2020A AMC 10 真题（P版）

1 答案 E

右边为 \(\dfrac{5}{12}-\dfrac13=\dfrac{5}{12}-\dfrac4{12}=\dfrac1{12}\)。所以 \(x=\dfrac34+\dfrac1{12}=\dfrac9{12}+\dfrac1{12}=\dfrac56\)。正确答案是 E。

English

The right side is \(\dfrac{5}{12}-\dfrac13=\dfrac{5}{12}-\dfrac4{12}=\dfrac1{12}\). Thus \(x=\dfrac34+\dfrac1{12}=\dfrac9{12}+\dfrac1{12}=\dfrac56\). Thus, E is the correct answer.

2 答案 C

五个数的总和为 \(5\cdot15=75\)。又 \(3+5+7=15\)，所以 \(a+b=75-15=60\)。因此 \(a\) 和 \(b\) 的平均数为 \(30\)。正确答案是 C。

English

The five numbers have total sum \(5\cdot15=75\). Since \(3+5+7=15\), we have \(a+b=75-15=60\), so the average of \(a\) and \(b\) is \(30\). Thus, C is the correct answer.

3 答案 A

将分母各因式改写为 \(5-c=-(c-5)\)、\(3-a=-(a-3)\) 和 \(4-b=-(b-4)\)。原式化为 \(\dfrac{(a-3)(b-4)(c-5)}{-(a-3)(b-4)(c-5)}=-1\)。正确答案是 A。

English

Rewrite the denominator factors as \(5-c=-(c-5)\), \(3-a=-(a-3)\), and \(4-b=-(b-4)\). The expression becomes \(\dfrac{(a-3)(b-4)(c-5)}{-(a-3)(b-4)(c-5)}=-1\). Thus, A is the correct answer.

4 答案 E

她行驶 \(2\cdot60=120\) 英里，所以收入为 \(120\cdot \$0.50=\$60\)。这次行程用汽油 \(\frac{120}{30}=4\) 加仑，花费 \(4\cdot\$2=\$8\)。净收入为 \(60-8=52\) 美元，用时 \(2\) 小时，所以每小时净报酬为 \(26\) 美元。正确答案是 E。

English

The driver travels \(2\cdot60=120\) miles, so she is paid \(120\cdot \$0.50=\$60\). The trip uses \(\frac{120}{30}=4\) gallons of gasoline, costing \(4\cdot\$2=\$8\). Her net pay is \(60-8=52\) dollars over \(2\) hours, or \(26\) dollars per hour. Thus, E is the correct answer.

5 答案 C

方程等价于 \(x^2-12x+34=2\) 或 \(x^2-12x+34=-2\)。第一式化为 \(x^2-12x+32=0\)，根为 \(4\)、\(8\)；第二式化为 \((x-6)^2=0\)，根为 \(6\)。所有实数解的和为 \(4+8+6=18\)。正确答案是 C。

English

The equation means \(x^2-12x+34=2\) or \(x^2-12x+34=-2\). The first gives \(x^2-12x+32=0\), with roots \(4\) and \(8\). The second gives \((x-6)^2=0\), with root \(6\). The sum of all real solutions is \(4+8+6=18\). Thus, C is the correct answer.

6 答案 B

末位只能是 \(0\)，因为这个数能被 \(5\) 整除，而且所有数字都是偶数。千位可为 \(2,4,6\) 或 \(8\)，共有四种；百位和十位各有 \(5\) 种选择。因此共有 \(4\cdot5\cdot5=100\) 个这样的整数。正确答案是 B。

English

The last digit must be \(0\), because the number is divisible by \(5\) and all digits are even. The thousands digit can be \(2,4,6,\) or \(8\), and each of the hundreds and tens digits has \(5\) choices. Thus there are \(4\cdot5\cdot5=100\) such integers. Thus, B is the correct answer.

