of \(x,y\) is the cube of the other. Let the smaller be \(u.\) Then \(u^4=xy=144,\) so \(u=2\sqrt3\) and \(u^3=24\sqrt3.\) Therefore \[ \frac{x+y}{2}=\frac{2\sqrt3+24\sqrt3}{2}=13\sqrt3. \]

Thus the correct answer is B.

24 答案 D

设亚当斯高中有 \(x\) 名男生和 \(y\) 名女生。由该校的平均分，\[ 71x+76y=74(x+y)\text{，}\]所以 \(y=\frac{3x}{2}\)。若贝克高中有 \(u\) 名男生和 \(v\) 名女生，则该校的平均分给出 \(u=2v\)。两校男生合在一起的平均分给出 \[ 71x+81u=79(x+u)\text{，}\]所以 \(u=4x\)，且 \(v=2x\)。因此，两校女生合在一起的平均分为 \[ \frac{76(\frac{3x}{2})+90(2x)}{\frac{3x}{2}+2x}=84\text{。}\]

所以正确答案是 D。

English

Let Adams have \(x\) boys and \(y\) girls. From its average, \[ 71x+76y=74(x+y), \] so \(y=\frac{3x}{2}.\) If Baker has \(u\) boys and \(v\) girls, its average gives \(u=2v.\) The combined boys’ average gives \[ 71x+81u=79(x+u), \] so \(u=4x\) and \(v=2x.\) Hence the combined girls’ average is \[ \frac{76(\frac{3x}{2})+90(2x)}{\frac{3x}{2}+2x}=84. \]

Thus the correct answer is D.

25 答案 B

若半径为 \(r\)，则一个角落球的球心为 \((r,r,r)\)，中心球的球心为 \((\frac{1}{2},\frac{1}{2},\frac{1}{2})\)。相切条件给出 \[ \sqrt3\left(\frac12-r\right)=2r\text{。}\]解方程并将分母有理化，\[ r=\frac{\sqrt3}{2(2+\sqrt3)} =\frac{2\sqrt3-3}{2}\text{。}\]

所以正确答案是 B。

English

If the radius is \(r,\) a corner sphere has center \((r,r,r)\) and the central sphere has center \((\frac{1}{2},\frac{1}{2},\frac{1}{2}).\) Tangency gives \[ \sqrt3\left(\frac12-r\right)=2r. \] Solving and rationalizing, \[ r=\frac{\sqrt3}{2(2+\sqrt3)} =\frac{2\sqrt3-3}{2}. \]

Thus the correct answer is B.

26 答案 A

按顺时针方向，将图中显示的平均数 \(a_0,a_1,\ldots,a_9\) 依次编号为 \(1,2,\ldots,10\)，并令 \(x_i\) 为相应位置所选的数。记 \(x_0=t,\ x_1=u\)。由 \(x_{i-1}+x_{i+1}=2a_i\)，依次得到 \[ \begin{aligned} x_2&=4-t, &x_3&=6-u,\\ x_4&=4+t, &x_5&=4+u,\\ x_6&=8-t, &x_7&=10-u,\\ x_8&=8+t, &x_9&=8+u \end{aligned}\text{。}\]最后两个闭合方程给出 \(t=6\) 和 \(u=-3\)。因此 \(x_5=4+u=1\)。

所以正确答案是 A。

English

Index the displayed averages \(a_0,a_1,\ldots,a_9\) clockwise as \(1,2,\ldots,10,\) and let \(x_i\) be the corresponding picked numbers. Write \(x_0=t,\ x_1=u.\) From \(x_{i-1}+x_{i+1}=2a_i,\) successive values are \[ \begin{aligned} x_2&=4-t, &x_3&=6-u,\\ x_4&=4+t, &x_5&=4+u,\\ x_6&=8-t, &x_7&=10-u,\\ x_8&=8+t, &x_9&=8+u. \end{aligned} \] The two closing equations give \(t=6\) and \(u=-3.\) Hence \(x_5=4+u=1.\)

Thus the correct answer is A.

27 答案 C

若三条高为 \(h_1,h_2,h_3\)，则对应的三条边长分别与 \(\frac{1}{h_1},\frac{1}{h_2},\frac{1}{h_3}\) 成正比。对于 \(5,12,13\)，\[ \frac15\gt\frac1{12}+\frac1{13}\text{，}\]所以它们的倒数不满足三角不等式。直接检验可知，其余各组数的倒数都满足所有严格的三角不等式。

所以正确答案是 C。

English

If the altitudes are \(h_1,h_2,h_3,\) then the corresponding sides are proportional to \(\frac{1}{h_1},\frac{1}{h_2},\frac{1}{h_3}.\) For \(5,12,13,\) \[ \frac15\gt\frac1{12}+\frac1{13}, \] so the reciprocals fail the triangle inequality. Direct checking shows that the reciprocals of each other listed triple satisfy all strict triangle inequalities.

