1980 AMC 12 真题（P版）

1 答案 C

这个数必须小于 \(\frac{100}{7}=14+\frac{2}{7}\)。小于该数的最大非负整数是 \(14\)。

因此，正确答案是 C。

English

The number must be less than \(\frac{100}{7}=14+\frac{2}{7}.\) The largest whole number below this value is \(14.\)

Therefore, the correct answer is C.

2 答案 D

第一个因式的次数为 \(2\cdot4=8\)，第二个因式的次数为 \(3\cdot3=9\)。因此乘积的次数为 \(8+9=17\)。

因此，正确答案是 D。

English

The first factor has degree \(2\cdot4=8,\) and the second has degree \(3\cdot3=9.\) Their product therefore has degree \(8+9=17.\)

Therefore, the correct answer is D.

3 答案 E

由题设可得 \[ \frac{2x-y}{x+y}=\frac23\text{。}\]因此 \(6x-3y=2x+2y\)，所以 \(4x=5y\)，从而 \(\frac{x}{y}=\frac{5}{4}\)。

因此，正确答案是 E。

English

The condition gives \[ \frac{2x-y}{x+y}=\frac23. \] Thus \(6x-3y=2x+2y,\) so \(4x=5y\) and \(\frac{x}{y}=\frac{5}{4}.\)

Therefore, the correct answer is E.

4 答案 C

在点 \(D\) 处，两个正方形的角各为 \(90^\circ\)，等边三角形的角为 \(60^\circ\)。因此 \[ \begin{aligned} \angle GDA &=360^\circ-90^\circ\\ &\quad-60^\circ-90^\circ\\ &=120^\circ \end{aligned}\text{。}\]

因此，正确答案是 C。

English

At \(D,\) the two square angles are \(90^\circ\) each and the equilateral-triangle angle is \(60^\circ.\) Hence \[ \begin{aligned} \angle GDA &=360^\circ-90^\circ\\ &\quad-60^\circ-90^\circ\\ &=120^\circ. \end{aligned} \]

Therefore, the correct answer is C.

5 答案 B

三角形 \(PQC\) 在 \(Q\) 处为直角，且 \(\angle P=60^\circ\)，所以它是一个 \(30^\circ\)-\(60^\circ\)-\(90^\circ\) 三角形。由于 \(QC=AQ\)，\[ \frac{PQ}{AQ}=\frac{PQ}{QC}=\frac1{\sqrt3} =\frac{\sqrt3}{3}\text{。}\]

因此，正确答案是 B。

English

Triangle \(PQC\) is right at \(Q\) and has \(\angle P=60^\circ,\) so it is a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle. Since \(QC=AQ,\) \[ \frac{PQ}{AQ}=\frac{PQ}{QC}=\frac1{\sqrt3} =\frac{\sqrt3}{3}. \]

Therefore, the correct answer is B.

6 答案 A

因为 \(x\gt0\)，两边除以 \(\sqrt x\) 不改变不等号方向：\[ 1\lt2\sqrt x\text{。}\]因此 \(\sqrt x\gt\frac{1}{2}\)，等价于 \(x\gt\frac{1}{4}\)。

因此，正确答案是 A。

English

Because \(x\gt0,\) division by \(\sqrt x\) preserves the inequality: \[ 1\lt2\sqrt x. \] Thus \(\sqrt x\gt\frac{1}{2},\) which is equivalent to \(x\gt\frac{1}{4}.\)

Therefore, the correct answer is A.

7 答案 B

三角形 \(ABC\) 是一个 \(3\)-\(4\)-\(5\) 直角三角形，所以 \(AC=5\)，其面积为 \(6\)。因为 \(5^2+12^2=13^2\)，三角形 \(ACD\) 在 \(C\) 处为直角，面积为 \((\frac{1}{2})(5)(12)=30\)。四边形的面积为 \(6+30=36\)。

因此，正确答案是 B。

English

Triangle \(ABC\) is a \(3\)-\(4\)-\(5\) right triangle, so \(AC=5\) and its area is \(6.\) Since \(5^2+12^2=13^2,\) triangle \(ACD\) is right at \(C\) and has area \((\frac{1}{2})(5)(12)=30.\) The quadrilateral’s area is \(6+30=36.\)

Therefore, the correct answer is B.

