2000 AMC 10 真题（P版）

1 答案 E

分解质因数得 \(2001 = 3 \cdot 23 \cdot 29\)。

若一个因数是 \(1\)，另两个因数可能为 \((3,667)\)，\((23,87)\)，或 \((29,69)\)。它们与 \(1\) 的和分别为 \(671\)，\(111\)，和 \(99\)。（数对 \((1,2001)\) 会重复因数 \(1\)。）

若没有因数是 \(1\)，三个质因数必须分别分给三个整数，只能得到 \(3,23,29\)，其和小得多。因此最大可能的和是 \(1 + 3 + 667 = 671\)。

所以正确答案是 E。

English

Factoring gives \(2001 = 3 \cdot 23 \cdot 29.\)

If one factor is \(1,\) the possible pairs for the other two factors are \((3,667),\) \((23,87),\) and \((29,69).\) Their corresponding sums with \(1\) are \(671,\) \(111,\) and \(99.\) (The pair \((1,2001)\) would repeat the factor \(1.\))

If no factor is \(1,\) all three prime factors must be split among the three integers, giving only \(3,23,29\) and a much smaller sum. Therefore, the largest possible sum is \(1 + 3 + 667 = 671.\)

Thus, the correct answer is E.

2 答案 A

将因数 \(2000\) 写成 \(2000^1\)。则 \[ \begin{aligned} 2000^1 \cdot 2000^{2000} &= 2000^{1 + 2000} \\ &= 2000^{2001} \end{aligned}\text{。}\]

其他每个选项都大于 \(2000^{2001}\)。

所以正确答案是 A。

English

Write the factor \(2000\) as \(2000^1.\) Then \[ \begin{aligned} 2000^1 \cdot 2000^{2000} &= 2000^{1 + 2000} \\ &= 2000^{2001}. \end{aligned} \]

Each of the other options is larger than \(2000^{2001}.\)

Thus, the correct answer is A.

3 答案 B

Jenny 每天吃掉 \(20\%\)，所以每天结束时剩下 \(80\%\)。

设原来有 \(x\) 颗，则 \[(0.8)^2 x = 0.64x = 32\text{，}\]因此 \(x = 50\)。

所以正确答案是 B。

English

Since Jenny eats \(20\%\) each day, \(80\%\) remain at the end of each day.

If \(x\) is the original number, then \[(0.8)^2 x = 0.64x = 32,\] so \(x = 50.\)

Thus, the correct answer is B.

4 答案 D

一月只有连接时间翻倍，所以账单增加的 \(\$17.54 - \$12.48 = \$5.06\) 就是十二月的连接时长费用。

因此固定月费是 \(\$12.48 - \$5.06 = \$7.42\)。

所以正确答案是 D。

English

January doubled only the connect time, so the increase \(\$17.54 - \$12.48 = \$5.06\) equals December’s connect-time cost.

The fixed monthly fee is therefore \(\$12.48 - \$5.06 = \$7.42.\)

Thus, the correct answer is D.

5 答案 B

由于 \(MN\) 是中位线，\(MN = \tfrac12 AB\)，所以它的长度不变。

底边 \(AB\) 固定，从 \(P\) 到 \(AB\) 的高也固定；当 \(P\) 沿平行线移动时，\(\triangle PAB\) 的面积不变。梯形 \(ABNM\) 是大三角形减去 \(\triangle PMN\)，面积也不变。

只有周长会改变，因为 \(PA\) 和 \(PB\) 会随着 \(P\) 的移动而变化。因此恰好有一个量会改变。

所以正确答案是 B。

English

Since \(MN\) is a midsegment, \(MN = \tfrac12 AB,\) which is fixed.

The base \(AB\) and the height from \(P\) to \(AB\) are both constant as \(P\) slides along the parallel line, so the area of \(\triangle PAB\) does not change. The trapezoid \(ABNM\) is the triangle minus \(\triangle PMN,\) both of whose areas are constant, so its area does not change either.

Only the perimeter changes, since \(PA\) and \(PB\) vary as \(P\) moves. So exactly one quantity changes.

Thus, the correct answer is B.