7 答案 C

从 \(-10\) 到 \(14\) 的整数总和为 \(25\cdot\dfrac{-10+14}{2}=50\)。若每行公共和为 \(S\)，则五行总和为 \(50\)，所以 \(5S=50\)，从而 \(S=10\)。正确答案是 C。

English

The sum of the integers from \(-10\) to \(14\) is \(25\cdot\dfrac{-10+14}{2}=50\). If every row has common sum \(S\), then the five row sums add to \(50\), so \(5S=50\) and \(S=10\). Thus, C is the correct answer.

8 答案 B

按四项一组：\((1+2+3-4)\) \(+(5+6+7-8)\) \(+\cdots\) \(+(197+198+199-200)\)。第 \(j\) 组的和为 \((4j-3)+(4j-2)\) \(+(4j-1)-4j=8j-6\)。

共有 \(50\) 组，所以和为 \(\sum_{j=1}^{50}(8j-6)=8\cdot\dfrac{50\cdot51}{2}\) \(-6\cdot50=9900\)。正确答案是 B。

English

Group the terms in blocks of four: \((1+2+3-4)\) \(+(5+6+7-8)\) \(+\cdots\) \(+(197+198+199-200)\). The \(j\)th block is \((4j-3)+(4j-2)\) \(+(4j-1)-4j=8j-6\).

There are \(50\) blocks, so the sum is \(\sum_{j=1}^{50}(8j-6)=8\cdot\dfrac{50\cdot51}{2}\) \(-6\cdot50=9900\). Thus, B is the correct answer.

9 答案 B

设成人和儿童各有 \(P\) 人。成人需要 \(\frac{P}{7}\) 个长凳单元，儿童需要 \(\frac{P}{11}\) 个长凳单元，因此 \(N=P\left(\dfrac17+\dfrac1{11}\right)=\dfrac{18P}{77}\)。

使其为整数的最小值是 \(P=77\)，此时 \(N=18\)。正确答案是 B。

English

If the equal number of adults and children is \(P\), then the adults use \(\frac{P}{7}\) bench sections and the children use \(\frac{P}{11}\) bench sections. Thus \(N=P\left(\dfrac17+\dfrac1{11}\right)=\dfrac{18P}{77}\).

The least positive integer occurs when \(P=77\), giving \(N=18\). Thus, B is the correct answer.

10 答案 B

这些立方体边长为 \(1,2,3,4,5,6,7\)，从大到小堆叠。单独表面积之和为 \(6(1^2+2^2+\cdots+7^2)\) \(=6\cdot140=840\)。

每个接触处隐藏两个正方形面，面积分别为 \(1^2,2^2,\ldots,6^2\)。所以总表面积为 \(840\) \(-2(1^2+2^2+\cdots+6^2)\) \(=840-182=658\)。正确答案是 B。

English

The cube side lengths are \(1,2,3,4,5,6,7\), stacked from largest on bottom to smallest on top. The sum of the surface areas of the separate cubes is \(6(1^2+2^2+\cdots+7^2)\) \(=6\cdot140=840\).

Each contact hides two square faces, with areas \(1^2,2^2,\ldots,6^2\). Subtracting these hidden faces gives \(840\) \(-2(1^2+2^2+\cdots+6^2)\) \(=840-182=658\). Thus, B is the correct answer.

11 答案 C

对中位数附近的一个数，排好序的列表中不超过它的项包括所有不超过它的普通整数和所有不超过它的平方数。因为 \(44^2=1936\) 且 \(45^2=2025\)，所以从 \(1936\) 到 \(2020\) 之间的任一数，都恰有 \(44\) 个平方数不超过它。

在 \(1975\) 处，不超过它的项有 \(1975+44=2019\) 个，其中普通整数截至 \(1975\)。在 \(1976\) 处，不超过该数的项有 \(2020\) 个，其中普通整数截至 \(1976\)，所以第 \(2020\) 项为 \(1976\)，下一项为 \(1977\)。中位数为 \(1976.5\)。正确答案是 C。

English

For a number near the median, the sorted list includes all ordinary integers up to that number and all squares up to that number. Since \(44^2=1936\) and \(45^2=2025\), there are \(44\) squares not exceeding any number from \(1936\) through \(2020\).

