2013B AMC 10 真题（P版）

\[ \begin{aligned} &\frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6} \\ \ &= \dfrac {12}9 - \dfrac 9{12} \\ &= \dfrac 43 - \dfrac 34 \\ &= \dfrac 7{12} \end{aligned} \]

所以正确答案是 C。

English

\[ \begin{aligned} &\frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6} \\ \ &= \dfrac {12}9 - \dfrac 9{12} \\ &= \dfrac 43 - \dfrac 34 \\ &= \dfrac 7{12} \end{aligned} \]

Thus, the correct answer is C.

花园尺寸为 \(2\cdot 15 = 30\) 英尺和 \(2\cdot 20=40\) 英尺，面积为 \(30\cdot 40 = 1200\) 平方英尺。

因此，预计收获 \[1200\cdot \dfrac 12 =600\] 磅土豆。

所以正确答案是 A。

English

The dimensions of the garden are \( 2\cdot 15 = 30\) feet by \(2\cdot 20=40\) feet. Thus, the square footage is \(30\cdot 40 = 1200.\)

Therefore, there are \[1200\cdot \dfrac 12 =600\] pounds of potatoes.

Thus, the correct answer is A .

设最低气温为 \(l\)。则最高气温为 \(l+16\)。

平均值满足 \[ \dfrac{l+l+16}2 = l+8 = 3\text{。}\] 因此 \(l = -5\)。

所以正确答案是 C。

English

Let \(l\) represent the low temperature. Then the high temperature is \(l+16.\)

The average satisfies \[ \dfrac{l+l+16}2 = l+8 = 3.\] Therefore, \(l = -5.\)

Thus, the correct answer is C.

倒数时，\(201\) 是第一个数，\(200\) 是第二个数；一般地，数 \(x\) 是第 \(202-x\) 个数。

因此 \(53\) 是第 \(202-53=149\) 个数。

所以正确答案是 D。

English

When counting backward, \(201\) is the first number, \(200\) is the second, and in general \(x\) is the \(202-x\)th number.

Thus \(53\) is the \(202-53=149\)th number counted.

Thus, the correct answer is D .

\(2a-ab=a(2-b)\)。

为了使它最小，取 \(b\) 尽可能大，使 \(2-b\) 最负，同时取 \(a\) 尽可能大。因为 \(a,b<6\)，可取 \(a=b=5\)。

此时 \(5(2-5)=-15\)。

所以正确答案是 B。

English

The expression is \(2a-ab=a(2-b)\).

To make it as small as possible, choose \(b\) as large as possible so that \(2-b\) is most negative, and choose \(a\) as large as possible. Since \(a,b<6\), take \(a=b=5\).

The minimum value is \(5(2-5)=-15\).

Thus, the correct answer is B .

所有五年级学生和家长的年龄总和为 \[33\cdot 11+55\cdot 33 = 66\cdot 33\text{。}\]

平均年龄等于总年龄除以总人数，所以 \[\dfrac{66\cdot 33}{88} = \dfrac 34 \cdot 33 = 24.75\text{。}\]

所以正确答案是 C。

English

The sum of the ages of all the fifth-graders and their parents is: \[33\cdot 11+55\cdot 33 = 66\cdot 33.\]

Then, as the average is the sum divided by the number of people, the average age must be: \[\dfrac{66\cdot 33}{88} = \dfrac 34 \cdot 33 = 24.75 .\]

Thus, the correct answer is C .

六个等间距点形成正六边形。只有角为 \(30^\circ,60^\circ,90^\circ\) 的三角形既不是等边也不是等腰三角形。

斜边是单位圆的直径，长为 \(2\)。这个 \(30\)-\(60\)-\(90\) 三角形的两条直角边为 \(1\) 和 \(\sqrt3\)。

面积为 \(\frac12\cdot1\cdot\sqrt3=\frac{\sqrt3}{2}\)。

所以正确答案是 B。

English

Six equally spaced points on the circle form a regular hexagon. A triangle using three of them is not equilateral or isosceles only when its angles are \(30^\circ,60^\circ,90^\circ\).

