1995 AMC 12 真题（P版）

1 答案 B

前三次成绩的总和为 \(87+83+88=258\)，所以平均分为 \(\frac{258}{3}=86\)。加上第四次成绩后，总分为 \(348\)，新的平均分为 \(\frac{348}{4}=87\)。平均分增加了 \(1\) 分，因此正确答案是 B。

English

The first three scores total \(87+83+88=258,\) so their average is \(\frac{258}{3}=86.\) With the fourth score, the total is \(348,\) and the new average is \(\frac{348}{4}=87.\) It increases by \(1,\) so the correct answer is B.

2 答案 D

平方得 \(2+\sqrt{x}=9\)，所以 \(\sqrt{x}=7\)。再次平方得 \(x=49\)。因此正确答案是 D。

English

Squaring gives \(2+\sqrt{x}=9,\) so \(\sqrt{x}=7.\) Squaring again gives \(x=49.\) Thus the correct answer is D.

3 答案 B

广告总价为 \(3(29.98)+9.98=99.92\) 美元。节省的钱为 \(99.99-99.92=0.07\) 美元，即 \(7\) 美分。因此正确答案是 B。

English

The advertised total is \(3(29.98)+9.98=99.92\) dollars. The savings is \(99.99-99.92=0.07\) dollars, or \(7\) cents. Thus the correct answer is B.

4 答案 B

有 \(M=(0.30)(0.20)P=0.06P\)，且 \(N=0.50P\)。因此 \(\frac{M}{N}=\frac{0.06}{0.50}=\frac{3}{25}\)。所以正确答案是 B。

English

We have \(M=(0.30)(0.20)P=0.06P\) and \(N=0.50P.\) Hence \(\frac{M}{N}=\frac{0.06}{0.50}=\frac{3}{25}.\) Thus the correct answer is B.

5 答案 C

田地面积为 \(300\cdot400=120{,}000\) 平方英尺，即 \(120{,}000\cdot144=17{,}280{,}000\) 平方英寸。估计蚂蚁总数为 \(3(17{,}280{,}000)=51{,}840{,}000\)，最接近 \(50\) 百万。因此正确答案是 C。

English

The area is \(300\cdot400=120{,}000\) square feet, or \(120{,}000\cdot144=17{,}280{,}000\) square inches. The estimate is \(3(17{,}280{,}000)=51{,}840{,}000,\) closest to \(50\) million. Thus the correct answer is C.

6 答案 C

将 \(A\) 固定为正面。折叠后，\(x\) 成为左面，\(B\) 成为上面。面 \(C\) 与 \(B\) 的右侧相连，随后折成右面，正好与 \(x\) 相对。因此正确答案是 C。

English

Hold \(A\) as the front face. Folding makes \(x\) the left face and \(B\) the top face. The face \(C,\) attached to the right of \(B,\) then folds to the right face, opposite \(x.\) Thus the correct answer is C.

7 答案 C

赤道长约为 \(2\pi(4000)=8000\pi\) 英里。飞行时间为 \(\frac{8000\pi}{500}=16\pi\approx50\) 小时。因此正确答案是 C。

English

The equator is approximately \(2\pi(4000)=8000\pi\) miles long. The flight time is \(\frac{8000\pi}{500}=16\pi\approx50\) hours. Thus the correct answer is C.

8 答案 C

大直角三角形中 \(AB=\sqrt{6^2+8^2}=10\)。由于 \(\overline{DE}\parallel\overline{AC}\)，三角形 \(BDE\) 与 \(BAC\) 相似，相似比为 \(\frac{DE}{AC}=\frac{4}{6}=\frac{2}{3}\)。因此 \(BD=(\frac{2}{3})(10)=\frac{20}{3}\)。所以正确答案是 C。

English

The large right triangle has \(AB=\sqrt{6^2+8^2}=10.\) Since \(\overline{DE}\parallel\overline{AC},\) triangles \(BDE\) and \(BAC\) are similar, with scale factor \(\frac{DE}{AC}=\frac{4}{6}=\frac{2}{3}.\) Therefore \(BD=(\frac{2}{3})(10)=\frac{20}{3}.\) Thus the correct answer is C.

