2018A AMC 10 真题（P版）

1 答案 B

由内向外逐步化简：\[ \begin{aligned} &\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 \\ &=\left(\left(\dfrac{1}{3}+1\right)^{-1}+1\right)^{-1}+1 \\ &=\left(\dfrac{3}{4}+1\right)^{-1}+1 \\ &=\dfrac{4}{7} + 1 \\ &=\dfrac{11}{7} \end{aligned} \]

因此正确答案是 B。

English

We can simplify this as follows. \[ \begin{aligned} &\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 \\ &=\left(\left(\dfrac{1}{3}+1\right)^{-1}+1\right)^{-1}+1 \\ &=\left(\dfrac{3}{4}+1\right)^{-1}+1 \\ &=\dfrac{4}{7} + 1 \\ &=\dfrac{11}{7} \end{aligned} \]

Thus, B is the correct answer.

2 答案 A

设 Jacqueline 有 \(x\) 加仑汽水，则 Alice 有 \(1.25x\) 加仑，Liliane 有 \(1.5x\) 加仑。

两人的汽水量之比为 \(\dfrac{1.5x}{1.25x}=1.2\)，所以 Liliane 的汽水比 Alice 多 \(20\%\)。

因此正确答案是 A。

English

Let \(x\) be the number of gallons of soda that Jacqueline has. Then Alice has \(1.25x\) gallons, and Liliane has \(1.5x\) gallons.

Therefore, the relationship can be found by dividing the amount of soda that each has to yield \(\dfrac{1.5x}{1.25x}=1.2,\) which means Liliane has \(20\%\) more soda than Alice.

Thus, A is the correct answer.

3 答案 E

把 \(10!\) 依次除以 \(60\)、\(60\) 和 \(24\)，即可求出血液经过多少天后过期。

第一次约去 \(6\) 和 \(10\)，第二次约去 \(3, 4\) 和 \(5\)，最后一次约去 \(8\)，并把 \(9\) 化为 \(3\)。

剩下 \(2, 7\) 和 \(3\)，乘积为 \(42\)。一月有 \(31\) 天，所以到二月 \(1\) 日时还剩 \(42 - 31 = 11\) 天。

从二月 \(1\) 日再过 \(11\) 天，血液将在二月 \(12\) 日过期。

因此正确答案是 E。

English

We can divide \(10!\) by \(60,\) \(60,\) and \(24\) to get the number of days that it takes for a unit of blood to expire.

The first division cancels a \(6\) and \(10.\) The second division cancels \(3, 4,\) and \(5.\) The final division cancels \(8\) and turns the \(9\) into a \(3.\)

This leaves \(2, 7,\) and a \(3,\) which multiply to \(42.\) There are \(31\) days in January, so by February \(1,\) the blood only has \(42 - 31 = 11\) days left.

\(11\) days from February \(1\) would make the blood expire on February \(12.\)

Thus, E is the correct answer.

4 答案 E

这 \(3\) 门数学课可以占据以下课时：\[(1, 3, 5)\text{，}\]\[(1, 3, 6)\text{，}\]\[ (1, 4, 6)\text{，}\]\[ (2, 4, 6)\text{。}\]

因此，选择数学课所在课时共有 \(4\) 种方法。

每种课时安排中，三门课有 \(3!\) 种顺序，所以总共有 \(6 \cdot 4 = 24\) 种课程表。

因此正确答案是 E。

English

The \(3\) classes can occupy the following periods: \[(1, 3, 5),\]\[(1, 3, 6),\]\[ (1, 4, 6),\]\[ (2, 4, 6).\]

This means that there are \(4\) ways to choose which periods the mathematics courses occur.

For each configuration, there are \(3!\) ways to determine the order of the courses, for a total of \(6 \cdot 4 = 24\) schedules.

Thus, E is the correct answer.