6 答案 C

只记录个位数，得到 \[ \begin{gathered} 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, \\ 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, \ldots \end{gathered} \]

依次查看每个数字第一次出现的位置，数字 \(6\) 是十个数字中最后出现的。

所以正确答案是 C。

English

Recording only the units digits gives the sequence \[ \begin{gathered} 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, \\ 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, \ldots \end{gathered} \]

Scanning for the first appearance of each digit, the digit \(6\) is the last of the ten digits to show up.

Thus, the correct answer is C.

7 答案 B

直角 \(\angle ADC = 90^\circ\) 被三等分成三个 \(30^\circ\) 角，所以 \(\angle ADP = 30^\circ\)，\(\angle ADB = 60^\circ\)。

在直角三角形 \(ADP\) 中，由 \(AD = 1\) 得 \(DP = \dfrac{1}{\cos 30^\circ} = \dfrac{2\sqrt3}{3}\)，\(AP = \tan 30^\circ = \dfrac{\sqrt3}{3}\)。

在直角三角形 \(ADB\) 中，由 \(AD = 1\) 得 \(DB = \dfrac{1}{\cos 60^\circ} = 2\)，\(AB = \tan 60^\circ = \sqrt3\)。

于是 \[ \begin{aligned} PB = AB - AP &= \sqrt3 - \dfrac{\sqrt3}{3} \\ &= \dfrac{2\sqrt3}{3} \end{aligned}\text{。}\]

\(\triangle BDP\) 的周长为 \[ \begin{gathered} DP + PB + DB \\ = \dfrac{2\sqrt3}{3} + \dfrac{2\sqrt3}{3} + 2 \\ = 2 + \dfrac{4\sqrt3}{3} \end{gathered}\text{。}\]

所以正确答案是 B。

English

The right angle \(\angle ADC = 90^\circ\) is trisected into three \(30^\circ\) angles, so \(\angle ADP = 30^\circ\) and \(\angle ADB = 60^\circ.\)

In right triangle \(ADP,\) with \(AD = 1,\) we get \(DP = \dfrac{1}{\cos 30^\circ} = \dfrac{2\sqrt3}{3}\) and \(AP = \tan 30^\circ = \dfrac{\sqrt3}{3}.\)

In right triangle \(ADB,\) with \(AD = 1,\) we get \(DB = \dfrac{1}{\cos 60^\circ} = 2\) and \(AB = \tan 60^\circ = \sqrt3.\)

Then \[ \begin{aligned} PB = AB - AP &= \sqrt3 - \dfrac{\sqrt3}{3} \\ &= \dfrac{2\sqrt3}{3}. \end{aligned} \]

The perimeter of \(\triangle BDP\) is \[ \begin{gathered} DP + PB + DB \\ = \dfrac{2\sqrt3}{3} + \dfrac{2\sqrt3}{3} + 2 \\ = 2 + \dfrac{4\sqrt3}{3}. \end{gathered} \]

Thus, the correct answer is B.

8 答案 D

设高一和高二学生总人数分别为 \(f\) 和 \(s\)。参加考试的人数相同，所以 \[\tfrac25 f = \tfrac45 s\text{。}\]

两边乘以 \(5\)，得 \(2f = 4s\)，所以 \(f = 2s\)。高一学生人数是高二学生人数的两倍。

所以正确答案是 D。

English

Let \(f\) and \(s\) be the numbers of freshmen and sophomores. The contestant counts are equal, so \[\tfrac25 f = \tfrac45 s.\]

Multiplying by \(5\) gives \(2f = 4s,\) so \(f = 2s.\) There are twice as many freshmen as sophomores.

Thus, the correct answer is D.

9 答案 C

因为 \(x \lt 2\)，所以 \(|x - 2| = 2 - x = p\)，于是 \(x = 2 - p\)。

因此 \[x - p = (2 - p) - p = 2 - 2p\text{。}\]

所以正确答案是 C。

English

Because \(x \lt 2,\) we have \(|x - 2| = 2 - x = p,\) so \(x = 2 - p.\)

Then \[x - p = (2 - p) - p = 2 - 2p.\]

Thus, the correct answer is C.