9 答案 D

最小的三角形有 \(8\) 个，每个由半条边、半条对角线和半条中位线围成。以正方形的一整条边为底、中心为顶点的三角形有 \(4\) 个；由对角线截出的半正方形三角形另有 \(4\) 个。总数为 \(8+4+4=16\)。因此正确答案是 D。

English

There are \(8\) smallest triangles, each bounded by a half-side, a half-diagonal, and a half-midline. There are \(4\) triangles whose base is a full side and vertex is the center, and \(4\) half-square triangles cut off by a diagonal. The total is \(8+4+4=16.\) Thus the correct answer is D.

10 答案 E

三个顶点是 \((0,0)\)、\((6,6)\) 和 \((-6,6)\)。水平底边长为 \(12\)，高为 \(6\)，所以面积是 \(\frac12(12)(6)=36\)。因此正确答案是 E。

English

The vertices are \((0,0),\) \((6,6),\) and \((-6,6).\) The horizontal base has length \(12,\) and the height is \(6,\) so the area is \(\frac12(12)(6)=36.\) Thus the correct answer is E.

11 答案 C

千位数字 \(a\) 为 \(4\) 或 \(5\)，而能被 \(5\) 整除意味着 \(d\) 为 \(0\) 或 \(5\)。有序对 \((b,c)\) 是从 \(3\)、\(4\)、\(5\)、\(6\) 中选两个数字并按递增顺序排列，共有 \(\binom42=6\) 种。因此总数为 \(2\cdot2\cdot6=24\)，正确答案是 C。

English

The thousands digit \(a\) is \(4\) or \(5,\) and divisibility by \(5\) makes \(d\) either \(0\) or \(5.\) The pair \((b,c)\) is obtained by choosing two of \(3,\) \(4,\) \(5,\) \(6\) in increasing order, giving \(\binom42=6\) pairs. Thus the count is \(2\cdot2\cdot6=24,\) and the correct answer is C.

12 答案 D

由 \(f(1)\le f(2)\) 可知斜率非负；由 \(f(3)\ge f(4)\) 可知斜率非正。因此斜率为 \(0\)，所以 \(f\) 是常函数。由于 \(f(5)=5\)，有 \(f(0)=5\)。因此正确答案是 D。

English

From \(f(1)\le f(2),\) the slope is nonnegative. From \(f(3)\ge f(4),\) it is nonpositive. Thus the slope is \(0,\) so \(f\) is constant. Since \(f(5)=5,\) we have \(f(0)=5.\) Thus the correct answer is D.

13 答案 C

个位、十位和百位的运算成立，对应的进位依次为 \(0\)、\(1\)、\(1\)。在千位上，把每个 \(2\) 都改为 \(6\)，便有 \(6+9+1=16\)，结果中的该位也正好是 \(6\)。接下来的两列分别给出 \(4+6+1=11\) 和 \(7+8+1=16\)，得到正确算式 \(746586+869430=1616016\)。因此 \(d+e=2+6=8\)，正确答案是 C。

English

The units, tens, and hundreds columns work with carries \(0,\) \(1,\) \(1.\) In the thousands column, replacing every \(2\) by \(6\) gives \(6+9+1=16,\) so the result digit is also \(6.\) The next columns then give \(4+6+1=11\) and \(7+8+1=16,\) producing the corrected equation \(746586+869430=1616016.\) Thus \(d+e=2+6=8,\) and the correct answer is C.

14 答案 E

当自变量分别为 \(3\) 与 \(-3\) 时，偶次幂项和常数项相同，而一次项由 \(-3\) 变为 \(3\)。因此 \(f(3)-f(-3)=6\)，所以 \(f(3)=2+6=8\)。正确答案是 E。

English

The even-power terms and constant are the same at \(3\) and \(-3,\) while the linear term changes from \(-3\) to \(3.\) Hence \(f(3)-f(-3)=6,\) so \(f(3)=2+6=8.\) Thus the correct answer is E.

15 答案 D

从 \(5\) 出发，依次落在 \(1,2,4,1,2,4,\ldots\)。因此循环 \(1,2,4\) 的长度为 \(3\)。由于 \(1995\) 能被 \(3\) 整除，第 \(1995\) 次的落点是 \(4\)。所以正确答案是 D。

English

Starting at \(5,\) the landing points are \(1,2,4,1,2,4,\ldots\) Thus the cycle \(1,2,4\) has length \(3.\) Since \(1995\) is divisible by \(3,\) the \(1995\)th landing point is \(4.\) Thus the correct answer is D.