The hypotenuse is a diameter of the unit circle, so it has length \(2\). The legs of the \(30\)-\(60\)-\(90\) triangle are \(1\) and \(\sqrt3\).

The area is \(\frac12\cdot1\cdot\sqrt3=\frac{\sqrt3}{2}\).

Thus, the correct answer is B .

令每辆车都行驶 \(40\) 英里，则总路程为 \(80\) 英里。

Ray 用 \(1\) 加仑，Tom 用 \(4\) 加仑，总共用 \(5\) 加仑。

合并油耗率为 \(80\div5=16\) 英里每加仑。

所以正确答案是 B。

English

Let each car drive \(40\) miles. Together they drive \(80\) miles.

Ray uses \(1\) gallon, while Tom uses \(4\) gallons, so together they use \(5\) gallons.

The combined rate is \(80\div5=16\) miles per gallon.

Thus, the correct answer is B .

分解后可知质因数只有 \(2\)、\(3\)、\(5\)。因为三个数两两互质，每个素数幂只能完整属于其中一个数。

又因为三个数都大于 \(1\)，所以每个数都必须含有其中一种质因数。

又因为 \[27000=2^3\cdot 3^3\cdot 5^3\text{，}\] 所以三个数必为 \(2^3,3^3,5^3\)，和为 \(160\)。

所以正确答案是 D。

English

Only one number is a multiple of \(2,\) only one is a multiple of \(3,\) and only one is a multiple of \(5.\)

Since each positive integer is greater than \(1,\) each must be a multiple of one of these primes.

Now \[27000=2^3\cdot 3^3\cdot 5^3. \] Therefore, the numbers must be \(2^3,3^3,5^3,\) and their sum is \(160.\)

Thus, the correct answer is D.

设三分球出手 \(t\) 次。

三分球命中 \(0.4 t\) 次，得分为 \(1.2t\)。

两分球出手 \(1.5t\) 次，命中 \(0.75t\) 次，得分为 \(1.5t\)。

因此总分满足 \[2.7t=54\text{，}\] 所以 \(t=20\)。

所以正确答案是 C。

English

Let \(t\) be the number of three-point attempts.

The number of made three-point shots is \(0.4 t,\) giving \(1.2t\) points.

The number of two-point attempts is \(1.5t,\) so the number made is \(0.75t,\) giving \(1.5t\) points.

Thus the total number of points satisfies \[2.7t=54,\] so \(t=20.\)

Thus, the correct answer is C.

方程可化为 \[x^2-10x+y^2+6y+34=0\text{。}\]

配方得 \[x^2-10x+25+y^2+6y+9 \] \[=(x-5)^2+(y+3)^2\] \[=0\text{。}\] 两个平方项都非负，而它们的和为 \(0\)，所以两者都等于 \(0\)，即 \[(x-5)^2 = (y+3)^2=0\text{。}\]

因此 \(x=5,y=-3\)，所以 \(x+y=2\)。

所以正确答案是 B。

English

This can be rewritten as \[x^2-10x+y^2+6y+34=0.\]

Completing the square yields \[x^2-10x+25+y^2+6y+9 \] \[=(x-5)^2+(y+3)^2\] \[=0.\] Since both squared terms are nonnegative and their sum is \(0,\) both must be \(0\). Thus \[(x-5)^2 = (y+3)^2=0.\]

Therefore, \(x=5,y=-3,\) so \(x+y=2.\)

Thus, the correct answer is B.

正五边形有 \(5\) 条边，长度相同；还有 \(5\) 条对角线，长度也相同。

选出第一条后，剩下 \(9\) 条线段，其中恰有 \(4\) 条与第一条同长。

因此概率为 \(\frac49\)。

所以正确答案是 B。

English

A regular pentagon has \(5\) sides of one length and \(5\) diagonals of another length.

After the first segment is chosen, there are \(9\) segments left, and exactly \(4\) of them have the same length as the first chosen segment.

Therefore the probability is \(\frac49\).

Thus, the correct answer is B .