At \(1975\), there are \(1975+44=2019\) list entries at most \(1975\). At \(1976\), there are \(2020\) entries at most \(1976\), so the \(2020\)th entry is \(1976\), and the next is \(1977\). The median is \(1976.5\). Thus, C is the correct answer.

12 答案 C

设重心为原点。重心把中线按 \(2:1\) 分割，所以可令中线 \(MV\) 水平，取 \(M=(8,0)\)、\(V=(-4,0)\)；中线 \(CU\) 竖直，取 \(C=(0,8)\)、\(U=(0,-4)\)。

因为 \(U\) 是 \(AM\) 的中点，所以 \(A=2U-M=(-8,-8)\)。\(\triangle AMC\) 的面积为 \(\dfrac12 |(16,8)\times(8,16)|\) \(=\dfrac12(256-64)=96\)。正确答案是 C。

English

Let the centroid be the origin. Since a centroid divides each median in a \(2:1\) ratio, we may place median \(MV\) horizontally with \(M=(8,0)\) and \(V=(-4,0)\), and median \(CU\) vertically with \(C=(0,8)\) and \(U=(0,-4)\).

Because \(U\) is the midpoint of \(AM\), we get \(A=2U-M=(-8,-8)\). The area of \(\triangle AMC\) is \(\dfrac12 |(16,8)\times(8,16)|\) \(=\dfrac12(256-64)=96\). Thus, C is the correct answer.

13 答案 B

设 \(p(x,y)\) 为从 \((x,y)\) 出发最终先碰到竖直边的概率。由对称性，令 \(a=p(1,1)=p(1,3)\)、\(b=p(2,1)=p(2,3)\)、\(c=p(1,2)\)，并令 \(d=p(2,2)\)。

每一步取四个相邻点概率的平均，得到 \(a=\dfrac{1+b+c}{4}\)、\(b=\dfrac{2a+d}{4}\)、\(c=\dfrac{1+2a+d}{4}\)，以及 \(d=\dfrac{b+c}{2}\)。解得 \(c=\dfrac58\)。起点为 \((1,2)\)，所以这就是所求概率。正确答案是 B。

English

Let \(p(x,y)\) be the probability of eventually hitting a vertical side first from point \((x,y)\). By symmetry, set \(a=p(1,1)=p(1,3)\), \(b=p(2,1)=p(2,3)\), \(c=p(1,2)\), and \(d=p(2,2)\).

The averaging equations are \(a=\dfrac{1+b+c}{4}\), \(b=\dfrac{2a+d}{4}\), \(c=\dfrac{1+2a+d}{4}\), and \(d=\dfrac{b+c}{2}\). Solving gives \(c=\dfrac58\), which is the desired probability from \((1,2)\). Thus, B is the correct answer.

14 答案 D

设 \(S_k=x^k+y^k\)。因为 \(x+y=4\) 且 \(xy=-2\)，数 \(x\) 和 \(y\) 是方程 \(t^2-4t-2=0\) 的两根，所以 \(S_k=4S_{k-1}+2S_{k-2}\)。

由 \(S_0=2\) 与 \(S_1=4\) 出发，得到 \(S_2=20\)、\(S_3=88\)、\(S_4=392\)，以及 \(S_5=1744\)。原式为 \(x+y+\dfrac{x^5+y^5}{x^2y^2}\) \(=4+\dfrac{1744}{4}=440\)。正确答案是 D。

English

Let \(S_k=x^k+y^k\). Since \(x+y=4\) and \(xy=-2\), the numbers \(x\) and \(y\) satisfy \(t^2-4t-2=0\), so \(S_k=4S_{k-1}+2S_{k-2}\).

