1982 AMC 12 真题（P版）

1 答案 E

模 \(x^2-2\) 时，\(x^2\equiv2\)，所以 \(x^3-2\equiv2x-2\)。这个一次多项式就是余式。

所以正确答案是 E。

English

Modulo \(x^2-2,\) we have \(x^2\equiv2,\) so \(x^3-2\equiv2x-2.\) This linear polynomial is the remainder.

Therefore, the correct answer is E.

2 答案 A

\(x\) 的八倍增加二后是 \(8x+2\)。它的四分之一为 \(\frac{8x+2}{4}=2x+\frac12\)。

所以正确答案是 A。

English

Eight times \(x,\) increased by two, is \(8x+2.\) One fourth is \(\frac{8x+2}{4}=2x+\frac12.\)

Therefore, the correct answer is A.

3 答案 C

当 \(x=2\) 时，\(x^x=2^2=4\)。因此 \((x^x)^{(x^x)}=4^4=256\)。

所以正确答案是 C。

English

At \(x=2,\) \(x^x=2^2=4.\) Hence \((x^x)^{(x^x)}=4^4=256.\)

Therefore, the correct answer is C.

4 答案 E

当 \(r\gt0\) 时，周长与面积相等给出 \(\pi r+2r=\frac{\pi r^2}{2}\)。两边除以 \(r\) 并求解，得到 \(r=\frac{2(\pi+2)}{\pi}=2+\frac{4}{\pi}\)。

所以正确答案是 E。

English

For \(r\gt0,\) equality of perimeter and area gives \(\pi r+2r=\frac{\pi r^2}{2}.\) Dividing by \(r\) and solving yields \(r=\frac{2(\pi+2)}{\pi}=2+\frac{4}{\pi}.\)

Therefore, the correct answer is E.

5 答案 C

写成 \(x=ka\) 和 \(y=kb\)。因为 \(a\lt b\)，所以 \(x\) 较小。由 \(k(a+b)=c\) 得 \(k=\frac{c}{a+b}\)，因此 \(x=\frac{ac}{a+b}\)。

所以正确答案是 C。

English

Write \(x=ka\) and \(y=kb.\) Since \(a\lt b,\) \(x\) is smaller. From \(k(a+b)=c,\) \(k=\frac{c}{a+b},\) so \(x=\frac{ac}{a+b}.\)

Therefore, the correct answer is C.

6 答案 D

内角和必须为 \((n-2)180^\circ\)，而缺少的角介于 \(0^\circ\) 与 \(180^\circ\) 之间。介于 \(2570^\circ\) 与 \(2750^\circ\) 之间唯一的 \(180^\circ\) 的倍数是 \(2700^\circ\)，所以缺少的角为 \(2700^\circ-2570^\circ=130^\circ\)。

所以正确答案是 D。

English

The total must be \((n-2)180^\circ\) and the missing angle lies between \(0^\circ\) and \(180^\circ.\) The only multiple of \(180^\circ\) between \(2570^\circ\) and \(2750^\circ\) is \(2700^\circ,\) leaving \(2700^\circ-2570^\circ=130^\circ.\)

Therefore, the correct answer is D.

7 答案 B

因为 \(x*y=x+y+xy\)，该运算满足交换律，单位元为 \(0\)，并且各数加 \(1\) 后会转化为普通乘法。因此它满足结合律，直接展开也能验证 C。当 \(x=1\) 且 \(y=z=0\) 时，命题 B 不成立：左边为 \(1*0=1\)，而右边为 \((1*0)+(1*0)=2\)。所以 B 错误。

所以正确答案是 B。

English

Since \(x*y=x+y+xy,\) the operation is commutative, has identity \(0,\) and becomes ordinary multiplication after adding \(1.\) Thus it is associative, and direct expansion also verifies C. Statement B fails when \(x=1\) and \(y=z=0:\) its left side is \(1*0=1,\) but its right side is \((1*0)+(1*0)=2.\) Hence B is false.

Therefore, the correct answer is B.

8 答案 B

等差条件为 \(2\binom n2=\binom n1+\binom n3\)。代入并约分得到 \(n^2-9n+14=0\)，即 \((n-2)(n-7)=0\)。因为 \(n\gt3\)，所以 \(n=7\)。

所以正确答案是 B。

English

The progression condition is \(2\binom n2=\binom n1+\binom n3.\) Substitution and cancellation give \(n^2-9n+14=0,\) or \((n-2)(n-7)=0.\) Since \(n\gt3,\) \(n=7.\)

Therefore, the correct answer is B.