8 答案 A

消去分母可得 \[ (a+b)^2=ab\text{，}\]即 \(a^2+ab+b^2=0\)。因为 \(b\ne0\)，令 \(t=\frac{a}{b}\)。于是 \(t^2+t+1=0\)，其判别式为 \(1-4=-3\)。因此没有实数对满足条件。

因此，正确答案是 A。

English

Clearing denominators gives \[ (a+b)^2=ab, \] or \(a^2+ab+b^2=0.\) Since \(b\ne0,\) let \(t=\frac{a}{b}.\) Then \(t^2+t+1=0,\) whose discriminant is \(1-4=-3.\) Thus there are no real pairs.

Therefore, the correct answer is A.

9 答案 E

取向东为水平方向的正方向。转弯后，\(3\) 英里的位移分量为 \((\frac{3\sqrt3}{2},-\frac{3}{2})\)。因此 \[ \left(\frac{3\sqrt3}{2}-x\right)^2 +\left(\frac32\right)^2=3\text{。}\]所以 \(\frac{3\sqrt3}{2}-x=\pm\frac{\sqrt3}{2}\)，得到 \(x=\sqrt3\) 或 \(x=2\sqrt3\)。因此该值不能唯一确定。

因此，正确答案是 E。

English

Take east as the positive horizontal direction. After the turn, the \(3\)-mile displacement has components \((\frac{3\sqrt3}{2},-\frac{3}{2}).\) Therefore \[ \left(\frac{3\sqrt3}{2}-x\right)^2 +\left(\frac32\right)^2=3. \] Hence \(\frac{3\sqrt3}{2}-x=\pm\frac{\sqrt3}{2},\) giving \(x=\sqrt3\) or \(x=2\sqrt3.\) The value is not unique.

Therefore, the correct answer is E.

10 答案 D

各齿轮的圆周长分别与 \(x\)、\(y\)、\(z\) 成正比，而它们的切向速度大小相等。因此它们的角速度之比为 \[ \frac1x:\frac1y:\frac1z\text{。}\]三项同时乘以 \(xyz\)，得到 \(yz:xz:xy\)。

因此，正确答案是 D。

English

The gears’ circumferences are proportional to \(x,\) \(y,\) and \(z,\) while their tangential speeds have equal magnitude. Their angular speeds are therefore proportional to \[ \frac1x:\frac1y:\frac1z. \] Multiplying all three terms by \(xyz\) gives \(yz:xz:xy.\)

Therefore, the correct answer is D.

11 答案 D

由两个已知的和可得 \[ \begin{aligned} 2a+9d&=20,\\ 2a+99d&=\frac15 \end{aligned}\text{。}\]两式相减得到 \(90d=-\frac{99}{5}\)，所以 \(d=-\frac{11}{50}\)，且 \(2a=\frac{1099}{50}\)。因此 \[ \begin{aligned} S_{110}&=55(2a+109d)\\ &=55(-2)\\ &=-110 \end{aligned}\text{。}\]

因此，正确答案是 D。

English

The two given sums yield \[ \begin{aligned} 2a+9d&=20,\\ 2a+99d&=\frac15. \end{aligned} \] Subtracting gives \(90d=-\frac{99}{5},\) so \(d=-\frac{11}{50}\) and \(2a=\frac{1099}{50}.\) Therefore \[ \begin{aligned} S_{110}&=55(2a+109d)\\ &=55(-2)\\ &=-110. \end{aligned} \]

Therefore, the correct answer is D.

12 答案 C

令 \(\tan\theta=n\)。则 \(m=\tan(2\theta)\)，并且 \(m=4n\)。因此 \[ 4n=\frac{2n}{1-n^2}\text{。}\]由非水平条件可知 \(n\ne0\)，所以 \(2(1-n^2)=1\)，即 \(n^2=\frac{1}{2}\)。因此 \(mn=4n^2=2\)。

因此，正确答案是 C。

English

Let \(\tan\theta=n.\) Then \(m=\tan(2\theta)\) and also \(m=4n.\) Hence \[ 4n=\frac{2n}{1-n^2}. \] The nonhorizontal condition gives \(n\ne0,\) so \(2(1-n^2)=1,\) or \(n^2=\frac{1}{2}.\) Thus \(mn=4n^2=2.\)

Therefore, the correct answer is C.