10 答案 D

由三角形不等式，\(x\) 和 \(y\) 都可以取严格介于 \(6 - 4 = 2\) 与 \(6 + 4 = 10\) 之间的任意数。

因此 \(|x - y|\) 可以取所有满足 \(0 \le |x - y| \lt 8\) 的值。

最小不能取到的正数是 \(10 - 2 = 8\)。

所以正确答案是 D。

English

By the triangle inequality, each of \(x\) and \(y\) can be any number strictly between \(6 - 4 = 2\) and \(6 + 4 = 10.\)

Then \(|x - y|\) can take any value with \(0 \le |x - y| \lt 8.\)

The smallest positive number not attainable is \(10 - 2 = 8.\)

Thus, the correct answer is D.

11 答案 C

在 \(4\) 和 \(18\) 之间的质数是 \(5\)、\(7\)、\(11\)、\(13\) 和 \(17\)。两个奇数的乘积是奇数、和是偶数，所以 \(xy - (x + y)\) 是奇数。

又因为 \(xy - (x + y)\) \(= (x - 1)(y - 1) - 1\) 会随着任一质数增大而增大，结果从 \(5 \cdot 7 - 12 = 23\) 到 \(13 \cdot 17 - 30 = 191\)。

选项中唯一在 \([23, 191]\) 内的奇数是 \(119 = 11 \cdot 13 - (11 + 13)\)。

所以正确答案是 C。

English

The primes between \(4\) and \(18\) are \(5,\) \(7,\) \(11,\) \(13,\) and \(17.\) The product of two of them is odd and the sum is even, so \(xy - (x + y)\) is odd.

Since \(xy - (x + y)\) \(= (x - 1)(y - 1) - 1\) increases as either prime increases, the result ranges from \(5 \cdot 7 - 12 = 23\) up to \(13 \cdot 17 - 30 = 191.\)

The only odd option in \([23, 191]\) is \(119 = 11 \cdot 13 - (11 + 13).\)

Thus, the correct answer is C.

12 答案 C

图形 \(n\) 可以看作前 \(n\) 个奇数之和加上前 \(n+1\) 个奇数之和，所以单位正方形个数为 \(n^2 + (n+1)^2\)。

对图形 \(100\) 有 \[ \begin{aligned} 100^2 + 101^2 &= 10000 + 10201 \\ &= 20201 \end{aligned}\text{。}\]

所以正确答案是 C。

English

Figure \(n\) can be split into the sum of the first \(n\) odd numbers and the first \(n+1\) odd numbers, giving \(n^2 + (n+1)^2\) unit squares.

For figure \(100,\) this is \[ \begin{aligned} 100^2 + 101^2 &= 10000 + 10201 \\ &= 20201. \end{aligned} \]

Thus, the correct answer is C.

13 答案 B

钉板有五行和五列。为了避免同一行或同一列中出现两个黄色钉，每行和每列都必须恰有一个黄色钉，因此五个黄色钉的位置被确定在长对角线上。

接着，四个红色钉必须分别放在第 \(2\) 行至第 \(5\) 行中，剩余的位置也迫使它们排在一条对角线上。继续放置绿色、蓝色和橙色钉时，每种颜色的位置也都被唯一确定。

因此恰好有一种符合条件的放法。

所以正确答案是 B。

English

The board has five rows and five columns. To avoid two yellow pegs in a row or column, there must be exactly one yellow peg in each row, forcing the yellow pegs onto the long diagonal.

The four red pegs must then each go in rows \(2\) through \(5,\) and the only positions left force them into a single diagonal as well. Continuing with green, blue, and orange, every color is forced into a unique position.

Hence there is exactly one valid arrangement.

Thus, the correct answer is B.