9 答案 B

三角形的面积为 \(4\)。当 \(0\le x\le1\) 时，该部分面积为 \[ \int_0^1\left(x-\frac x9\right)\,dx=\frac49\text{。}\]对于截线 \(x=t\ge1\)，其左侧面积为 \(\frac49+\int_1^t(1-\frac{x}{9})\,dx\)。令其等于 \(2\)，得到 \(t-\frac{t^2}{18}=\frac{5}{2}\)，相关的解为 \(t=3\)。

所以正确答案是 B。

English

The triangle has area \(4.\) The portion with \(0\le x\le1\) has area \[ \int_0^1\left(x-\frac x9\right)\,dx=\frac49. \] For a cut \(x=t\ge1,\) the area to its left is therefore \(\frac49+\int_1^t(1-\frac{x}{9})\,dx.\) Setting this equal to \(2\) gives \(t-\frac{t^2}{18}=\frac{5}{2},\) whose relevant solution is \(t=3.\)

Therefore, the correct answer is B.

10 答案 A

边长为 \(12\)、\(18\)、\(24\) 的三角形半周长为 \(27\)，面积为 \[ \sqrt{27\cdot15\cdot9\cdot3}=27\sqrt{15}\text{，}\]所以内切圆半径为 \(r=\sqrt{15}\)。以 \(BC\) 为底，高为 \(h=\frac{2(27\sqrt{15})}{24}=\frac{9\sqrt{15}}{4}\)。因此，从 \(\triangle ABC\) 到 \(\triangle AMN\) 的相似比为 \(1-\frac{r}{h}=1-\frac{4}{9}=\frac{5}{9}\)。它的周长为 \((\frac{5}{9})(12+18+24)=30\)。

所以正确答案是 A。

English

The sides \(12,\) \(18,\) \(24\) have semiperimeter \(27\) and area \[ \sqrt{27\cdot15\cdot9\cdot3}=27\sqrt{15}, \] so the inradius is \(r=\sqrt{15}.\) Taking \(BC\) as base, the altitude is \(h=\frac{2(27\sqrt{15})}{24}=\frac{9\sqrt{15}}{4}.\) Thus the similarity scale from \(\triangle ABC\) to \(\triangle AMN\) is \(1-\frac{r}{h}=1-\frac{4}{9}=\frac{5}{9}.\) Its perimeter is \((\frac{5}{9})(12+18+24)=30.\)

Therefore, the correct answer is A.

11 答案 C

首位数字依次取 \(1,\ldots,9\) 时，相差 \(2\) 的末位数字个数为 \(1+1+2+2+2\) 加上 \(2+2+1+1\)，共 \(14\) 个。此外，首位为 \(2\) 时末位还可为 \(0\)，因而共有 \(15\) 个首末数字有序对。中间两位有 \(8\cdot7\) 种有序选法。因此总数为 \(15\cdot56=840\)。

因此，正确答案为 C。

English

For leading digits \(1,\ldots,9,\) the number of possible last digits differing by \(2\) is \(1+1+2+2+2\) plus \(2+2+1+1,\) or \(14.\) In addition, leading digit \(2\) may end in \(0,\) giving \(15\) endpoint pairs. The middle digits can then be chosen in \(8\cdot7\) ordered ways. Thus the count is \(15\cdot56=840.\)

Therefore, the correct answer is C.

12 答案 A

令 \(g(x)=ax^7+bx^3+cx\)，则它是奇函数。由 \(f(-7)=g(-7)-5=7\)，得 \(g(-7)=12\)。因此 \(g(7)=-12\)，从而 \(f(7)=-12-5=-17\)。

因此，正确答案为 A。

English

Let \(g(x)=ax^7+bx^3+cx,\) which is odd. Since \(f(-7)=g(-7)-5=7,\) \(g(-7)=12.\) Therefore \(g(7)=-12,\) and \(f(7)=-12-5=-17.\)

Therefore, the correct answer is A.