5 答案 D

Alice 的话不正确，说明 \(d \lt 6\)。Bob 的话不正确，说明 \(d \gt 5\)。Charlie 的话不正确，说明 \(d \gt 4\)。

合并这些条件可得 \(5 \lt d\) 且 \(d \lt 6\)，因此 \(d\) 位于区间 \((5, 6)\)。

因此正确答案是 D。

English

Alice’s statement tells us that \(d \lt 6.\) Bob’s statement tells us that \(d \gt 5.\) Charlie’s statement tells us that \(d \gt 4.\)

Combining all of these tells us that \(5 \lt d\) and \(d \lt 6,\) which means \(d\) is in the interval \((5, 6).\)

Thus, D is the correct answer.

6 答案 B

喜欢票占 \(65\%\)，所以不喜欢票占 \(35\%\)。因此视频分数等于总票数的 \(65\% - 35\% = 30\%\)。

已知分数为 \(90\)，所以总票数为 \(90 \div 30\% = 300\)。

因此正确答案是 B。

English

If \(65\%\) of votes were like votes, then \(35\%\) of votes are dislike votes. Then Sangho’s score is \(65\% - 35\% = 30\%\) the total number of votes.

We know that Sangho’s score is \(90,\) so the total number of votes is \(90 \div 30\% = 300.\)

Thus, B is the correct answer.

7 答案 E

原式可改写为 \[ (2^5 \cdot 5^3) \cdot \left(\dfrac{2}{5}\right)^n = 2^{5 + n} \cdot 5^{3 - n}\text{。}\]

要使它成为整数，两个指数都必须非负，即 \[ \begin{aligned} 5 + n \geq 0 &\Rightarrow n \geq -5 \\ 3 - n \geq 0 &\Rightarrow n \leq 3 \end{aligned}\text{。}\]

所以 \(n\) 共有 \(5 + 3 + 1 = 9\) 个取值。

因此正确答案是 E。

English

We can rewrite the expression as \[ (2^5 \cdot 5^3) \cdot \left(\dfrac{2}{5}\right)^n = 2^{5 + n} \cdot 5^{3 - n}. \]

For this to be an integer, both exponents must be nonnegative. This means that \[ \begin{aligned} 5 + n \geq 0 &\Rightarrow n \geq -5 \\ 3 - n \geq 0 &\Rightarrow n \leq 3. \end{aligned} \]

This gives us \(5 + 3 + 1 = 9\) values for \(n.\)

Thus, E is the correct answer.

8 答案 C

设 Joe 有 \(x\) 枚 \(5\) 分硬币，则他有 \(x + 3\) 枚 \(10\) 分硬币。

因此，他有 \[ 23 - x - (x + 3) = 20 - 2x \] 枚 \(25\) 分硬币。

这些硬币的总价值为 \[ 5x + 10(x + 3) + 25(20 - 2x) \]\[= 530 - 35x\text{。} \]

由总价值可得 \[ 530 - 35x = 320 \Rightarrow x = 6\text{。} \]

所以 Joe 有 \(20 - 2 \cdot 6 = 8\) 枚 \(25\) 分硬币；他的 \(25\) 分硬币比 \(5\) 分硬币多 \(8 - 6 = 2\) 枚。

因此正确答案是 C。

English

Let \(x\) be the number of \(5\)-cent coins that Joe has. Then the number of \(10\)-cent coins he has is \(x + 3.\)

Therefore, Joe has \[ 23 - x - (x + 3) = 20 - 2x \] \(25\)-cent coins.

The total value of all these coins is \[ 5x + 10(x + 3) + 25(20 - 2x) \]\[= 530 - 35x. \]

We know that \[ 530 - 35x = 320 \Rightarrow x = 6. \]

This means that Joe has \(20 - 2 \cdot 6 = 8\) \(25\)-cent coins. Therefore, he has \(8 - 6 = 2\) more \(25\)-cent coins than \(5\)-cent coins.

Thus, C is the correct answer.