14 答案 C

\(71, 76, 80, 82, 91\) 除以 \(3\) 的余数分别是 \(2, 1, 2, 1, 1\)。前三个成绩的和必须能被 \(3\) 整除，唯一可行的三元组是 \(76 + 82 + 91 = 249\)。因此第三个输入的成绩是 \(91\)，前两个是 \(76\) 和 \(82\)。

因为 \(249\) 比 \(4\) 的倍数多一，第四个成绩必须比 \(4\) 的倍数多三，只有 \(71\) 符合；剩下 \(80\) 作为第五个成绩。

确实，部分和 \(76, 158, 249, 320, 400\) 分别能被 \(1, 2, 3, 4, 5\) 整除。

所以正确答案是 C。

English

The residues of \(71, 76, 80, 82, 91\) modulo \(3\) are \(2, 1, 2, 1, 1.\) The sum of the first three scores must be divisible by \(3,\) and the only such triple is \(76 + 82 + 91 = 249,\) so the third score entered is \(91\) and the first two are \(76\) and \(82.\)

Since \(249\) is one more than a multiple of \(4,\) the fourth score must be three more than a multiple of \(4,\) which only \(71\) satisfies. That leaves \(80\) as the fifth and last score.

Indeed \(76, 158, 249, 320, 400\) are divisible by \(1, 2, 3, 4, 5.\)

Thus, the correct answer is C.

15 答案 E

通分到分母 \(ab\) 得 \[\dfrac{a}{b} + \dfrac{b}{a} - ab = \dfrac{a^2 + b^2 - (ab)^2}{ab}\text{。}\]

代入 \(ab = a - b\)，得到 \[\dfrac{a^2 + b^2 - (a - b)^2}{ab} = \dfrac{2ab}{ab} = 2\text{。}\]

所以正确答案是 E。

English

Over the common denominator \(ab,\) \[\dfrac{a}{b} + \dfrac{b}{a} - ab = \dfrac{a^2 + b^2 - (ab)^2}{ab}.\]

Substituting \(ab = a - b\) gives \[\dfrac{a^2 + b^2 - (a - b)^2}{ab} = \dfrac{2ab}{ab} = 2.\]

Thus, the correct answer is E.

16 答案 B

取坐标 \(A = (0, 3)\)、\(B = (6, 0)\)、\(C = (4, 2)\)、\(D = (2, 0)\)。

直线 \(AB\) 是 \(x + 2y = 6\)，直线 \(CD\) 是 \(x - y = 2\)。联立解得 \(E = \left(\dfrac{10}{3}, \dfrac{4}{3}\right)\)。

因此 \[ \begin{aligned} AE &= \sqrt{\left(\dfrac{10}{3}\right)^2 + \left(\dfrac{4}{3} - 3\right)^2} \\ &= \sqrt{\dfrac{100}{9} + \dfrac{25}{9}} \\ &= \dfrac{5\sqrt5}{3} \end{aligned}\text{。}\]

所以正确答案是 B。

English

Place the points at \(A = (0, 3),\) \(B = (6, 0),\) \(C = (4, 2),\) \(D = (2, 0).\)

Line \(AB\) is \(x + 2y = 6\) and line \(CD\) is \(x - y = 2.\) Solving simultaneously gives \(E = \left(\dfrac{10}{3}, \dfrac{4}{3}\right).\)

Then \[ \begin{aligned} AE &= \sqrt{\left(\dfrac{10}{3}\right)^2 + \left(\dfrac{4}{3} - 3\right)^2} \\ &= \sqrt{\dfrac{100}{9} + \dfrac{25}{9}} \\ &= \dfrac{5\sqrt5}{3}. \end{aligned} \]

Thus, the correct answer is B.

17 答案 D

二十五分硬币换五分硬币，或五分硬币换一分硬币，都保持总金额不变。只有一分硬币换五枚二十五分硬币时，金额增加 \(5 \cdot 25 - 1 = 124\) 美分。

从 \(1\) 美分开始，Boris 的总金额总是 \(1 + 124n\) 美分，其中 \(n\) 是非负整数。

只有 \(\$7.45\) 符合，因为 \(745 = 1 + 124 \cdot 6\)。

所以正确答案是 D。

English

Trading a quarter for five nickels or a nickel for five pennies does not change the total value. Only trading a penny for five quarters changes it, adding \(5 \cdot 25 - 1 = 124\) cents.

Starting from \(1\) cent, Boris always has \(1 + 124n\) cents for some nonnegative integer \(n.\)

Only \(\$7.45\) has this form, since \(745 = 1 + 124 \cdot 6.\)

Thus, the correct answer is D.