Using \(S_0=2\) and \(S_1=4\), we get \(S_2=20\), \(S_3=88\), \(S_4=392\), and \(S_5=1744\). The expression is \(x+y+\dfrac{x^5+y^5}{x^2y^2}\) \(=4+\dfrac{1744}{4}=440\). Thus, D is the correct answer.

15 答案 E

\(12!\) 的质因数分解为 \(2^{10}3^5 5^2 7^1 11^1\)。因此 \(12!\) 的正因数总数为 \((10+1)(5+1)\) \(\cdot(2+1)(1+1)(1+1)=792\)。

平方因数的所有质数指数必须为偶数，因此共有 \(6\cdot3\cdot2\cdot1\cdot1=36\) 个。所求概率为 \(\frac{36}{792}=\frac{1}{22}\)，所以 \(m+n=1+22=23\)。正确答案是 E。

English

The prime factorization of \(12!\) is \(2^{10}3^5 5^2 7^1 11^1\). Therefore \(12!\) has \((10+1)(5+1)\) \(\cdot(2+1)(1+1)(1+1)=792\) positive divisors.

A square divisor must use only even exponents, giving \(6\cdot3\cdot2\cdot1\cdot1=36\) square divisors. The probability is \(\frac{36}{792}=\frac{1}{22}\), so \(m+n=1+22=23\). Thus, E is the correct answer.

16 答案 B

对于 \(d<\dfrac12\)，每个单位正方形中，距离某个格点不超过 \(d\) 的部分由四个四分之一圆组成，总面积为 \(\pi d^2\)。大正方形由单位正方形铺成，所以所求概率为 \(\pi d^2\)。

令 \(\pi d^2=\dfrac12\)，得 \(d=\sqrt{\dfrac{1}{2\pi}}\approx0.399\)，四舍五入到十分位为 \(0.4\)。正确答案是 B。

English

For \(d<\dfrac12\), the points within \(d\) of lattice points occupy, in each unit square, four quarter-circles whose total area is \(\pi d^2\). The enormous square is tiled by unit squares, so the desired probability is \(\pi d^2\).

Setting \(\pi d^2=\dfrac12\) gives \(d=\sqrt{\dfrac{1}{2\pi}}\approx0.399\), which rounds to \(0.4\). Thus, B is the correct answer.

17 答案 E

多项式在每个平方数 \(1^2,2^2,\ldots,100^2\) 处变号，且首项系数为正。因此满足 \(P(n)\le0\) 的整数落在区间 \([1^2,2^2]\)、\([3^2,4^2]\)、\(\ldots\)、\([99^2,100^2]\) 中。

对奇数 \(k\)，区间 \([k^2,(k+1)^2]\) 中的整数个数为 \((k+1)^2-k^2+1=2k+2\)。对 \(k=1,3,\ldots,99\) 求和，得到 \(2(1+3+\cdots+99)\) \(+2\cdot50=5000+100\) \(=5100\)。正确答案是 E。

English

The polynomial changes sign at each square \(1^2,2^2,\ldots,100^2\), and its leading coefficient is positive. Thus \(P(n)\le0\) for integers in the intervals \([1^2,2^2]\), \([3^2,4^2]\), \(\ldots\), \([99^2,100^2]\).

For odd \(k\), the interval \([k^2,(k+1)^2]\) contains \((k+1)^2-k^2+1=2k+2\) integers. Summing over odd \(k=1,3,\ldots,99\) gives \(2(1+3+\cdots+99)\) \(+2\cdot50=5000+100\) \(=5100\). Thus, E is the correct answer.

18 答案 C

只需考虑模 \(2\) 的奇偶性。条件是 \(ad-bc\) 为 \(1\)，也就是说矩阵 \(\begin{pmatrix}a&b\\ c&d\end{pmatrix}\) 在 \(\mathbb F_2\) 上可逆。

可逆矩阵共有 \((4-1)(4-2)=6\) 个 \(2\times2\) 矩阵，它们都定义在 \(\mathbb F_2\) 上。每一种奇偶模式可提升为 \(2^4=16\) 个取自 \(\{0,1,2,3\}\) 的四元组。因此共有 \(6\cdot16=96\) 个四元组。正确答案是 C。

English

Only parity matters. Modulo \(2\), the condition is that \(ad-bc\) is \(1\), meaning the matrix \(\begin{pmatrix}a&b\\ c&d\end{pmatrix}\) is invertible over \(\mathbb F_2\).