13 答案 B

把位移向量表示为复数，依次为 \[ 1,\ \frac i2,\ \left(\frac i2\right)^2,\ldots\text{。}\]它们的和是 \[ \frac1{1-\frac{i}{2}} =\frac{1+\frac{i}{2}}{1+\frac{1}{4}} =\frac45+\frac25i\text{。}\]因此小虫趋近于 \((\frac{4}{5},\frac{2}{5})\)。

因此，正确答案是 B。

English

As complex numbers, the successive displacement vectors are \[ 1,\ \frac i2,\ \left(\frac i2\right)^2,\ldots. \] Their sum is \[ \frac1{1-\frac{i}{2}} =\frac{1+\frac{i}{2}}{1+\frac{1}{4}} =\frac45+\frac25i. \] Thus the bug approaches \((\frac{4}{5},\frac{2}{5}).\)

Therefore, the correct answer is B.

14 答案 A

直接复合可得 \[ f(f(x)) =\frac{c^2x}{(2c+6)x+9}\text{。}\]要使其恒等于 \(x\)，分母必须为常数 \(c^2\)。因此 \(2c+6=0\)，且 \(c^2=9\)，两者共同给出 \(c=-3\)。

因此，正确答案是 A。

English

Direct composition gives \[ f(f(x)) =\frac{c^2x}{(2c+6)x+9}. \] For this to equal \(x\) identically, the denominator must be the constant \(c^2.\) Thus \(2c+6=0\) and \(c^2=9,\) both of which give \(c=-3.\)

Therefore, the correct answer is A.

15 答案 B

设价格为 \(p\) 分，则 \[ \frac{104}{100}p=100n\text{，}\]所以 \(13p=1250n\)。由于 \(13\) 与 \(1250\) 互质，\(13\) 必须整除 \(n\)。最小的正整数取值是 \(n=13\)。

因此，正确答案是 B。

English

Let the price be \(p\) cents. Then \[ \frac{104}{100}p=100n, \] so \(13p=1250n.\) Since \(13\) and \(1250\) are relatively prime, \(13\) must divide \(n.\) The smallest positive possibility is \(n=13.\)

Therefore, the correct answer is B.

16 答案 B

设立方体棱长为 \(s\)。正四面体的每条棱都是立方体的面对角线，所以长度为 \(s\sqrt2\)。正四面体的表面积为 \[ 4\left(\frac{\sqrt3}{4}(s\sqrt2)^2\right) =2\sqrt3\,s^2\text{。}\]立方体的表面积为 \(6s^2\)，所以所求比为 \(\frac{6}{2\sqrt3}=\sqrt3\)。

因此，正确答案是 B。

English

Let the cube edge be \(s.\) Each tetrahedron edge is a cube face diagonal, so it has length \(s\sqrt2.\) The tetrahedron’s surface area is \[ 4\left(\frac{\sqrt3}{4}(s\sqrt2)^2\right) =2\sqrt3\,s^2. \] The cube’s surface area is \(6s^2,\) so the ratio is \(\frac{6}{2\sqrt3}=\sqrt3.\)

Therefore, the correct answer is B.

17 答案 D

展开得 \[ \begin{aligned} (n+i)^4 &=n^4-6n^2+1\\ &\quad+4n(n^2-1)i \end{aligned}\text{。}\]当且仅当 \(n=0\)、\(n=1\) 或 \(n=-1\) 时，虚部为零。对这三个整数，实部都是整数，所以共有 \(3\) 个取值。

因此，正确答案是 D。

English

Expansion gives \[ \begin{aligned} (n+i)^4 &=n^4-6n^2+1\\ &\quad+4n(n^2-1)i. \end{aligned} \] The imaginary part vanishes exactly when \(n=0,\) \(n=1,\) or \(n=-1.\) For each of these three integers the real part is an integer, so there are \(3\) values.

Therefore, the correct answer is D.

18 答案 D

由 \(\sin x=b^a\) 可得 \[ \cos x=\sqrt{1-b^{2a}}\text{。}\]因此 \[ \log_b\cos x =\frac12\log_b(1-b^{2a})\text{。}\]

因此，正确答案是 D。

English

We have \(\sin x=b^a,\) so \[ \cos x=\sqrt{1-b^{2a}}. \] Therefore \[ \log_b\cos x =\frac12\log_b(1-b^{2a}). \]

Therefore, the correct answer is D.