报完以 \(n\) 结尾的这一轮时，说出的数的总个数为 \[ 1+2+\cdots+n=\frac{n(n+1)}2\text{。}\]

报完以 \(9\) 结尾的这一轮后，恰好说了 \(\frac{9\cdot10}{2}=45\) 个数。下一轮报的是 \(1,2,\ldots,10\)，其中第八项就是总序列中的第 \(53\) 个数，所以这个数是 \(8\)。

所以正确答案是 E。

English

Through the turn ending with \(n\), the total number of numbers spoken is \[ 1+2+\cdots+n=\frac{n(n+1)}2. \]

After the turn ending with \(9\), exactly \(\frac{9\cdot10}{2}=45\) numbers have been spoken. The next turn is the list \(1,2,\ldots,10\), so its eighth entry is the \(53\)rd number spoken. That entry is \(8\).

Thus, the correct answer is E .

条件 \(x\clubsuit y=y\clubsuit x\) 展开为 \(x^2y-xy^2=y^2x-yx^2\)。

合并同类项并因式分解得 \(2x^2y-2xy^2=2xy(x-y)=0\)。

因此 \(x=0\)、\(y=0\) 或 \(x=y\)，这是三条直线。

所以正确答案是 E。

English

The condition is \(x\clubsuit y=y\clubsuit x\), so \(x^2y-xy^2=y^2x-yx^2\).

Combining like terms gives \(2x^2y-2xy^2=2xy(x-y)=0\).

Thus \(x=0\), \(y=0\), or \(x=y\). These are three lines.

Thus, the correct answer is E .

设等边三角形的边长为 \(s\)，面积为 \(A\)。则 \(a=3s\)。

边长为 \(s\) 的正六边形由 \(6\) 个边长为 \(s\) 的等边三角形组成，面积为 \(6A\)。

若正六边形面积也为 \(A\)，它的边长应为 \(\dfrac{s}{\sqrt 6} \)，所以 \[b = 6\cdot \dfrac{s}{\sqrt 6} = s\sqrt 6\text{。}\]

因此 \[\dfrac ab = \dfrac{3s}{s\sqrt 6} = \dfrac{3 \sqrt 6}6 = \dfrac{\sqrt 6} 2\text{。}\]

所以正确答案是 B。

English

Let \(s\) be the side length of the equilateral triangle and \(A\) its area. Then \(a=3s.\)

A regular hexagon of side length \(s\) consists of \(6\) equilateral triangles of side length \(s,\) so its area is \(6A.\)

Therefore, a regular hexagon of area \(A\) has side length \(\dfrac{s}{\sqrt 6} .\) Hence \[b = 6\cdot \dfrac{s}{\sqrt 6} = s\sqrt 6 .\]

It follows that \[\dfrac ab = \dfrac{3s}{s\sqrt 6} = \dfrac{3 \sqrt 6}6 = \dfrac{\sqrt 6} 2.\]

Thus, the correct answer is B.

因为 \(PE:PD:DE=1.5:2:2.5\) \(=3:4:5\)，所以三角形 \(DPE\) 在 \(P\) 处为直角，即两条中线 \(AD\) 和 \(CE\) 互相垂直。

重心把每条中线按 \(2:1\) 分割，所以 \(CE=3\cdot PE=4.5\)，\(AD=3\cdot PD=6\)。

四边形 \(AEDC\) 的对角线 \(AD\) 与 \(CE\) 垂直，面积为 \(\frac12(AD)(CE)=\frac12\cdot6\cdot4.5=13.5\)。

所以正确答案是 B。

English

Since \(PE:PD:DE=1.5:2:2.5\) \(=3:4:5\), triangle \(DPE\) is right at \(P\). Thus medians \(AD\) and \(CE\) are perpendicular.

The centroid divides each median in a \(2:1\) ratio, so \(CE=3\cdot PE=4.5\) and \(AD=3\cdot PD=6\).

Quadrilateral \(AEDC\) has perpendicular diagonals \(AD\) and \(CE\), so its area is \(\frac12(AD)(CE)=\frac12\cdot6\cdot4.5=13.5\).

Thus, the correct answer is B .