There are \((4-1)(4-2)=6\) invertible \(2\times2\) matrices over \(\mathbb F_2\). Each parity pattern lifts to \(2^4=16\) choices from \(\{0,1,2,3\}\), so there are \(6\cdot16=96\) quadruples. Thus, C is the correct answer.

19 答案 E

离开顶面后，先选择 \(5\) 个顶部环面之一。由于不允许从底部环回到顶部环，每条有效路径都由顶部环阶段、一次向下移动、底部环阶段组成。

固定第一个顶部环面。在顶部 \(5\) 环上，可以不重复地沿环走若干步后停止；选择不动有一种，选择一个方向后可走一到四步，所以共有 \(1+2\cdot4=9\) 种顶部环路径。停止处有 \(2\) 种向下移动选择，因此顶部部分有 \(18\) 种。

到达底部环后，同理有 \(1+2\cdot4=9\) 种方式在底部 \(5\) 环上不重复移动并进入底面。总数为 \(5\cdot18\cdot9=810\)。正确答案是 E。

English

After leaving the top face, choose one of the \(5\) top-ring faces. Because moves from the bottom ring to the top ring are forbidden, every valid path has a top-ring phase, then one move down to the bottom ring, then a bottom-ring phase.

Fix the first top-ring face. On the top ring, the path can move around the \(5\)-cycle without revisiting a face and then stop at any point: there are \(1+2\cdot4=9\) possible top-ring paths. From the stopping face, there are \(2\) possible downward moves to the bottom ring, so the top part has \(18\) choices.

Once in the bottom ring, the path can move around the bottom \(5\)-cycle without revisiting a face and then enter the bottom face; this gives \(1+2\cdot4=9\) choices. The total is \(5\cdot18\cdot9=810\). Thus, E is the correct answer.

20 答案 D

取 \(A=(0,0)\)，\(C=(20,0)\)。因为 \(\angle ACD=90^\circ\) 且 \(CD=30\)，可取 \(D=(20,30)\)。点 \(E\) 的坐标是 \((5,0)\)，所以直线 \(BD\) 的方程为 \(y=2(x-5)\)。

因为 \(\angle ABC=90^\circ\)，点 \(B\) 在以 \(AC\) 为直径的圆上：\((x-10)^2+y^2=100\)。与 \(y=2(x-5)\) 联立，得 \(x=2\) 或 \(10\)。凸四边形对应 \(B=(2,-6)\)。

于是 \([ACD]=\dfrac12\cdot20\cdot30=300\)，且 \([ABC]=\dfrac12\cdot20\cdot6=60\)。总面积为 \(360\)。正确答案是 D。

English

Place \(A=(0,0)\) and \(C=(20,0)\). Since \(\angle ACD=90^\circ\) and \(CD=30\), take \(D=(20,30)\). The point \(E\) is \((5,0)\), so line \(BD\) has equation \(y=2(x-5)\).

Because \(\angle ABC=90^\circ\), point \(B\) lies on the circle with diameter \(AC\): \((x-10)^2+y^2=100\). Intersecting with \(y=2(x-5)\) gives \(x=2\) or \(10\). The convex quadrilateral uses \(B=(2,-6)\).

Then \([ACD]=\dfrac12\cdot20\cdot30=300\), and \([ABC]=\dfrac12\cdot20\cdot6=60\). The total area is \(360\). Thus, D is the correct answer.