13 答案 D

令 \(t=\log_ba\)，则 \(a=b^t\)，且 \(p=\frac{\log_bt}{t}\)。于是 \(a^p=(b^t)^{\frac{\log_bt}{t}}\)，即 \(b^{\log_bt}=t\)。因此 \(a^p=\log_ba\)。

因此，正确答案为 D。

English

Set \(t=\log_ba,\) so \(a=b^t\) and \(p=\frac{\log_bt}{t}.\) Then \(a^p=(b^t)^{\frac{\log_bt}{t}},\) which equals \(b^{\log_bt}=t.\) Therefore \(a^p=\log_ba.\)

Therefore, the correct answer is D.

14 答案 C

这里 \(AP=75\)、\(PG=15\)，且 \(AG\perp PG\)。点 \(N\) 到 \(AG\) 的距离按 \(\frac{AN}{AP}=\frac{45}{75}\) 成比例，因此该距离为 \(15(\frac{45}{75})=9\)。所以半径为 \(15\) 的圆中，这条弦的长度为 \(2\sqrt{15^2-9^2}=2\sqrt{144}=24\)。

因此，正确答案为 C。

English

Here \(AP=75,\) \(PG=15,\) and \(AG\perp PG.\) The distance from \(N\) to \(AG\) scales with \(\frac{AN}{AP}=\frac{45}{75},\) so it is \(15(\frac{45}{75})=9.\) Therefore the chord in the radius-\(15\) circle has length \(2\sqrt{15^2-9^2}=2\sqrt{144}=24.\)

Therefore, the correct answer is C.

15 答案 D

对于非整数 \(x\)，令 \(n=\lfloor x\rfloor\)；则 \(\lfloor x-2\rfloor=n-2\)。因而 \(2n+3=3(n-2)+5\)，所以 \(n=4\)，且 \(y=11\)。由于 \(4\lt x\lt5\)，可得 \(15\lt x+y\lt16\)。

因此，正确答案为 D。

English

For nonintegral \(x,\) let \(n=\lfloor x\rfloor;\) then \(\lfloor x-2\rfloor=n-2.\) Thus \(2n+3=3(n-2)+5,\) so \(n=4\) and \(y=11.\) Since \(4\lt x\lt5,\) we have \(15\lt x+y\lt16.\)

Therefore, the correct answer is D.

16 答案 B

六个外表面的面积为 \(6(3^2-1^2)=48\)。三条长为 \(3\) 的正方形通道各有四个内壁，但每个内壁中央长为一个单位的部分被垂直通道挖去，因此每个内壁留下的面积为 \(2\)。所以内部表面积为 \(3\cdot4\cdot2=24\)。总面积为 \(48+24=72\)。

因此，正确答案为 B。

English

The six outer faces contribute \(6(3^2-1^2)=48.\) Each of the three length-\(3\) square tunnels has four inner walls, but the central unit segment of every wall is removed by a perpendicular tunnel, leaving area \(2\) per wall. Thus the inside contributes \(3\cdot4\cdot2=24.\) The total is \(48+24=72.\)

Therefore, the correct answer is B.

17 答案 C

令 \(t=3^x\gt0\)。则 \(9t^2-28t+3=0\)，其两根为 \(t=3\) 和 \(t=\frac{1}{9}\)。两根均为正数，且各自对应唯一的实数 \(x\)，因此共有 \(2\) 个解。

因此，正确答案为 C。

English

Let \(t=3^x\gt0.\) Then \(9t^2-28t+3=0,\) whose roots are \(t=3\) and \(t=\frac{1}{9}.\) Both are positive and each corresponds to one real \(x,\) so there are \(2\) solutions.

Therefore, the correct answer is C.

18 答案 D

令 \(H=(0,0,0)\)、\(F=(u,0,0)\)、\(G=(0,v,0)\)，且 \(C=(0,0,w)\)。则 \(B=(u,0,w)\)，且 \(D=(0,v,w)\)。由 \(60^\circ\) 的条件得 \(w=\sqrt3u\)，而由 \(45^\circ\) 的条件得 \(v=w\)。因此 \[ \begin{aligned} \cos\angle BHD &=\frac{w^2} {\sqrt{u^2+w^2}\sqrt{2w^2}}\\ &=\frac{\sqrt6}{4} \end{aligned}\text{。}\]

因此，正确答案为 D。

English

Let \(H=(0,0,0),\) \(F=(u,0,0),\) \(G=(0,v,0),\) and \(C=(0,0,w).\) Then \(B=(u,0,w)\) and \(D=(0,v,w).\) The \(60^\circ\) condition gives \(w=\sqrt3u,\) while the \(45^\circ\) condition gives \(v=w.\) Hence \[ \begin{aligned} \cos\angle BHD &=\frac{w^2} {\sqrt{u^2+w^2}\sqrt{2w^2}}\\ &=\frac{\sqrt6}{4}. \end{aligned} \]

Therefore, the correct answer is D.