设红色摊位交换 \(m\) 次，蓝色摊位交换 \(n\) 次。

最后红色和蓝色代币数分别为 \(75-2m+n\) 和 \(75+m-3n\)。结束时必须红色少于 \(2\) 个、蓝色少于 \(3\) 个。

求解这些终止情况，只得到两组候选终态：\((1,2)\)，对应 \((m,n)=(59,44)\)；或 \((0,0)\)，对应 \((m,n)=(60,45)\)。

终态 \((0,0)\) 不可能，因为最后一次交换一定会产生一个蓝色或一个红色代币。

终态 \((1,2)\) 可以达到。从红、蓝代币数 \((75,75)\) 开始，先在蓝色摊位交换 \(25\) 次，再在红色摊位交换 \(50\) 次，接着依次在蓝色摊位交换 \(16\) 次、红色摊位交换 \(8\) 次、蓝色摊位交换 \(3\) 次，最后在红色摊位交换 \(1\) 次。每个阶段的红、蓝代币数依次为 \[ \begin{aligned} &(100,0),(0,50),(16,2),\\ &(0,10),(3,1),(1,2) \end{aligned}\text{。}\]

因此 Alex 最后有 \(59+44=103\) 个银色代币，正确答案是 E。

English

Suppose Alex makes \(m\) exchanges at the red-token booth and \(n\) exchanges at the blue-token booth.

He then has \(75-2m+n\) red tokens and \(75+m-3n\) blue tokens. At the end he must have fewer than \(2\) red tokens and fewer than \(3\) blue tokens.

Solving these terminal possibilities gives only two candidate final token counts: \((1,2)\), which comes from \((m,n)=(59,44)\), or \((0,0)\), which comes from \((m,n)=(60,45)\).

The final count \((0,0)\) is impossible, because the last exchange would always create either one blue token or one red token.

The final count \((1,2)\) is attainable. Starting from \((75,75)\) red and blue tokens, make \(25\) blue-booth exchanges, then \(50\) red-booth exchanges, then \(16\) blue-booth exchanges, then \(8\) red-booth exchanges, then \(3\) blue-booth exchanges, and finally \(1\) red-booth exchange. The red-blue counts become \[ \begin{aligned} &(100,0),(0,50),(16,2),\\ &(0,10),(3,1),(1,2). \end{aligned} \]

Therefore Alex ends with \(59+44=103\) silver tokens, and the correct answer is E .

给定前三位数字时，如果它们的和不超过 \(9\)，就能确定一个满足条件的数。

按千位数字分类讨论。

如果千位是 \(1\)，那么百位和十位数字之和必须不超过 \(8\)。设其和为 \(s\)，则有 \(s+1\) 种取法，因为百位数字可以从 \(0\) 到 \(s\)。

因此这一情形给出第九个三角数： \[\dfrac{9\cdot 10}2 =45\text{。}\]

如果千位是 \(2\)，那么小于 \(2013\) 的数中只有 \(2002\) 满足条件，所以总数为 \(46\)。

所以正确答案是 D。

English

Once the first three digits are given, their sum determines the units digit, provided that the sum is at most \(9.\)

We separate the possibilities according to the thousands digit.

If the thousands digit is \(1,\) then the sum of the hundreds and tens digits must be at most \(8.\) For each possible sum \(s,\) there are \(s+1\) choices, because the hundreds digit can range from \(0\) to \(s.\)

Thus, this case gives the ninth triangular number: \[\dfrac{9\cdot 10}2 =45.\]

If the thousands digit is \(2,\) the only qualifying number below \(2013\) is \(2002,\) giving \(46\) cases in total.

Thus, the correct answer is D.

设公差为 \(d\)，则 \(c=b-d\)、\(a=b+d\)。

二次式恰好有一个实根，所以判别式为 \(0\)，即 \(b^2-4ac=0\)。代入得 \(b^2=4(b+d)(b-d)=4b^2-4d^2\)，因此 \(4d^2=3b^2\)。

由于 \(b,d\ge0\)，由上式得 \(2d=\sqrt3 b\)。重根为 \(\frac{-b}{2a}=\frac{-b}{2(b+d)}=\frac{-1}{2+\sqrt3}=-2+\sqrt3\)。

所以正确答案是 D。

English

Let the common difference be \(d\), so \(c=b-d\) and \(a=b+d\).