19 答案 D

设半径为 \(r\)，长为 \(20\) 的弦到圆心的距离为 \(u\)，相邻弦之间的共同间距为 \(v\)。则 \[ \begin{aligned} r^2&=u^2+10^2,\\ r^2&=(u+v)^2+8^2,\\ r^2&=(u+2v)^2+4^2 \end{aligned}\text{。}\]相邻两式依次相减，得到 \(2uv+v^2=36\) 和 \(2uv+3v^2=48\)，所以 \(v^2=6\)，且 \(u=\frac{15}{\sqrt6}\)。因此 \[ r^2=u^2+100=\frac{275}{2}\text{，}\]从而 \(r=\frac{5\sqrt{22}}{2}\)。

因此，正确答案是 D。

English

Let \(r\) be the radius, \(u\) the distance to the \(20\)-unit chord, and \(v\) the common spacing. Then \[ \begin{aligned} r^2&=u^2+10^2,\\ r^2&=(u+v)^2+8^2,\\ r^2&=(u+2v)^2+4^2. \end{aligned} \] Consecutive subtraction gives \(2uv+v^2=36\) and \(2uv+3v^2=48,\) so \(v^2=6\) and \(u=\frac{15}{\sqrt6}.\) Hence \[ r^2=u^2+100=\frac{275}{2}, \] and \(r=\frac{5\sqrt{22}}{2}.\)

Therefore, the correct answer is D.

20 答案 C

共有 \(\binom{12}{6}=924\) 种选法。总值至少为 \(50\) 美分的情形包括：六枚都是十美分硬币；五枚十美分硬币加任意一枚其他硬币；或四枚十美分硬币加两枚五美分硬币。有利选法数为 \[ \begin{aligned} &\binom66\\ &\quad+\binom65\binom61\\ &\quad+\binom64\binom42\\ &=1+36+90\\ &=127 \end{aligned}\text{。}\]因此概率为 \(\frac{127}{924}\)。

因此，正确答案是 C。

English

There are \(\binom{12}{6}=924\) selections. A value of at least \(50\) cents occurs with six dimes; with five dimes and any one other coin; or with four dimes and two nickels. The favorable count is \[ \begin{aligned} &\binom66\\ &\quad+\binom65\binom61\\ &\quad+\binom64\binom42\\ &=1+36+90\\ &=127. \end{aligned} \] Thus the probability is \(\frac{127}{924}.\)

Therefore, the correct answer is C.

21 答案 A

把 \(\triangle ABC\) 的总面积定为 \(1\)。因为 \(BD:DC=1:2\)，所以 \([ABD]=\frac{1}{3}\)，且 \([ACD]=\frac{2}{3}\)。中点和三等分条件给出 \(AF:FD=3:1\)。因此 \[ [BDF]=\frac14[ABD]=\frac1{12}\text{。}\]此外，\[ \begin{aligned} [AEF] &=\frac{AE}{AC}\frac{AF}{AD}[ACD]\\ &=\frac12\cdot\frac34\cdot\frac23\\ &=\frac14 \end{aligned}\text{。}\]所以 \([FDCE]=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}\)，所求比为 \(\frac{\frac{1}{12}}{\frac{5}{12}}=\frac{1}{5}\)。

因此，正确答案是 A。

English

Scale the total area of \(\triangle ABC\) to \(1.\) Since \(BD:DC=1:2,\) we have \([ABD]=\frac{1}{3}\) and \([ACD]=\frac{2}{3}.\) The midpoint and trisection conditions give \(AF:FD=3:1.\) Hence \[ [BDF]=\frac14[ABD]=\frac1{12}. \] Also \[ \begin{aligned} [AEF] &=\frac{AE}{AC}\frac{AF}{AD}[ACD]\\ &=\frac12\cdot\frac34\cdot\frac23\\ &=\frac14. \end{aligned} \] Therefore \([FDCE]=\frac{2}{3}-\frac{1}{4}=\frac{5}{12},\) and the requested ratio is \(\frac{\frac{1}{12}}{\frac{5}{12}}=\frac{1}{5}.\)

Therefore, the correct answer is A.