9 答案 E

由相似三角形可知，小三角形的边长是大三角形对应边长的 \(\sqrt{\frac{1}{40}}\) 倍。

因此，\(\triangle ADE\) 的边长是大三角形对应边长的 \(4\sqrt{\frac{1}{40}}\) 倍。

两者的面积比为 \[ \left(4\sqrt{\dfrac{1}{40}}\right)^2 = 16 \cdot \dfrac{1}{40} = \dfrac{2}{5}\text{。}\]

所以 \(\triangle ADE\) 的面积为 \(\frac{2}{5} \cdot 40 = 16\)，梯形的面积为 \(40 - 16 = 24\)。

因此正确答案是 E。

English

We know that the side length of the smaller triangles is \(\sqrt{\frac{1}{40}}\) times the length of the larger triangle from similar triangles.

Then the side length of \(\triangle ADE\) is \(4\sqrt{\frac{1}{40}}\) times the length of the side length of the larger triangle.

This makes the ratio of the areas \[ \left(4\sqrt{\dfrac{1}{40}}\right)^2 = 16 \cdot \dfrac{1}{40} = \dfrac{2}{5}. \]

Therefore, the area of \(\triangle ADE\) is \(\frac{2}{5} \cdot 40 = 16.\) The area of the trapezoid is then \(40 - 16 = 24.\)

Thus, E is the correct answer.

10 答案 A

题中给出的差与所求的和互为共轭式，将两式相乘即可消去根号。

乘积为 \[ 49 - x^2 - 25 + x^2 = 24\text{。}\]

所以两式的乘积是 \(24\)，所求的和为 \(24 \div 3 = 8\)。

因此正确答案是 A。

English

Note that the left hand side of the equation and the desired expression are conjugates. Multiplying them would remove the square roots.

Multiplying them yields \[ 49 - x^2 - 25 + x^2 = 24. \]

This means that the product of the values of the expressions is equal to \(24.\) The desired value is therefore \(24 \div 3 = 8.\)

Thus, A is the correct answer.

11 答案 E

用隔板法求 \(n\)：这等价于把 \(10\) 个球放入 \(7\) 个盒子，并要求每个盒子至少有一个球。

这类问题的公式为 \[ \binom{n - 1}{k - 1}\text{，}\] 其中 \(n\) 是球数，\(k\) 是盒子数。

七个正点数之和为 \(10\) 时，任何一个骰子的点数都不会超过 \(4\)，所以每个面最大为 \(6\) 这一上限不会带来额外限制。因此所求答案为 \[ \binom{9}{6} = \binom{9}{3} = 84\text{。}\]

因此正确答案是 E。

English

We can use stars and bars to find \(n.\) It is the same as finding the number of ways to put \(10\) balls into \(7\) boxes, where each box has at least one ball.

The formula for such a scenario is \[ \binom{n - 1}{k - 1}, \] where \(n\) is the number of balls and \(k\) is the number of boxes.

No die can exceed \(4\) in a sum of \(10\) from seven positive rolls, so the upper bound of \(6\) creates no additional restriction. The desired answer is therefore \[ \binom{9}{6} = \binom{9}{3} = 84. \]

Thus, E is the correct answer.

12 答案 C

第二个方程等价于 \(|x|-|y|=1\) 或 \(|x|-|y|=-1\)，因此只需检查四种线性关系 \(x=y\pm1\) 和 \(x=-y\pm1\)。

分别与 \(x+3y=3\) 联立，得到 \((x,y)=\left(\dfrac32,\dfrac12\right)\)、\((0,1)\)、再次得到 \((0,1)\)，以及 \((-3,2)\)。

其中有三个互不相同的有序对，而且都满足原绝对值方程。因此正确答案是 C。

English

The second equation says \(|x|-|y|=1\) or \(|x|-|y|=-1\), so it is enough to check the four linear possibilities \(x=y\pm1\) and \(x=-y\pm1\).

Combining these with \(x+3y=3\) gives \((x,y)=\left(\dfrac32,\dfrac12\right)\), \((0,1)\), \((0,1)\) again, and \((-3,2)\).