16 答案 E

亚特兰大的实际观众人数 \(A\) 满足 \(45{,}000\le A\le55{,}000\)。对波士顿的实际观众人数 \(B\)，条件为 \(|60{,}000-B|\le0.1B\)，所以 \(\frac{60{,}000}{1.1}\le B\le\frac{60{,}000}{0.9}\)。因此最大差值为 \[ \frac{60{,}000}{0.9}-45{,}000 =21{,}666.\overline6\text{，}\]四舍五入后为 \(22{,}000\)。所以正确答案是 E。

English

Atlanta’s actual attendance \(A\) satisfies \(45{,}000\le A\le55{,}000.\) For Boston’s actual attendance \(B,\) the condition is \(|60{,}000-B|\le0.1B,\) so \(\frac{60{,}000}{1.1}\le B\le\frac{60{,}000}{0.9}.\) The largest difference is therefore \[ \frac{60{,}000}{0.9}-45{,}000 =21{,}666.\overline6, \] which rounds to \(22{,}000.\) Thus the correct answer is E.

17 答案 E

设 \(O\) 为圆心。沿五边形经过每个顶点时，方向改变 \(72^\circ\)，所以直线 \(AB\) 与 \(DC\) 的锐角夹角为 \(36^\circ\)。通向两个切点的半径分别垂直于这两条直线；对图示圆而言，小圆心角 \(\angle AOD\) 是其补角 \(180^\circ-36^\circ=144^\circ\)。因此小弧 \(\overset{\frown}{AD}\) 的度数为 \(144^\circ\)，正确答案是 E。

English

Let \(O\) be the circle’s center. The direction changes by \(72^\circ\) at each vertex of the pentagon, so the acute angle between lines \(AB\) and \(DC\) is \(36^\circ.\) The radii to their tangency points are perpendicular to these lines; for the circle shown, the minor central angle \(\angle AOD\) is the supplementary angle \(180^\circ-36^\circ=144^\circ.\) Therefore minor arc \(\overset{\frown}{AD}\) measures \(144^\circ,\) and the correct answer is E.

18 答案 D

设 \(OA=x\)，\(OB=y\)。由余弦定理得 \[ 1=x^2+y^2-\sqrt3xy\text{。}\]把它看作关于 \(x\) 的二次方程，其判别式为 \(3y^2-4(y^2-1)=4-y^2\)。要使 \(x\) 为实数，必须有 \(y\le2\)；当 \(x=\sqrt3\) 时可以取到等号。因此最大值为 \(2\)，正确答案是 D。

English

Let \(OA=x\) and \(OB=y.\) The Law of Cosines gives \[ 1=x^2+y^2-\sqrt3xy. \] As a quadratic in \(x,\) this has discriminant \(3y^2-4(y^2-1)=4-y^2.\) A real \(x\) requires \(y\le2,\) and equality is attained when \(x=\sqrt3.\) Hence the maximum is \(2,\) and the correct answer is D.

19 答案 C

令 \(ABC\) 的边长为 \(1\)，并取 \(B=(0,0)\)、\(C=(1,0)\)。若内部三角形边长为 \(s\)，设 \(D=(d,0)\)、\(E=(d,s)\)。由于 \(E\) 位于 \(AC\) 上，有 \(s=\sqrt3(1-d)\)。内部三角形的第三个顶点为 \(F=(d-\frac{\sqrt3}{2}s,\frac12s)\)，而 \(F\) 位于 \(AB\) 上，所以 \[ \frac{s}{2}=\sqrt3\left(d-\frac{\sqrt3}{2}s\right)\text{，}\]从而 \(d=\frac{2s}{\sqrt3}\)。代入可得 \(s=\frac{1}{\sqrt3}\)。等边三角形的面积比等于边长比的平方，因此所求比为 \(s^2=\frac{1}{3}\)。所以正确答案是 C。

English

Let \(ABC\) have side \(1,\) with \(B=(0,0)\) and \(C=(1,0).\) If the inner side is \(s,\) write \(D=(d,0)\) and \(E=(d,s).\) Since \(E\) lies on \(AC,\) \(s=\sqrt3(1-d).\) The third inner vertex is \(F=(d-\frac{\sqrt3}{2}s,\frac12s),\) and \(F\) lies on \(AB,\) so \[ \frac{s}{2}=\sqrt3\left(d-\frac{\sqrt3}{2}s\right), \] giving \(d=\frac{2s}{\sqrt3}.\) Substitution yields \(s=\frac{1}{\sqrt3}.\) Areas of equilateral triangles scale as the square of their sides, so the ratio is \(s^2=\frac{1}{3}.\) Thus the correct answer is C.