19 答案 B

在 \([2,3]\) 上，\(f(x)=2x-4\)；在 \([3,4]\) 上，\(f(x)=8-2x\)；而在 \([4,8]\) 上，\(f(x)=0\)。因而最小值为 \(0\)，最大值为 \(2\)，两者之和为 \(2\)。

因此，正确答案为 B。

English

On \([2,3],\) \(f(x)=2x-4;\) on \([3,4],\) \(f(x)=8-2x;\) and on \([4,8],\) \(f(x)=0.\) Thus the minimum is \(0\) and the maximum is \(2,\) whose sum is \(2.\)

Therefore, the correct answer is B.

20 答案 D

对每个正整数 \(k\)，取 \(x=k^2+1\)，且 \(y=k(k^2+1)=kx\)。则 \(y^2=k^2x^2=x^2(x-1)\)，所以 \(x^2+y^2=x^3\)。这样就得到无穷多个正整数对。

因此，正确答案为 D。

English

For every positive integer \(k,\) take \(x=k^2+1\) and \(y=k(k^2+1)=kx.\) Then \(y^2=k^2x^2=x^2(x-1),\) so \(x^2+y^2=x^3.\) This supplies infinitely many positive-integer pairs.

Therefore, the correct answer is D.

21 答案 E

取 \(C=(0,0)\)、\(B=(0,s)\)，且 \(A=(t,0)\)。则 \(M=(\frac{t}{2},\frac{s}{2})\)，且 \(N=(\frac{t}{2},0)\)。由垂直关系得 \((\frac{t}{2},\frac{s}{2})\cdot(\frac{t}{2},-s)=0\)，从而 \(t^2=2s^2\)。因此 \[ \begin{aligned} BN&=\sqrt{(\frac{t}{2})^2+s^2}\\ &=\sqrt{\frac{s^2}{2}+s^2}\\ &=\frac{s\sqrt6}{2} \end{aligned}\text{。}\]

因此，正确答案为 E。

English

Set \(C=(0,0),\) \(B=(0,s),\) and \(A=(t,0).\) Then \(M=(\frac{t}{2},\frac{s}{2})\) and \(N=(\frac{t}{2},0).\) Perpendicularity gives \((\frac{t}{2},\frac{s}{2})\cdot(\frac{t}{2},-s)=0,\) hence \(t^2=2s^2.\) Therefore \[ \begin{aligned} BN&=\sqrt{(\frac{t}{2})^2+s^2}\\ &=\sqrt{\frac{s^2}{2}+s^2}\\ &=\frac{s\sqrt6}{2}. \end{aligned} \]

Therefore, the correct answer is E.

22 答案 E

点 \(P\) 到两面墙的水平距离分别为 \(a\cos75^\circ\) 和 \(a\cos45^\circ\)，所以 \(w=a(\cos75^\circ+\cos45^\circ)\)。由和差化积公式，\(\cos75^\circ+\cos45^\circ=\sin75^\circ\)。又因为 \(h=a\sin75^\circ\)，所以 \(w=h\)。

因此，正确答案为 E。

English

The horizontal distances from \(P\) to the walls are \(a\cos75^\circ\) and \(a\cos45^\circ,\) so \(w=a(\cos75^\circ+\cos45^\circ).\) The sum-to-product identity gives \(\cos75^\circ+\cos45^\circ=\sin75^\circ.\) Since \(h=a\sin75^\circ,\) \(w=h.\)

Therefore, the correct answer is E.