These are three distinct ordered pairs, and each satisfies the original absolute-value equation. Thus, C is the correct answer.

13 答案 D

折痕是 \(\overline{AB}\) 的垂直平分线。设折痕为 \(\overline{DE}\)。

由 \(AA\) 相似判定，\(\triangle ADE\sim\triangle ACB\)。因此 \[\dfrac{BC}{AC}=\dfrac{DE}{AD}\text{。}\] 代入各边长得 \[\dfrac34=\dfrac{DE}{\frac{5}{2}}\text{，}\] 所以 \(DE=\dfrac{15}{8}\)。

因此正确答案是 D。

English

The crease is the perpendicular bisector of \(\overline{AB}.\) Let \(\overline{DE}\) be the crease.

By \(AA\) similarity, \(\triangle ADE\sim\triangle ACB.\) Therefore, \[\dfrac{BC}{AC}=\dfrac{DE}{AD}.\] Plugging in the side lengths gives \[\dfrac34=\dfrac{DE}{\frac{5}{2}},\] so \(DE=\dfrac{15}{8}.\)

Thus, D is the correct answer.

14 答案 A

令 \(a=3^{96}\)，\(b=2^{96}\)。原式为 \(\dfrac{81a+16b}{a+b}=16+\dfrac{65a}{a+b}\)，所以小于 \(16+65=81\)。

要证明其最大整数为 \(80\)，还要证明原式大于 \(80\)。这等价于 \(81a+16b>80a+80b\)，也就是 \(a>64b\)。

因为 \(\left(\dfrac32\right)^2=\dfrac94>2\)，所以 \(\dfrac{a}{b}=\left(\dfrac32\right)^{96}>2^{48}>64\)。于是原式大于 \(80\) 且小于 \(81\)。因此正确答案是 A。

English

Let \(a=3^{96}\) and \(b=2^{96}\). The expression is \(\dfrac{81a+16b}{a+b}=16+\dfrac{65a}{a+b}\), so it is less than \(16+65=81\).

To show the floor is \(80\), we also need the expression to be greater than \(80\). This is equivalent to \(81a+16b>80a+80b\), or \(a>64b\).

Because \(\left(\dfrac32\right)^2=\dfrac94>2,\) we have \(\dfrac{a}{b}=\left(\dfrac32\right)^{96}>2^{48}>64.\) Hence the expression is greater than \(80\) and less than \(81.\) Thus, A is the correct answer.

15 答案 D

设大圆的圆心为 \(X\)，两个小圆的圆心分别为 \(Y,Z\)。则 \(XY=XZ=13-5=8\)，且 \(YZ=10\)。

切点处的半径共线，因此 \(XA\) 与 \(XY\) 共线，\(XB\) 与 \(XZ\) 共线，于是 \(\triangle XAB\sim \triangle XYZ\)。所以 \(\dfrac{AB}{YZ}=\dfrac{XA}{XY}=\dfrac{13}{8}\)。

因此 \(AB=10\cdot\dfrac{13}{8}=\dfrac{65}{4}\)，从而 \(m+n=65+4=69\)。所以正确答案是 D。

English

Let \(X\) be the center of the large circle and let \(Y,Z\) be the centers of the two smaller circles. Then \(XY=XZ=13-5=8\) and \(YZ=10\).

The radii to tangent points make \(XA\) collinear with \(XY\) and \(XB\) collinear with \(XZ\), so \(\triangle XAB\sim \triangle XYZ\). Thus \(\dfrac{AB}{YZ}=\dfrac{XA}{XY}=\dfrac{13}{8}\).

Hence \(AB=10\cdot\dfrac{13}{8}=\dfrac{65}{4}\), so \(m+n=65+4=69\). Thus, D is the correct answer.