A quadratic with exactly one real root has discriminant \(0\), so \(b^2-4ac=0\). Substituting gives \(b^2=4(b+d)(b-d)=4b^2-4d^2\), hence \(4d^2=3b^2\).

Since \(b,d\ge0\), \(2d=\sqrt3 b\). The double root is \(\frac{-b}{2a}=\frac{-b}{2(b+d)}=\frac{-1}{2+\sqrt3}=-2+\sqrt3\).

Thus, the correct answer is D .

\(2013=3\cdot11\cdot61\)，所以分子必须含因子 \(61\)，从而 \(a_1\ge61\)。

但 \(61!\) 还含有素因子 \(59\)，而 \(2013\) 中没有这个因子，所以分母必须含因子 \(59\)，从而 \(b_1\ge59\)。

因此 \(a_1+b_1\ge120\)。这个下界可以达到：\(2013=\frac{61!\,11!\,3!}{59!\,10!\,5!}\)。

所以 \(|a_1-b_1|=61-59=2\)，正确答案是 B。

English

The prime factorization is \(2013=3\cdot11\cdot61\), so the numerator must contain a factor of \(61\). Hence \(a_1\ge61\).

But \(61!\) also contains the prime factor \(59\), which is not in \(2013\), so the denominator must contain a factor of \(59\). Hence \(b_1\ge59\).

The lower bound \(a_1+b_1\ge120\) is attainable because \(2013=\frac{61!\,11!\,3!}{59!\,10!\,5!}\).

Thus \(|a_1-b_1|=61-59=2\), and the correct answer is B .

若数列前两项为 \(u,v\)，则第七项为 \(5u+8v\)。

设两个数列前两项分别为 \((a_1,a_2)\) 和 \((b_1,b_2)\)，且 \(a_1\lt b_1\)。因为两者第七项相同，\(5a_1+8a_2=5b_1+8b_2\)，所以 \(5(b_1-a_1)=8(a_2-b_2)\)。

由于 \(5\) 与 \(8\) 互质，\(b_1-a_1\) 至少为 \(8\)。非递减要求 \(b_2\ge b_1\ge a_1+8\)。

取 \(a_1=0\)、\(b_1=b_2=8\)、\(a_2=13\) 可达到最小值，得到 \(N=5\cdot0+8\cdot13=104\)。

所以正确答案是 C。

English

A sequence starting with \(u,v\) has seventh term \(5u+8v\).

For two sequences \((a_1,a_2)\) and \((b_1,b_2)\) with different first terms, assume \(a_1\lt b_1\). Then \(5a_1+8a_2=5b_1+8b_2\), so \(5(b_1-a_1)=8(a_2-b_2)\).

Since \(5\) and \(8\) are relatively prime, \(b_1-a_1\) is at least \(8\), and then nondecreasing order gives \(b_2\ge b_1\ge a_1+8\).

The smallest construction is \(a_1=0\), \(b_1=b_2=8\), and \(a_2=13\). This gives \(N=5\cdot0+8\cdot13=104\).

Thus, the correct answer is C .

设每条线的公共和为 \(S\)。\[ \begin{aligned} S &= A+J+E \\ &=B+J+F \\ &=C+J+G \\ &=D+J+H \end{aligned} \] \[ \begin{aligned} 4S &= A+B+C+D+E \\ &\quad+F+G+H+4J \end{aligned} \] \[4S = 45+3J\] \[45+3J \equiv 0 \mod 4\] \[3J \equiv 3 \mod 4\] \[J \equiv 1 \mod 4\]

四条线的和相加后可知中心数字只能为 \(J=1,5,9\)。先看 \(J=1\) 的情况。

当 \(J=1\) 时，满足条件的数对为 \[2+9 = 3+8 = 4+7 = 5+6\] 四对可分配给四条直线，有 \(4!\) 种，每对内部可交换，有 \(2^4\) 种，其中 \(2+9\iff 9+2\) 表示同一对的两种顺序。

对 \(J=5\) 和 \(J=9\) 也同理，因此总数为 \[3\cdot 4! \cdot 2^4 = 1152\] 种。

所以正确答案是 C。

English

Let \(S\) be defined as: \[ \begin{aligned} S &= A+J+E \\ &=B+J+F \\ &=C+J+G \\ &=D+J+H \end{aligned} \] \[ \begin{aligned} 4S &= A+B+C+D+E \\ &\quad+F+G+H+4J \end{aligned} \] \[4S = 45+3J\] \[45+3J \equiv 0 \mod 4\] \[3J \equiv 3 \mod 4\] \[J \equiv 1 \mod 4\]

This means that \(J=1,5,9.\) From here, let’s assume \(J=1.\) We will see that the other cases are similar enough to omit.