22 答案 E

下包络线先取 \(4x+1\)，直到它与 \(x+2\) 于 \(x=\frac{1}{3}\) 相交；随后取 \(x+2\)，直到它与 \(-2x+4\) 于 \(x=\frac{2}{3}\) 相交；此后则取 \(-2x+4\)。因此最大值在 \(x=\frac{2}{3}\) 处取得，等于 \[ \frac23+2=\frac83\text{。}\]

因此，正确答案是 E。

English

The lower envelope is \(4x+1\) until it meets \(x+2\) at \(x=\frac{1}{3},\) then is \(x+2\) until that line meets \(-2x+4\) at \(x=\frac{2}{3},\) and thereafter is \(-2x+4.\) Thus its maximum occurs at \(x=\frac{2}{3}\) and equals \[ \frac23+2=\frac83. \]

Therefore, the correct answer is E.

23 答案 C

设斜边端点为 \((0,0)\) 和 \((h,0)\)，直角顶点为 \((u,v)\)。直角条件给出 \(u^2+v^2=hu\)。到 \((\frac{h}{3},0)\) 与 \((\frac{2h}{3},0)\) 的距离平方之和为 \[ \begin{aligned} &(u-\frac{h}{3})^2+v^2\\ &\quad+(u-\frac{2h}{3})^2+v^2 =\frac{5h^2}{9} \end{aligned}\text{。}\]另一方面，这个和也等于 \(\sin^2x+\cos^2x=1\)。因此 \(h=\frac{3}{\sqrt5}=\frac{3\sqrt5}{5}\)。

因此，正确答案是 C。

English

Let the hypotenuse have endpoints \((0,0)\) and \((h,0),\) and let the right-angle vertex be \((u,v).\) The right-angle condition gives \(u^2+v^2=hu.\) The squared distances to \((\frac{h}{3},0)\) and \((\frac{2h}{3},0)\) sum to \[ \begin{aligned} &(u-\frac{h}{3})^2+v^2\\ &\quad+(u-\frac{2h}{3})^2+v^2 =\frac{5h^2}{9}. \end{aligned} \] This also equals \(\sin^2x+\cos^2x=1.\) Hence \(h=\frac{3}{\sqrt5}=\frac{3\sqrt5}{5}.\)

Therefore, the correct answer is C.

24 答案 D

该多项式可分解为 \[ \begin{aligned} &8x^3-4x^2-42x+45\\ &\qquad=(2x-3)^2(2x+5) \end{aligned}\text{。}\]因此重根为 \(r=\frac{3}{2}=1.5\)，所列数中最接近它的是 \(1.52\)。

因此，正确答案是 D。

English

The polynomial factors as \[ \begin{aligned} &8x^3-4x^2-42x+45\\ &\qquad=(2x-3)^2(2x+5). \end{aligned} \] Thus the repeated root is \(r=\frac{3}{2}=1.5,\) and the nearest listed number is \(1.52.\)

Therefore, the correct answer is D.

25 答案 C

\(2m-1\) 最后一次出现的位置是 \[ 1+3+\cdots+(2m-1)=m^2\text{。}\]因此 \[ \begin{aligned} a_n&=2\left\lceil\sqrt n\right\rceil-1\\ &=2\left\lfloor\sqrt{n-1}\right\rfloor+1 \end{aligned}\text{。}\]所以 \(b=2\)、\(c=-1\)、\(d=1\)，从而 \(b+c+d=2\)。

因此，正确答案是 C。

English

The final occurrence of \(2m-1\) is in position \[ 1+3+\cdots+(2m-1)=m^2. \] Therefore \[ \begin{aligned} a_n&=2\left\lceil\sqrt n\right\rceil-1\\ &=2\left\lfloor\sqrt{n-1}\right\rfloor+1. \end{aligned} \] Thus \(b=2,\) \(c=-1,\) and \(d=1,\) so \(b+c+d=2.\)

Therefore, the correct answer is C.

26 答案 E

四个球心组成一个棱长为 \(2\) 的正四面体。它的内切球半径为 \[ d=\frac{2\sqrt6}{12}=\frac{\sqrt6}{6}\text{。}\]外切正四面体的每个面都与球心四面体的相应面平行，并且比该面到共同中心的距离多一个单位。由相似关系，\[ \frac{s}{2}=\frac{d+1}{d}\text{。}\]因此 \(s=2+\frac{2}{d}=2+2\sqrt6\)。

因此，正确答案是 E。

English

The ball centers form a regular tetrahedron of edge \(2.\) Its inradius is \[ d=\frac{2\sqrt6}{12}=\frac{\sqrt6}{6}. \] Each face of the circumscribed tetrahedron is parallel to the corresponding face of this center tetrahedron and lies one unit farther from the common center. By similarity, \[ \frac{s}{2}=\frac{d+1}{d}. \] Hence \(s=2+\frac{2}{d}=2+2\sqrt6.\)

Therefore, the correct answer is E.