16 答案 D

设 \(P\) 是从 \(B\) 向 \(\overline{AC}\) 所作高的垂足。由勾股定理得 \(AC=29\)。用两种方式计算面积，得到 \[\dfrac{29\cdot PB}{2}=\dfrac{20\cdot21}{2}\text{，}\] 所以 \(PB=\dfrac{420}{29}\)，它介于 \(14\) 和 \(15\) 之间。

当端点从 \(A\) 移到 \(P\) 时，它到 \(B\) 的距离从 \(20\) 连续减小到 \(PB\)。因此长度分别为 \(15,16,17,18,19,20\) 的线段各有一条。当端点从 \(P\) 移到 \(C\) 时，这个距离从 \(PB\) 连续增大到 \(21\)，所以长度分别为 \(15,16,17,18,19,20,21\) 的线段各有一条。这样共有 \(6+7=13\) 条不同的线段。

所以正确答案是 D。

English

Let \(P\) be the foot of the altitude from \(B\) to \(\overline{AC}.\) The Pythagorean Theorem gives \(AC=29.\) Computing the area in two ways gives \[\dfrac{29\cdot PB}{2}=\dfrac{20\cdot21}{2},\] so \(PB=\dfrac{420}{29},\) which lies between \(14\) and \(15.\)

As the endpoint moves from \(A\) to \(P,\) its distance from \(B\) decreases continuously from \(20\) to \(PB.\) Thus there is one segment of each integer length \(15,16,17,18,19,20.\) As the endpoint moves from \(P\) to \(C,\) the distance increases continuously from \(PB\) to \(21,\) giving one segment of each integer length \(15,16,17,18,19,20,21.\) These are \(6+7=13\) distinct segments.

Thus, D is the correct answer.

17 答案 C

分情况讨论 \(S\) 中的最小元素：

最小元素不可能是 \(1\)，因为这样集合中不能再放入任何其他数。

最小元素也不可能是 \(2\)，因为这时必须选入除 \(1\) 以外的所有奇数，而其中 \(3\) 和 \(9\) 会违反条件。

若最小元素是 \(3\)，可以选 \(7\) 和 \(11\)，再从 \(4\) 与 \(8\) 中选一个，并从 \(5\) 与 \(10\) 中选一个。

无论怎样选，最多只能得到 \(5\) 个元素，所以最小元素不能是 \(3\)。

从 \(4\) 开始时，可以选入 \(6, 7, 9\) 和 \(11\)，再从 \(5\) 与 \(10\) 中选一个，便得到一个含 \(6\) 个元素的集合。

因此正确答案是 C。

English

We proceed by casing on possible values for \(S:\)

\(1\) cannot be the smallest element since that would mean that no other number can be in the set.

\(2\) cannot be the smallest element since we would have to include every odd number except \(1.\) This would make \(3\) and \(9\) violate the rule.

Let \(3\) be the smallest element. Then we can include \(7\) and \(11.\) We can finally include either \(4\) or \(8\) and \(5\) or \(10.\)

Either way, the maximum number of elements that we can include is \(5,\) so \(3\) cannot be the smallest element.

Starting with \(4,\) we can include \(6, 7, 9\) and \(11.\) Finally, we can add either \(5\) or \(10,\) creating a \(6\)-element set.

Thus, C is the correct answer.

18 答案 D

每一组系数得到的和必为正数、负数或零。

由于对称性，正数与负数的数量相等：把所有 \(1\) 换成 \(-1\)，同时把所有 \(-1\) 换成 \(1\) 即可一一配对。

只有当所有系数都为 \(0\) 时，和才等于 \(0\)。

系数的组合共有 \(3^8 = 6561\) 种。因为 \(3\) 的每个幂都大于它前面各个幂之和，所以不同的系数组合给出不同的值。更具体地说，在两组系数最高的不同位上，差的绝对值至少为 \(3^k\)，而所有更低的位加起来最多只能抵消 \(2(1+3+\cdots+3^{k-1})=3^k-1\)。

因此，不同的非负整数共有 \[ \dfrac{6561 - 1}{2} + 1 = 3281 \] 个。

因此正确答案是 D。

English

Note that every number formed by this sum is either positive, negative, or zero.