20 答案 B

可选的奇数有 \(3\) 个，偶数有 \(2\) 个。乘积 \(ab\) 为奇数的概率是 \(\frac{9}{25}\)，为偶数的概率是 \(\frac{16}{25}\)。因此 \[ \begin{aligned} P(ab+c\text{ 为偶数}) &=\frac9{25}\cdot\frac35\\ &\quad{}+\frac{16}{25}\cdot\frac25\\ &=\frac{59}{125} \end{aligned}\text{。}\]所以正确答案是 B。

English

There are \(3\) odd and \(2\) even choices. The product \(ab\) is odd with probability \(\frac{9}{25}\) and even with probability \(\frac{16}{25}.\) Therefore \[ \begin{aligned} P(ab+c\text{ even}) &=\frac9{25}\cdot\frac35\\ &\quad{}+\frac{16}{25}\cdot\frac25\\ &=\frac{59}{125}. \end{aligned} \] Thus the correct answer is B.

21 答案 E

已知对角线的中点是 \((0,0)\)，半长为 \(5\)。因此另一条对角线的端点必须是 \((u,v)\) 和 \((-u,-v)\)，其中 \(u^2+v^2=25\)。反过来，中点相同且长度相等的两条对角线构成一个长方形。这个圆上共有 \(12\) 个整数点：\[ \begin{gathered} (\pm5,0),(0,\pm5),\\ (\pm3,\pm4),(\pm4,\pm3) \end{gathered}\text{。}\]一对对径点确定同一条对角线，因此有 \(6\) 种可能。其中一种就是已知对角线本身，会产生退化图形，所以还剩 \(6-1=5\) 个长方形。因此正确答案是 E。

English

The given diagonal has midpoint \((0,0)\) and half-length \(5.\) The other diagonal must therefore have endpoints \((u,v)\) and \((-u,-v)\) with \(u^2+v^2=25.\) Conversely, two equal diagonals with the same midpoint form a rectangle. There are \(12\) integer points on this circle: \[ \begin{gathered} (\pm5,0),(0,\pm5),\\ (\pm3,\pm4),(\pm4,\pm3). \end{gathered} \] Antipodal points determine the same diagonal, giving \(6\) possibilities. One is the given diagonal itself, which is degenerate, so \(6-1=5\) rectangles remain. Thus the correct answer is E.

22 答案 E

长方形的完整边长必须大于同一边上剪后剩余部分的长度。唯一能使两个差与剩下的切边构成直角三角形的安排，是边长为 \(25\) 和 \(31\) 的长方形，对应剩余长度为 \(20\) 和 \(19\)。被剪去的角所成直角三角形的两条直角边为 \(25-20=5\) 和 \(31-19=12\)，斜边正好是给定边长 \(13\)。因此五边形面积为 \[ \begin{aligned} 25\cdot31-\frac12(5)(12) &=775-30\\ &=745 \end{aligned}\text{。}\]所以正确答案是 E。

English

The full rectangle sides must be longer than the two remnants on those same sides. The only assignment whose two differences and remaining cut side form a right triangle is the rectangle \(25\) by \(31,\) with remnants \(20\) and \(19.\) The removed corner then has legs \(25-20=5\) and \(31-19=12,\) whose hypotenuse is the listed side \(13.\) Thus the pentagon’s area is \[ \begin{aligned} 25\cdot31-\frac12(5)(12) &=775-30\\ &=745. \end{aligned} \] Hence the correct answer is E.

23 答案 D

由三角形不等式得 \(5\le k\le25\)。当 \(k\le15\) 时，最长边为 \(15\)，三角形为钝角三角形的条件是 \[ 15^2\gt11^2+k^2\text{，}\]从而 \(5\le k\le10\)，共有 \(6\) 个值。当 \(k\gt15\) 时，钝角条件为 \(k^2\gt11^2+15^2=346\)，得到 \(19\le k\le25\)，共有 \(7\) 个值。总数为 \(6+7=13\)，所以正确答案是 D。

English

The triangle inequality gives \(5\le k\le25.\) For \(k\le15,\) the longest side is \(15,\) and the triangle is obtuse when \[ 15^2\gt11^2+k^2, \] which gives \(5\le k\le10,\) or \(6\) values. For \(k\gt15,\) it is obtuse when \(k^2\gt11^2+15^2=346,\) giving \(19\le k\le25,\) or \(7\) values. The total is \(6+7=13,\) so the correct answer is D.