23 答案 A

设连续的三边长为 \(n\)、\(n+1\)、\(n+2\)，其对角分别为 \(\theta\)、\(\phi\)、\(2\theta\)。由正弦定理，\[ \frac{n+2}{n}=\frac{\sin2\theta}{\sin\theta}=2\cos\theta\text{。}\]对最小角使用余弦定理还可得 \[ \cos\theta=\frac{n^2+6n+5}{2(n+1)(n+2)}\text{。}\]令两式相等，得到 \(n^2-3n-4=0\)，所以 \(n=4\)。因此三边长为 \(4\)、\(5\)、\(6\)，且 \(\cos\theta=\frac{6}{2\cdot4}=\frac{3}{4}\)。

因此，正确答案为 A。

English

Let the consecutive sides be \(n,\) \(n+1,\) \(n+2,\) opposite angles \(\theta,\) \(\phi,\) \(2\theta.\) By the law of sines, \[ \frac{n+2}{n}=\frac{\sin2\theta}{\sin\theta}=2\cos\theta. \] The law of cosines at the smallest angle also gives \[ \cos\theta=\frac{n^2+6n+5}{2(n+1)(n+2)}. \] Equating these expressions yields \(n^2-3n-4=0,\) so \(n=4.\) Hence the sides are \(4,\) \(5,\) \(6\) and \(\cos\theta=\frac{6}{2\cdot4}=\frac{3}{4}.\)

Therefore, the correct answer is A.

24 答案 A

首先，由点 \(A\) 的幂可得 \(AH(AH+7)=AG\cdot AF\)。后一个乘积为 \(2\cdot15=30\)，所以 \(AH=3\)、\(BJ=6\)，且 \(BH=13\)。令 \(BD=u\)、\(DE=x\)、\(EC=v\)。对从 \(B\) 和 \(C\) 引出的割线使用点幂定理，得 \[ \begin{aligned} u(u+x)&=BH\cdot BJ\\ &=13\cdot6=78,\\ v(v+x)&=CF\cdot CG\\ &=1\cdot14=14 \end{aligned}\text{。}\]另外，\(u+x+v=16\)。将两个点幂方程相减，并利用该和式，得到 \(u-v=4\)。因而 \(u=10-\frac{x}{2}\)，且 \(v=6-\frac{x}{2}\)。代入 \(v(v+x)=14\) 得 \(x^2=88\)，所以 \(DE=2\sqrt{22}\)。

因此，正确答案为 A。

English

First, power of \(A\) gives \(AH(AH+7)=AG\cdot AF.\) The latter product is \(2\cdot15=30,\) so \(AH=3,\) \(BJ=6,\) and \(BH=13.\) Let \(BD=u,\) \(DE=x,\) \(EC=v.\) From the secants at \(B\) and \(C,\) \[ \begin{aligned} u(u+x)&=BH\cdot BJ\\ &=13\cdot6=78,\\ v(v+x)&=CF\cdot CG\\ &=1\cdot14=14. \end{aligned} \] Also \(u+x+v=16.\) Subtracting the power equations and using the sum gives \(u-v=4.\) Hence \(u=10-\frac{x}{2}\) and \(v=6-\frac{x}{2}.\) Substitution into \(v(v+x)=14\) gives \(x^2=88,\) so \(DE=2\sqrt{22}.\)

Therefore, the correct answer is A.

25 答案 D

若在第三次向东走之前已经向南走了 \(i\) 次，则到达 \(C\) 的最后一步是向东走，而此前的 \(2+i\) 步中有两步向东。因此 \[ \begin{aligned} \Pr(C) &=\sum_{i=0}^3\binom{2+i}{2}2^{-(3+i)}\\ &=\frac18+\frac3{16}+\frac6{32}\\ &\quad+\frac{10}{64}\\ &=\frac{21}{32} \end{aligned}\text{。}\]

因此，正确答案为 D。

English

If \(i\) south moves occur before the third east move, the final step to \(C\) is east and the preceding \(2+i\) steps contain two east moves. Thus \[ \begin{aligned} \Pr(C) &=\sum_{i=0}^3\binom{2+i}{2}2^{-(3+i)}\\ &=\frac18+\frac3{16}+\frac6{32}\\ &\quad+\frac{10}{64}\\ &=\frac{21}{32}. \end{aligned} \]

Therefore, the correct answer is D.

26 答案 B

最后两位八进制数字 \(3c_8\) 表示 \(24+c\pmod{64}\)。在 \(24,\ldots,31\) 范围内，模 \(64\) 的平方数只有 \(25\)。因此 \(24+c=25\)，所以 \(c=1\)。

因此，正确答案为 B。

English

The last two octal digits \(3c_8\) represent \(24+c\pmod{64}.\) Squares modulo \(64\) in the range \(24,\ldots,31\) include only \(25.\) Thus \(24+c=25,\) so \(c=1.\)

Therefore, the correct answer is B.