The number of positive numbers equals the number of negative numbers due to symmetry (flip the \(1\) s to \(-1\) s and \(-1\) s to \(1\) s).

The only way for the sum to be \(0\) is if all the coefficients are \(0.\)

The total number of numbers is \(3^8 = 6561.\) Because each power of \(3\) is larger than the sum of all previous powers of three, each combination of coefficients yields a different value. More explicitly, at the highest place where two combinations differ, the difference has magnitude at least \(3^k,\) while all lower places together can cancel at most \(2(1+3+\cdots+3^{k-1})=3^k-1.\)

Therefore, there are \[ \dfrac{6561 - 1}{2} + 1 = 3281 \] distinct nonnegative integers.

Thus, D is the correct answer.

19 答案 E

只需考虑 \(m\) 的个位数字。在 \(20\) 个连续的可能指数中，每个模 \(4\) 的余数出现 \(5\) 次，并且有 \(10\) 个偶数指数。个位为 \(1\) 的底数对全部 \(20\) 个指数都成功；个位为 \(3\) 或 \(7\) 的底数在能被 \(4\) 整除的那 \(5\) 个指数上成功；个位为 \(5\) 的底数从不成功；个位为 \(9\) 的底数在 \(10\) 个偶数指数上成功。

因此，在 \(5\cdot20=100\) 个等可能数对中，有 \(20+5+0+5+10=40\) 个符合条件，概率为 \(\dfrac{40}{100}=\dfrac25\)。

所以正确答案是 E。

English

Only the units digit of \(m\) matters. Among the \(20\) consecutive possible exponents, each residue modulo \(4\) occurs \(5\) times, and \(10\) exponents are even. A base ending in \(1\) succeeds for all \(20\) exponents; bases ending in \(3\) or \(7\) succeed for the \(5\) exponents divisible by \(4;\) a base ending in \(5\) never succeeds; and a base ending in \(9\) succeeds for the \(10\) even exponents.

Thus \(20+5+0+5+10=40\) of the \(5\cdot20=100\) equally likely pairs work, and the probability is \(\dfrac{40}{100}=\dfrac25.\)

Thus, E is the correct answer.

20 答案 B

把行和列从 \(-3\) 编号到 \(3\)，中心为 \((0,0)\)。旋转和反射可以改变坐标的符号，也可以交换两个坐标，所以 \((x,y)\) 所在的轨道由 \((|x|,|y|)\) 从小到大排序后得到的数对决定。这样的数对是满足 \(0\le u\le v\le3\) 的 \((u,v)\)，共有 \(4+3+2+1=10\) 个。

每个轨道选定一个小方格的颜色后，该轨道内其余方格的颜色都由对称性确定。

因此，在考虑两种颜色都必须出现的条件之前，共有 \(2^{10}\) 种对称涂法。

全黑和全白两种涂法不合要求，所以有效的对称扫描码共有 \(2^{10}-2=1022\) 种。因此正确答案是 B。

English

Number rows and columns from \(-3\) through \(3,\) with the center at \((0,0).\) Rotations and reflections can change signs and exchange coordinates, so the orbit of \((x,y)\) is determined by the ordered pair obtained by sorting \((|x|,|y|).\) These pairs are \((u,v)\) with \(0\le u\le v\le3,\) of which there are \(4+3+2+1=10.\)

Once one square in each orbit is colored, symmetry forces the colors of all other squares in that orbit.

There are therefore \(2^{10}\) symmetric colorings before the condition about using both colors. The all-black and all-white colorings are not allowed.

The total number of valid symmetric scanning codes is \(2^{10}-2=1022\). Thus, B is the correct answer.