21 答案 C

令 \(X=2^{17}\)。则 \[ \begin{gathered} \dfrac{2^{289}+1}{2^{17}+1} = \dfrac{X^{17}+1}{X+1} \\ = X^{16}-X^{15}+X^{14} \\ {}-\cdots-X+1 \end{gathered} \]

把相邻项配成 \(X^{16}-X^{15}\)、\(X^{14}-X^{13}\)、\(\ldots\)、\(X^2-X\)，最后剩下 \(+1\)。每一对都形如 \(2^{17m}(2^{17}-1)\)，在二进制表示中贡献 \(17\) 个一。共有 \(8\) 对，再加最后的 \(1\)，所以 \(k=8\cdot17+1=137\)。正确答案是 C。

English

Let \(X=2^{17}\). Then \[ \begin{gathered} \dfrac{2^{289}+1}{2^{17}+1} = \dfrac{X^{17}+1}{X+1} \\ = X^{16}-X^{15}+X^{14} \\ {}-\cdots-X+1 \end{gathered} \].

Pair consecutive terms: \(X^{16}-X^{15}\), \(X^{14}-X^{13}\), \(\ldots\), \(X^2-X\), and then the final \(+1\). Each pair is \(2^{17m}(2^{17}-1)\), contributing \(17\) ones in binary. There are \(8\) such pairs plus the final \(1\), so \(k=8\cdot17+1=137\). Thus, C is the correct answer.

22 答案 A

写作 \(1000=qn+r\)，其中 \(0\le r<n\)。则 \(\left\lfloor\dfrac{1000}{n}\right\rfloor=q\)。另外两个取整值通常也是 \(q\)，只有当减去 \(1\) 或 \(2\) 后，从 \(1000\) 跨过了 \(n\) 的倍数时，也就是 \(r\) 较小时，才会减少。

对 \(n>1\)，三个取整值之和不能被 \(3\) 整除当且仅当 \(r=0\) 或 \(r=1\)。\(r=0\) 对应 \(1000\) 的因数，除去 \(1\) 后有 \(15\) 个。\(r=1\) 对应 \(999\) 的因数且不为 \(1\)，这样的因数有 \((3+1)(1+1)-1=7\) 个。总数为 \(22\)。正确答案是 A。

English

Write \(1000=qn+r\), where \(0\le r<n\). Then \(\left\lfloor\dfrac{1000}{n}\right\rfloor=q\). The other two floors are usually also \(q\), except that subtracting \(1\) or \(2\) from \(1000\) crosses a multiple of \(n\) when \(r\) is small.

For \(n>1\), the sum is not divisible by \(3\) exactly when \(r=0\) or \(r=1\). The case \(r=0\) gives divisors of \(1000\), excluding \(1\), for \(15\) values. The case \(r=1\) gives divisors of \(999\), excluding \(1\), for \((3+1)(1+1)-1=7\) values. The total is \(22\). Thus, A is the correct answer.

23 答案 A

令 \(R\) 表示 \(90^\circ\) 旋转，允许的旋转为 \(R,R^2,R^3\)。令 \(X\) 和 \(Y\) 分别表示关于两个坐标轴的反射。前两个变换确定后，第三个变换必须是它们乘积的逆，才能使总效果为恒等变换。

在两个旋转的组合中，有 \(6\) 个有序对的乘积是非恒等旋转。一个旋转与一个反射的乘积是允许的坐标轴反射，当且仅当该旋转为 \(R^2\)，这给出 \(4\) 个有序对。最后，两个不同的坐标轴反射可以按任一顺序出现，又给出 \(2\) 个有序对。总共有 \(6+4+2=12\) 个有效序列。正确答案是 A。

English

Let \(R\) be a \(90^\circ\) rotation, so the allowed rotations are \(R,R^2,R^3\). Let \(X\) and \(Y\) be the reflections across the coordinate axes. Once the first two transformations are chosen, the third is forced to be the inverse of their product.