27 答案 C

对方程取共轭后，每个 \(i\) 都变为 \(-i\)，且 \(z\) 变为 \(\bar z\)。由于自变量取相反数时，奇次幂项也会改变符号，所以这个共轭方程恰好就是原多项式在 \(-\bar z\) 处取值为零。因此 \(-\bar z=-a+bi\) 必定是一个根。

因此，正确答案为 C。

English

Conjugating the equation changes each \(i\) to \(-i\) and \(z\) to \(\bar z.\) Because the odd-powered terms also change sign when the input is negated, this conjugated equation is precisely the original polynomial evaluated at \(-\bar z.\) Thus \(-\bar z=-a+bi\) must be a root.

Therefore, the correct answer is C.

28 答案 B

新的平均数 \(\frac{602}{17}\) 小于原平均数 \(\frac{n+1}{2}\)，但两者之差小于 \(\frac{n}{n-1}\)，因而 \(n=69\) 或 \(70\)。又因为剩余各数之和 \((n-1)(\frac{602}{17})\) 必须为整数，所以 \(n-1\) 是 \(17\) 的倍数，从而 \(n=69\)。擦去的数为 \[ \begin{aligned} 2415-68\cdot\frac{602}{17} &=2415-2408\\ &=7 \end{aligned}\text{。}\]

因此，正确答案为 B。

English

The new average \(\frac{602}{17}\) is below the original average \(\frac{n+1}{2}\) but differs from it by less than \(\frac{n}{n-1},\) forcing \(n=69\) or \(70.\) Since the remaining sum \((n-1)(\frac{602}{17})\) must be integral, \(n-1\) is divisible by \(17,\) so \(n=69.\) The erased number is \[ \begin{aligned} 2415-68\cdot\frac{602}{17} &=2415-2408\\ &=7. \end{aligned} \]

Therefore, the correct answer is B.

29 答案 A

将三者排序为 \(x\le y\le z\)。取得最小值时三者的差异最大，所以 \(z=2x\)，且 \(y=1-3x\)。该顺序要求 \(\frac{1}{5}\le x\le\frac{1}{4}\)。因而 \(xyz=2x^2(1-3x)\)。它唯一的区间内部临界点是最大值点，所以比较两个端点：相应的值分别为 \(\frac{4}{125}\) 和 \(\frac{1}{32}\)。最小值为 \(\frac{1}{32}\)。

因此，正确答案为 A。

English

Order \(x\le y\le z.\) At a minimum the spread is maximal, so \(z=2x\) and \(y=1-3x.\) The ordering requires \(\frac{1}{5}\le x\le\frac{1}{4}.\) Thus \(xyz=2x^2(1-3x).\) Its only interior critical point is a maximum, so compare endpoints: the values are \(\frac{4}{125}\) and \(\frac{1}{32},\) respectively. The minimum is \(\frac{1}{32}.\)

Therefore, the correct answer is A.

30 答案 D

令 \(\alpha=15+\sqrt{220}\)，且 \(\beta=15-\sqrt{220}\)，则 \(0\lt\beta\lt1\)。整数 \(S_n=\alpha^n+\beta^n\) 满足 \(S_n=30S_{n-1}-5S_{n-2}\)，因此对每个 \(n\ge1\)，都有 \(S_n\equiv0\pmod{10}\)。于是 \[ \begin{aligned} \alpha^{19}+\alpha^{82} &=S_{19}+S_{82}\\ &\quad-(\beta^{19}+\beta^{82}) \end{aligned} \] 等于 \(10\) 的某个倍数减去一个小于 \(1\) 的正数。因此它的整数部分以 \(9\) 为个位数字。

因此，正确答案为 D。

English

Let \(\alpha=15+\sqrt{220}\) and \(\beta=15-\sqrt{220},\) so \(0\lt\beta\lt1.\) The integers \(S_n=\alpha^n+\beta^n\) satisfy \(S_n=30S_{n-1}-5S_{n-2},\) hence \(S_n\equiv0\pmod{10}\) for every \(n\ge1.\) Therefore \[ \begin{aligned} \alpha^{19}+\alpha^{82} &=S_{19}+S_{82}\\ &\quad-(\beta^{19}+\beta^{82}) \end{aligned} \] is a multiple of \(10\) minus a positive number less than \(1.\) Its integer part therefore ends in \(9.\)

Therefore, the correct answer is D.