If \(J=1,\) then we know that the pairs of numbers that satisfy the equality above are: \[2+9 = 3+8 = 4+7 = 5+6\] There are \(4!\) ways to distribute the pairs over the four groups, and then \(2^4\) ways for these groups to swap elements (i.e. \(2+9\iff 9+2\)).

Now, if we look at the \(J=5\) and \(J=9\) cases, we see a similar pattern in the number of groupings and swaps. As such, we have: \[3\cdot 4! \cdot 2^4 = 1152\] possibilities.

Thus, the correct answer is C .

设 \(BD=x\)，则 \(CD=14-x\)。分别对直角三角形 \(ABD\) 和 \(ACD\) 应用勾股定理，得到 \[ 13^2-x^2=15^2-(14-x)^2\text{。}\]因此 \(BD=5\)，\(CD=9\)，且 \(AD=12\)。

得到下图：

因为 \(AD\perp BC\) 且 \(DE\perp AC\)，所以 \(\angle ADE=\angle ACD\)。又有 \(\angle AFB=\angle ADB=90^\circ\)，故 \(A,B,D,F\) 四点共圆。因此 \[ \angle ABF=\angle ADF=\angle ACD\text{。}\]在直角三角形 \(ACD\) 中，\(\cos(\angle ACD)=\frac35\)，且 \(\sin(\angle ACD)=\frac45\)。所以在直角三角形 \(ABF\) 中，\[ \begin{aligned} BF&=13\cdot\frac35,\\ AF&=13\cdot\frac45 \end{aligned}\text{。}\]

由托勒密定理，\[ \begin{aligned} &AB \cdot DF + DB \cdot AF \\ &= BF\cdot AD \end{aligned}\text{。}\]因此，\[ \begin{aligned} 13\cdot DF&+5\left(13\cdot\frac45\right)\\ &=12\left(13\cdot\frac35\right) \end{aligned}\text{。}\]

两边除以 \(13\)，得 \(DF+4=\frac{36}{5}\)，所以 \(DF=\frac{16}{5}\)。于是 \(m+n=21\)，正确答案是 B。

English

Let \(BD=x\), so \(CD=14-x\). Applying the Pythagorean Theorem to right triangles \(ABD\) and \(ACD\) gives \[ 13^2-x^2=15^2-(14-x)^2. \] Thus \(BD=5\), \(CD=9\), and \(AD=12\).

This yields the following diagram:

Because \(AD\perp BC\) and \(DE\perp AC\), we have \(\angle ADE=\angle ACD\). Also, \(\angle AFB=\angle ADB=90^\circ\), so \(A,B,D,F\) are concyclic. Hence \[ \angle ABF=\angle ADF=\angle ACD. \] In right triangle \(ACD\), \(\cos(\angle ACD)=\frac35\) and \(\sin(\angle ACD)=\frac45\). Therefore, in right triangle \(ABF\), \[ \begin{aligned} BF&=13\cdot\frac35,\\ AF&=13\cdot\frac45. \end{aligned} \]

By Ptolemy’s Theorem, we get \[ \begin{aligned} &AB \cdot DF + DB \cdot AF \\ &= BF\cdot AD . \end{aligned} \] Therefore, \[ \begin{aligned} 13\cdot DF&+5\left(13\cdot\frac45\right)\\ &=12\left(13\cdot\frac35\right). \end{aligned} \]

Dividing by \(13\) gives \(DF+4=\frac{36}{5}\), so \(DF=\frac{16}{5}\). Thus \(m+n=21\), and the correct answer is B .