24 答案 A

合并对数得 \[ \log_{200}(5^A2^B)=C\text{，}\]所以 \(5^A2^B=200^C=5^{2C}2^{3C}\)。因此 \(A=2C\)，且 \(B=3C\)。没有公因数的正整数三元组为 \((A,B,C)=(2,3,1)\)，其和为 \(6\)。所以正确答案是 A。

English

Combining logarithms gives \[ \log_{200}(5^A2^B)=C, \] so \(5^A2^B=200^C=5^{2C}2^{3C}.\) Hence \(A=2C\) and \(B=3C.\) The relatively prime positive triple is \((A,B,C)=(2,3,1),\) whose sum is \(6.\) Thus the correct answer is A.

25 答案 B

将排序后的数列写成 \(m,8,8,d,m+18\)。（若第四个数也是 \(8\)，总和与排序条件将无法同时满足。）由于总和为 \(5(12)=60\)，有 \[ m+8+8+d+(m+18)=60\text{，}\]所以 \(d=26-2m\)。条件 \(8\le d\le m+18\) 和 \(m\le8\) 给出 \(3\le m\le8\)。这六个取值分别得到 \(d=20\)、\(18\)、\(16\)、\(14\)、\(12\)、\(10\)，均有效。因此共有 \(6\) 种可能，正确答案是 B。

English

Write the sorted list as \(m,8,8,d,m+18.\) (If the fourth entry were \(8,\) the sum and ordering conditions would be impossible.) Since the total is \(5(12)=60,\) \[ m+8+8+d+(m+18)=60, \] so \(d=26-2m.\) The conditions \(8\le d\le m+18\) and \(m\le8\) give \(3\le m\le8.\) These six values yield \(d=20,\) \(18,\) \(16,\) \(14,\) \(12,\) \(10,\) all valid. Thus there are \(6\) possibilities, and the correct answer is B.

26 答案 C

设半径为 \(r\)，且 \(OE=x\)。由相交弦定理得 \[ (r+x)(r-x)=6\cdot2\text{，}\]所以 \(r^2-x^2=12\)。取 \(O=(0,0)\)、\(D=(0,-r)\)、\(E=(x,0)\)。由于 \(DE:EF=3:1\)，有 \[ \begin{aligned} F&=D+\frac43(E-D)\\ &=\left(\frac{4x}{3},\frac r3\right) \end{aligned}\text{。}\]因为 \(F\) 在圆上，\(\frac{16x^2}{9}+\frac{r^2}{9}=r^2\)，所以 \(r^2=2x^2\)。因此 \(x^2=12\)，且 \(r^2=24\)。圆的面积为 \(24\pi\)，所以正确答案是 C。

English

Let the radius be \(r\) and \(OE=x.\) Intersecting chords gives \[ (r+x)(r-x)=6\cdot2, \] so \(r^2-x^2=12.\) Put \(O=(0,0),\) \(D=(0,-r),\) and \(E=(x,0).\) Since \(DE:EF=3:1,\) \[ \begin{aligned} F&=D+\frac43(E-D)\\ &=\left(\frac{4x}{3},\frac r3\right). \end{aligned} \] Because \(F\) lies on the circle, \(\frac{16x^2}{9}+\frac{r^2}{9}=r^2,\) so \(r^2=2x^2.\) Therefore \(x^2=12\) and \(r^2=24.\) The area is \(24\pi,\) so the correct answer is C.

27 答案 E

上一行的每个数都会贡献给下一行的两个数，而新出现的两个边界数又额外贡献 \(2\)。因此 \(f(n)=2f(n-1)+2\)，且 \(f(1)=0\)。所以 \[ f(n)=2^n-2\text{。}\]由于 \(2^{20}\equiv76\pmod{100}\)，且 \(76^2\equiv76\pmod{100}\)，可得 \(2^{100}\equiv76\pmod{100}\)。因此 \(f(100)\equiv74\pmod{100}\)，正确答案是 E。

English

Every previous entry contributes to two entries of the next row, and the two new boundary entries contribute an additional \(2.\) Thus \(f(n)=2f(n-1)+2,\) with \(f(1)=0.\) Hence \[ f(n)=2^n-2. \] Since \(2^{20}\equiv76\pmod{100}\) and \(76^2\equiv76\pmod{100},\) we have \(2^{100}\equiv76\pmod{100}.\) Therefore \(f(100)\equiv74\pmod{100},\) and the correct answer is E.