21 答案 E

把 \(y=x^2-a\) 代入 \(x^2+y^2=a^2\)，得到 \(x^2+(x^2-a)^2=a^2\)，所以 \(x^2(x^2-(2a-1))=0\)。

因子 \(x^2=0\) 总会给出一个交点 \((0,-a)\)。另一个因子恰好在 \(2a-1>0\) 时给出两个额外的实交点。

因此，恰有三个交点的条件是 \(a>\dfrac12\)。所以正确答案是 E。

English

Substitute \(y=x^2-a\) into \(x^2+y^2=a^2\). This gives \(x^2+(x^2-a)^2=a^2\), so \(x^2(x^2-(2a-1))=0\).

The factor \(x^2=0\) always gives the single point \((0,-a)\). The other factor gives two additional real points exactly when \(2a-1>0\).

There are exactly three intersection points when \(a>\dfrac12\). Thus, E is the correct answer.

22 答案 D

由 \(\gcd(a,b)=24=2^3\cdot3\) 和 \(\gcd(b,c)=36=2^2\cdot3^2\) 可知，\(a\) 能被 \(2^3\cdot3\) 整除，但不能被 \(3^2\) 整除。

由 \(\gcd(b,c)=36\) 和 \(\gcd(c,d)=54=2\cdot3^3\) 可知，\(d\) 能被 \(2\cdot3^3\) 整除，但不能被 \(2^2\) 整除。因此 \(\gcd(d,a)=2\cdot3\cdot n\)，其中 \(n\) 不含因子 \(2\) 或 \(3\)。

由 \(70<6n<100\) 得 \(12\le n\le16\)。这个范围内不含因子 \(2\) 或 \(3\) 的整数 \(n\) 只有 \(13\)，所以 \(13\) 一定是 \(a\) 的因数。因此正确答案是 D。

English

From \(\gcd(a,b)=24=2^3\cdot3\) and \(\gcd(b,c)=36=2^2\cdot3^2\), the number \(a\) is divisible by \(2^3\cdot3\) but not by \(3^2\).

From \(\gcd(b,c)=36\) and \(\gcd(c,d)=54=2\cdot3^3\), the number \(d\) is divisible by \(2\cdot3^3\) but not by \(2^2\). Therefore \(\gcd(d,a)=2\cdot3\cdot n\), where \(n\) has no factor \(2\) or \(3\).

Since \(70<6n<100,\) the integer \(n\) satisfies \(12\le n\le16.\) The only value in this range with no factor \(2\) or \(3\) is \(13,\) so \(13\) must divide \(a.\) Thus, D is the correct answer.

23 答案 D

设 \(S\) 的边长为 \(x\)，可将田地分成下图所示的几个部分。

田地的面积可用两种方式表示：\[ \dfrac{3 \cdot 4}{2} = x^2 + \dfrac{x(3 - x)}{2}\] \[ + \dfrac{x(4 - x)}{2} + \dfrac{2 \cdot 5}{2}\text{。}\]

化简得 \[ 6 = \dfrac{7x}{2} + 5 \] \[ x = \dfrac{2}{7}\text{。}\]

所求比例为 \[ \dfrac{6 - x^2}{6} = \dfrac{6 - \frac{4}{49}}{6} = \dfrac{145}{147}\text{。}\]

因此正确答案是 D。

English

Let \(x\) be the side length of \(S.\) Then we can split the field up into the following shapes.

We can express the area of the field in two ways: \[ \dfrac{3 \cdot 4}{2} = x^2 + \dfrac{x(3 - x)}{2}\]\[ + \dfrac{x(4 - x)}{2} + \dfrac{2 \cdot 5}{2}. \]

Simplifying yields \[ 6 = \dfrac{7x}{2} + 5 \]\[ x = \dfrac{2}{7}. \]

The desired fraction is \[ \dfrac{6 - x^2}{6} = \dfrac{6 - \frac{4}{49}}{6} = \dfrac{145}{147}. \]

Thus, D is the correct answer.