恰好有四个正因数的数是 \(p^3\) 或 \(pq\)，其中 \(p\) 和 \(q\) 是不同素数。

若 \(m=p^3\)，因数和为 \(1+p+p^2+p^3\)。\(p=11\) 与 \(p=13\) 的对应值分别低于 \(2010\) 和高于 \(2019\)，所以这种情况没有解。

若 \(m=pq\)，因数和为 \(1+p+q+pq=(p+1)(q+1)\)。若其中一个素数为 \(2\)，和可被 \(3\) 整除；候选只有 \(2010\) 和 \(2016\)，但 \(\frac{2010}{3}-1=669\)、\(\frac{2016}{3}-1=671\) 都不是质数。

若两个素数都是奇数，和可被 \(4\) 整除，候选为 \(2012\) 和 \(2016\)。\(2012=4\cdot503\) 会给出 \(3\) 和 \(502\)，不可行；而 \(2016=4\cdot504\) \(=(3+1)(503+1)\) 可行。

所以恰有一个好数，正确答案是 A。

English

An integer with exactly four positive divisors is either \(p^3\) or \(pq\), where \(p\) and \(q\) are distinct primes.

If \(m=p^3\), the divisor sum is \(1+p+p^2+p^3\). The values for \(p=11\) and \(p=13\) fall below and above the interval \(2010\) to \(2019\), so this case gives none.

In the \(m=pq\) case, the divisor sum is \(1+p+q+pq=(p+1)(q+1)\). If one prime is \(2\), the sum is divisible by \(3\); only \(2010\) and \(2016\) qualify, but \(\frac{2010}{3}-1=669\) and \(\frac{2016}{3}-1=671\) are not prime.

If both primes are odd, then the sum is divisible by \(4\), leaving \(2012\) and \(2016\). The factorization \(2012=4\cdot503\) would give primes \(3\) and \(502\), impossible, while \(2016=4\cdot504\) \(=(3+1)(503+1)\) works.

Thus exactly one number is nice, and the correct answer is A .

只需在模 \(900=\operatorname{lcm}(25,36,100)\) 下分析，因为 \(5\) 进制、\(6\) 进制和以十为底表示时的末两位都以此为周期。

设 \(5\) 进制末两位为 \(a_1a_0\)，\(6\) 进制末两位为 \(b_1b_0\)。个位条件给出 \(a_0+b_0\equiv2N\equiv2a_0\pmod{10}\)，所以 \(a_0=b_0\)。

写作 \(N=150N_3+30a_1+a_0\)。\(6\) 进制十位条件给出 \(N_3\equiv a_1+b_1\pmod6\)，所以 \(N=900N_4+180a_1\) \(+150b_1+a_0\)。

比较 \(S\) 和 \(2N\) 的十位条件，化简得 \(5a_1\equiv b_1\pmod{10}\)。在 \(0\le a_1\le4\)、\(0\le b_1\le5\) 下，可行对为 \((0,0),(2,0),(4,0),(1,5),(3,5)\)。

又 \(a_0=b_0\) 有 \(5\) 种选择，所以共有 \(5\cdot5=25\) 个 \(N\)。

所以正确答案是 E。

English

It is enough to work modulo \(900=\operatorname{lcm}(25,36,100)\), because the last two base-\(5\), base-\(6\), and decimal digits repeat with that period.

Let the last two base-\(5\) digits be \(a_1a_0\) and the last two base-\(6\) digits be \(b_1b_0\). The units digit condition gives \(a_0+b_0\equiv2N\equiv2a_0\pmod{10}\), so \(a_0=b_0\).

Writing \(N=150N_3+30a_1+a_0\), the base-\(6\) tens digit condition gives \(N_3\equiv a_1+b_1\pmod6\), so \(N=900N_4+180a_1\) \(+150b_1+a_0\).

The tens digit condition for \(S\) and \(2N\) reduces to \(5a_1\equiv b_1\pmod{10}\). With \(0\le a_1\le4\) and \(0\le b_1\le5\), the valid pairs are \((0,0),(2,0),(4,0),(1,5),(3,5)\).

There are \(5\) choices for \(a_0=b_0\), so there are \(5\cdot5=25\) choices of \(N\).

Thus, the correct answer is E .