Among two rotations, \(6\) ordered pairs have a nonidentity rotation as their product. A rotation and a reflection have an allowed axis-reflection as their product exactly when the rotation is \(R^2\), giving \(4\) ordered pairs. Finally, the two different axis-reflections can occur in either order, giving \(2\) more pairs. Altogether there are \(6+4+2=12\) valid sequences. Thus, A is the correct answer.

24 答案 C

第一条给出 \(n+120\equiv0\pmod{21}\)，即 \(n\equiv6\pmod{21}\)，但 \(n+120\) 不能被 \(63\) 整除。第二条给出 \(n+63\equiv0\pmod{60}\)，即 \(n\equiv57\pmod{60}\)，但 \(n+63\) 不能被 \(120\) 整除。

联立 \(n\equiv6\pmod{21}\) 和 \(n\equiv57\pmod{60}\)，得 \(n\equiv237\pmod{420}\)。大于 \(1000\) 的候选为 \(1077,1497,1917,\ldots\)。第一个不满足第一条最大公因数条件，第二个不满足第二条，\(1917\) 满足。数字和为 \(18\)。正确答案是 C。

English

The first gcd condition gives \(n+120\equiv0\pmod{21}\), so \(n\equiv6\pmod{21}\), but \(n+120\) must not be divisible by \(63\). The second gives \(n+63\equiv0\pmod{60}\), so \(n\equiv57\pmod{60}\), but \(n+63\) must not be divisible by \(120\).

Solving \(n\equiv6\pmod{21}\) and \(n\equiv57\pmod{60}\) gives \(n\equiv237\pmod{420}\). The candidates above \(1000\) are \(1077,1497,1917,\ldots\). The first fails the first gcd condition, the second fails the second gcd condition, and \(1917\) works. The digit sum is \(18\). Thus, C is the correct answer.

25 答案 A

对任意初始结果，Jason 比较重新掷 \(0,1,2\) 或 \(3\) 枚骰子的最佳概率。重新掷三枚的概率为 \(\frac{15}{216}=\frac{5}{72}\)。若保留两枚、重新掷一枚，获胜概率可达 \(\frac{1}{6}\)，条件是某对保留骰子点数之和不超过 \(6\)。

若恰好重新掷两枚，则只保留一枚。保留点数 \(1,2,3,4,5,6\) 时，获胜结果数分别为 \(5,4,3,2,1,0\)，总结果数为 \(36\)。这种选择只有在两个最小点数之和至少为 \(7\)，且最小点数为 \(1,2\) 或 \(3\) 时才可能最优。

满足条件的有序结果排序后为 \((1,6,6)\)、\((2,5,5),(2,5,6),(2,6,6)\)，以及 \((3,4,4)\)、\((3,4,5)\)、\((3,4,6)\)、\((3,5,5)\)、\((3,5,6)\)、\((3,6,6)\)。按排列数计数，共有 \(3+12+27=42\) 个结果，总结果数为 \(216\)，所以概率为 \(\frac{42}{216}=\frac{7}{36}\)。正确答案是 A。

English

For any initial roll, Jason compares the best probabilities from rerolling \(0,1,2,\) or \(3\) dice. Rerolling all three dice has probability \(\frac{15}{216}=\frac{5}{72}\). Rerolling one die has probability \(\frac{1}{6}\) whenever some pair of kept dice has sum at most \(6\).

If he rerolls exactly two dice, he keeps one die. Keeping a die showing \(1,2,3,4,5,6\) gives probabilities \(5,4,3,2,1,0\) out of \(36\), respectively. This can be optimal only when the two smallest dice sum at least \(7\) and the smallest die is \(1,2,\) or \(3\).

The sorted rolls satisfying this are \((1,6,6)\), \((2,5,5),(2,5,6),(2,6,6)\), and \((3,4,4),\) \((3,4,5),\) \((3,4,6),\) \((3,5,5),\) \((3,5,6),\) \((3,6,6)\). Counting permutations gives \(3+12+27=42\) rolls out of \(216\), so the probability is \(\frac{42}{216}=\frac{7}{36}\). Thus, A is the correct answer.