24 答案 D

设 \(BC = a, BG = x, GC = y\)，并设过 \(A\) 的高为 \(h\)。

由角平分线定理，\(BG:GC=AB:AC=5:1\)，所以 \(BG=\dfrac{5a}{6}\)。因为 \(DE\) 是中位线，\(DE=\dfrac a2\)。在 \(\triangle ADE\) 中再次应用该定理，得 \(DF:FE=AD:AE=5:1\)，所以 \(DF=\dfrac56DE=\dfrac{5a}{12}\)。

梯形的高为 \(\frac{h}{2}\)，且 \(\frac{ah}{2}=120\)。它的平均底长为 \(\dfrac12\left(\dfrac{5a}{12}+\dfrac{5a}{6}\right)=\dfrac{5a}{8}\)。因此面积为 \[\dfrac{5a}{8}\cdot\dfrac h2=\dfrac58\cdot\dfrac{ah}{2}=75\text{。}\]所以正确答案是 D。

English

Let \(BC = a, BG = x, GC = y,\) and \(h\) be the length of the altitude through \(A.\)

By the Angle Bisector Theorem, \(BG:GC=AB:AC=5:1,\) so \(BG=\dfrac{5a}{6}.\) Because \(DE\) is a midsegment, \(DE=\dfrac a2.\) Applying the same theorem in \(\triangle ADE\) gives \(DF:FE=AD:AE=5:1,\) so \(DF=\dfrac56DE=\dfrac{5a}{12}.\)

The trapezoid’s height is \(\frac{h}{2},\) and \(\frac{ah}{2}=120.\) Its average base length is \(\dfrac12\left(\dfrac{5a}{12}+\dfrac{5a}{6}\right)=\dfrac{5a}{8}.\) Therefore, its area is \[\dfrac{5a}{8}\cdot\dfrac h2=\dfrac58\cdot\dfrac{ah}{2}=75.\] Thus, D is the correct answer.

25 答案 D

各位数字相同的数可写成 \[\begin{aligned} A_n&=a\dfrac{10^n-1}{9},\\ B_n&=b\dfrac{10^n-1}{9},\\ C_n&=c\dfrac{10^{2n}-1}{9} \end{aligned}\text{。}\]

把这些表达式代入 \(C_n-B_n=A_n^2\)，再除以 \(\frac{10^n-1}{9}\)，得到 \[c(10^n+1)-b=a^2\dfrac{10^n-1}{9}\text{。}\]

整理可得 \[(9c-a^2)10^n=9b-9c-a^2\text{。}\]若它对两个不同的 \(n\) 值都成立，两式相减可知 \(9c-a^2=0\)。原方程又迫使 \(9b-9c-a^2=0\)。

因此 \(c=\frac{a^2}{9}\)，且 \(b=\frac{2a^2}{9}\)。因为 \(a,b,c\) 都是非零数字，候选三元组为 \((a,b,c)=(3,2,1)\)、\((6,8,4)\) 和 \((9,18,9)\)。最后一个三元组无效，因为 \(18\) 不是一位数字。最大的有效和为 \(6+8+4=18\)。所以正确答案是 D。

English

The repeated-digit numbers can be written as \[\begin{aligned} A_n&=a\dfrac{10^n-1}{9},\\ B_n&=b\dfrac{10^n-1}{9},\\ C_n&=c\dfrac{10^{2n}-1}{9}. \end{aligned}\]

Substituting these expressions into \(C_n-B_n=A_n^2\) and dividing by \(\frac{10^n-1}{9}\) gives \[c(10^n+1)-b=a^2\dfrac{10^n-1}{9}.\]

Rearranging yields \[(9c-a^2)10^n=9b-9c-a^2.\] If this holds for two different values of \(n,\) subtracting the two equations shows that \(9c-a^2=0.\) The displayed equation then also forces \(9b-9c-a^2=0.\)

Hence \(c=\frac{a^2}{9}\) and \(b=\frac{2a^2}{9}.\) Because \(a,b,c\) are nonzero digits, the candidates are \((a,b,c)=(3,2,1),\) \((6,8,4),\) and \((9,18,9).\) The last triple is invalid because \(18\) is not a digit. The greatest valid sum is \(6+8+4=18.\) Thus, D is the correct answer.