Thus the correct answer is C.

28 答案 B

设从连续四个顶点引出的切线段长为 \(u,v,w,z\)。则 \[ \begin{aligned} u+v&=70, &v+w&=90,\\ w+z&=130, &z+u&=110 \end{aligned}\text{。}\]因此 \(v=70-u\)、\(w=20+u\)、\(z=110-u\)。若内切圆半径为 \(r\)，则角 \(A\) 所在顶点的切线段长为 \(r\cot(\frac{A}{2})\)。两个对角互补，所以 \(uw=vz\)。因此 \[ u(20+u)=(70-u)(110-u)\text{，}\]解得 \(u=38.5\)。所以长度为 \(130\) 的边上的两段分别为 \(w=58.5\) 和 \(z=71.5\)，两者之差为 \(13\)。

所以正确答案是 B。

English

Let the tangent lengths from the four consecutive vertices be \(u,v,w,z.\) Then \[ \begin{aligned} u+v&=70, &v+w&=90,\\ w+z&=130, &z+u&=110. \end{aligned} \] Thus \(v=70-u,\) \(w=20+u,\) and \(z=110-u.\) If the inradius is \(r,\) a vertex with angle \(A\) has tangent length \(r\cot(\frac{A}{2}).\) Opposite angles are supplementary, so \(uw=vz.\) Therefore \[ u(20+u)=(70-u)(110-u), \] giving \(u=38.5.\) Hence the two segments of the \(130\)-side are \(w=58.5\) and \(z=71.5,\) whose difference is \(13.\)

Thus the correct answer is B.

29 答案 D

将这些整数分成形如 \(m,3m,9m,\ldots\) 的链，其中 \(m\) 不是 \(3\) 的倍数。在每条链中，两个相邻项不能同时选取，所以最大选择方案是从 \(m\) 开始隔项选取。等价地，选择质因数 \(3\) 的指数为偶数的整数。这样的整数有 \[ \begin{aligned} &\left(100-\left\lfloor\frac{100}{3}\right\rfloor\right)\\ &\quad+\left(\left\lfloor\frac{100}{9}\right\rfloor -\left\lfloor\frac{100}{27}\right\rfloor\right)\\ &\quad+\left\lfloor\frac{100}{81}\right\rfloor\\ &=67+8+1=76 \end{aligned}\text{。}\]上述链的论证也证明了不可能选取更多元素。

所以正确答案是 D。

English

Partition the integers into chains \(m,3m,9m,\ldots,\) with \(m\) not a multiple of \(3.\) In each chain, no two adjacent terms may both be selected, so a maximum selection takes alternating terms starting with \(m.\) Equivalently, select the integers whose exponent of \(3\) is even. There are \[ \begin{aligned} &\left(100-\left\lfloor\frac{100}{3}\right\rfloor\right)\\ &\quad+\left(\left\lfloor\frac{100}{9}\right\rfloor -\left\lfloor\frac{100}{27}\right\rfloor\right)\\ &\quad+\left\lfloor\frac{100}{81}\right\rfloor\\ &=67+8+1=76. \end{aligned} \] The chain argument also proves no larger selection is possible.

Thus the correct answer is D.

30 答案 E

因为 \(a,b\) 是方程 \(t^2-6t+1=0\) 的根，\[ R_n=6R_{n-1}-R_{n-2}\text{。}\]从 \(R_0=1,\ R_1=3\) 开始，个位数字依次为 \[ 1,3,7,9,7,3,1,\ldots\text{，}\]周期为 \(6\)。因为 \(12345\equiv3\pmod6\)，所求个位数字与 \(R_3\) 的个位数字相同，即 \(9\)。

所以正确答案是 E。

English

Because \(a,b\) are roots of \(t^2-6t+1=0,\) \[ R_n=6R_{n-1}-R_{n-2}. \] Starting with \(R_0=1,\ R_1=3,\) the units digits are \[ 1,3,7,9,7,3,1,\ldots, \] with period \(6.\) Since \(12345\equiv3\pmod6,\) the units digit is the same as that of \(R_3,\) namely \(9.\)

Thus the correct answer is E.