27 答案 E

设两个实立方根分别为 \(u\) 和 \(v\)。则 \[ \begin{aligned} uv&=\sqrt[3]{25-52}=-3,\\ u^3+v^3&=10 \end{aligned}\text{。}\]若 \(t=u+v\)，则 \(t^3=10-9t\)，所以 \[ \begin{gathered} t^3+9t-10\\ =(t-1)(t^2+t+10)\\ =0 \end{gathered}\text{。}\]唯一的实数解是 \(t=1\)，但它不在选项中。

因此，正确答案是 E。

English

Let the two real cube roots be \(u\) and \(v.\) Then \[ \begin{aligned} uv&=\sqrt[3]{25-52}=-3,\\ u^3+v^3&=10. \end{aligned} \] If \(t=u+v,\) then \(t^3=10-9t,\) so \[ \begin{gathered} t^3+9t-10\\ =(t-1)(t^2+t+10)\\ =0. \end{gathered} \] The only real solution is \(t=1,\) which is not listed.

Therefore, the correct answer is E.

28 答案 C

令 \(\omega^2+\omega+1=0\)。由于 \(1+\omega=-\omega^2\)，把 \(\omega\) 代入多项式可得 \[ \omega^{2n}+1+\omega^{4n}\text{。}\]该式通常为 \(0\)；仅当 \(\omega^{2n}=1\) 时，它才等于 \(3\)。后一种情形恰好在 \(n\) 是 \(3\) 的倍数时发生。选项中只有 \(21\) 能被 \(3\) 整除，所以此时整除性不成立。

因此，正确答案是 C。

English

Let \(\omega^2+\omega+1=0.\) Since \(1+\omega=-\omega^2,\) the polynomial evaluated at \(\omega\) is \[ \omega^{2n}+1+\omega^{4n}. \] This is \(0\) unless \(\omega^{2n}=1,\) in which case it is \(3.\) The latter occurs exactly when \(n\) is divisible by \(3.\) Among the choices only \(21\) is divisible by \(3,\) so that is the value for which divisibility fails.

Therefore, the correct answer is C.

29 答案 A

三个方程相加得到 \[ \begin{aligned} x^2-2xy+2y^2 &{}+6yz+9z^2\\ &=175 \end{aligned}\text{，}\]即 \[ (x-y)^2+(y+3z)^2=175\text{。}\]一个平方同余于 \(0\) 或 \(1\pmod4\)，所以两个平方之和不可能同余于 \(3\pmod4\)。但 \(175\equiv3\pmod4\)，矛盾。因此不存在整数有序三元组。

因此，正确答案是 A。

English

Adding the three equations gives \[ \begin{aligned} x^2-2xy+2y^2 &{}+6yz+9z^2\\ &=175, \end{aligned} \] or \[ (x-y)^2+(y+3z)^2=175. \] A square is congruent to \(0\) or \(1\pmod4,\) so a sum of two squares cannot be congruent to \(3\pmod4.\) But \(175\equiv3\pmod4,\) a contradiction. Thus there are no integer triples.

Therefore, the correct answer is A.

30 答案 B

每个两位数块都必须属于 \[ \{16,25,36,49,64,81\}\text{。}\]先用前两位限制平方根，再只保留最后两位合格的平方数，可得下列可能的中间两位：

只有中间两位 \(64\) 合格，由此得到 \[ \begin{aligned} 166464&=408^2,\\ 646416&=804^2 \end{aligned}\text{。}\]因此共有 \(2\) 个准平方数。

因此，正确答案是 B。

English

Each two-digit block must belong to \[ \{16,25,36,49,64,81\}. \] Restricting the square root by the first block and retaining only squares with an allowed final block leaves the following possible middle blocks:

Only the middle block \(64\) is allowed, producing \[ \begin{aligned} 166464&=408^2,\\ 646416&=804^2. \end{aligned} \] Thus there are \(2\) squarish numbers.

Therefore, the correct answer is B.