28 答案 E

设圆心到长为 \(14\) 和 \(10\) 的弦的有向距离分别为 \(u\) 和 \(v\)，且 \(u-v=6\)。则 \[ r^2-u^2=7^2,\qquad r^2-v^2=5^2\text{。}\]因此 \(v^2-u^2=24\)，所以 \((v-u)(v+u)=24\)。由于 \(v-u=-6\)，可得 \(u+v=-4\)，进而 \(u=1\)，且 \(v=-5\)。中间那条弦的有向距离为 \(-2\)，而 \(r^2=49+1=50\)。它的长度平方为 \(4(50-4)=184\)，所以 \(a=184\)。因此正确答案是 E。

English

Let the signed distances from the center to the \(14\)- and \(10\)-chords be \(u\) and \(v,\) with \(u-v=6.\) Then \[ r^2-u^2=7^2,\qquad r^2-v^2=5^2. \] Hence \(v^2-u^2=24,\) so \((v-u)(v+u)=24.\) Since \(v-u=-6,\) we get \(u+v=-4,\) and therefore \(u=1\) and \(v=-5.\) The midway chord is at signed distance \(-2,\) while \(r^2=49+1=50.\) Its squared length is \(4(50-4)=184,\) so \(a=184.\) Thus the correct answer is E.

29 答案 C

因为 \(2310=2\cdot3\cdot5\cdot7\cdot11\)，每个质因数恰好属于三个因数中的一个。若三个因数都大于 \(1\)，它们对应的无序质因数组就是把五个对象分成三个非空组，其数量为 \[ S(5,3)=\frac{3^5-3\cdot2^5+3}{6}=25\text{。}\]若有一个因数为 \(1\)，则把这些质因数分成两个非空组，得到 \(S(5,2)=2^4-1=15\)。由于互不相交的质因数集合不同，所得因数也互不相同。因此总数为 \(25+15=40\)，正确答案是 C。

English

Since \(2310=2\cdot3\cdot5\cdot7\cdot11,\) each prime belongs to exactly one of the three factors. If all factors exceed \(1,\) their unordered prime groups form a partition of five objects into three nonempty blocks, counted by \[ S(5,3)=\frac{3^5-3\cdot2^5+3}{6}=25. \] If one factor is \(1,\) the primes are partitioned into two nonempty blocks, giving \(S(5,2)=2^4-1=15.\) The factors are distinct because their disjoint prime sets differ. Thus the total is \(25+15=40,\) and the correct answer is C.

30 答案 D

将大立方体取为 \([0,3]^3\)。所求平面为 \(x+y+z=\frac{9}{2}\)。一个下方顶点为 \((i,j,k)\) 的单位立方体，其中 \(i,j,k\in\{0,1,2\}\)，其内部 \(x+y+z\) 的取值从 \(s=i+j+k\) 到 \(s+3\)。它与该平面相交，当且仅当 \(s=2\)、\(s=3\) 或 \(s=4\)。在 \[ (1+x+x^2)^3 \] 中，\(x^2\)、\(x^3\)、\(x^4\) 的系数分别为 \(6\)、\(7\)、\(6\)。因此该平面与 \(6+7+6=19\) 个单位立方体相交，正确答案是 D。

English

Take the large cube as \([0,3]^3.\) The plane is \(x+y+z=\frac{9}{2}.\) A unit cube with lower corner \((i,j,k),\) where \(i,j,k\in\{0,1,2\},\) contains values of \(x+y+z\) from \(s=i+j+k\) to \(s+3.\) It intersects the plane exactly when \(s=2,\) \(s=3,\) or \(s=4.\) The coefficients of \(x^2,\) \(x^3,\) \(x^4\) in \[ (1+x+x^2)^3 \] are \(6,\) \(7,\) \(6.\) Thus the plane intersects \(6+7+6=19\) unit cubes, and the correct answer is D.