18 答案 C

在正方形内部，她能看到除去中央边长为 \(5 - 2 = 3\) 的正方形以外的所有区域，面积为 \(25 - 9 = 16\) 平方千米。

在正方形外部，可见区域是四个 \(5 \times 1\) 的长方形加四个半径为 \(1\) 的四分之一圆，面积为 \(4 \cdot 5 + \pi = 20 + \pi\) 平方千米。

总面积为 \(36 + \pi \approx 39\) 平方千米。

所以正确答案是 C。

English

Inside the square, Charlyn sees everything except a central square of side \(5 - 2 = 3,\) an area of \(25 - 9 = 16\) square kilometers.

Outside the square, the region is four rectangles each \(5 \times 1\) plus four quarter circles of radius \(1,\) an area of \(4 \cdot 5 + \pi = 20 + \pi\) square kilometers.

The total area is \(36 + \pi \approx 39\) square kilometers.

Thus, the correct answer is C.

19 答案 D

设正方形边长为 \(1\)。一个小三角形的直角边为 \(1\) 和 \(r\)，面积为 \(\tfrac12 r = m\)，所以 \(r = 2m\)。

两个小三角形相似，因此另一个小三角形的直角边为 \(1\) 和 \(\tfrac1r\)，面积为 \(\tfrac12 \cdot \tfrac1r = \tfrac{1}{4m}\)。

由于正方形面积为 \(1\)，所求比值是 \(\dfrac{1}{4m}\)。

所以正确答案是 D。

English

Let the square have side \(1.\) One small triangle has legs \(1\) and \(r,\) with area \(\tfrac12 r = m,\) so \(r = 2m.\)

The two small triangles are similar, so the other has legs \(1\) and \(\tfrac1r,\) with area \(\tfrac12 \cdot \tfrac1r = \tfrac{1}{4m}.\)

Since the square has area \(1,\) the desired ratio is \(\dfrac{1}{4m}.\)

Thus, the correct answer is D.

20 答案 C

注意 \[ \begin{gathered} A \cdot M \cdot C + AM + MC + CA \\ = (A+1)(M+1)(C+1) \\ {}- (A + M + C) - 1 \\ = (A+1)(M+1)(C+1) - 11 \end{gathered}\text{。}\]

现在要最大化三个正整数的乘积，它们的和为 \(13\)。最平均的分配是 \(4, 4, 5\)，乘积为 \(4 \cdot 4 \cdot 5 = 80\)。

因此原式的最大值为 \(80 - 11 = 69\)。

所以正确答案是 C。

English

Notice that \[ \begin{gathered} A \cdot M \cdot C + AM + MC + CA \\ = (A+1)(M+1)(C+1) \\ {}- (A + M + C) - 1 \\ = (A+1)(M+1)(C+1) - 11. \end{gathered} \]

We maximize a product of three positive integers summing to \(13.\) The most balanced split is \(4, 4, 5,\) giving \(4 \cdot 4 \cdot 5 = 80.\)

The maximum is \(80 - 11 = 69.\)

Thus, the correct answer is C.

21 答案 B

有些爬行生物是短吻鳄，而所有短吻鳄都是凶猛生物，所以这些对象既是爬行生物，也是凶猛生物。因此有些凶猛生物是爬行生物，即 II 一定为真。

陈述 I 不一定成立，因为并非每只短吻鳄都必须是爬行生物；陈述 III 也不一定成立，因为所有短吻鳄都有可能都是爬行生物。因此只有 II 一定成立。

所以正确答案是 B。

English

Some creepy crawlers are alligators, and all alligators are ferocious, so those creatures are both creepy crawlers and ferocious. Hence some ferocious creatures are creepy crawlers, making II true.

Statement I fails because not every alligator need be a creepy crawler, and III fails because it is possible that all alligators are creepy crawlers. Only II must hold.

Thus, the correct answer is B.

22 答案 C

设共有 \(n\) 人，总共喝了 \(8n\) 盎司，其中牛奶总量为 \(M\)，咖啡总量为 \(C\)。Angela 只喝了其中一杯，所以 \[\tfrac14 M + \tfrac16 C = \tfrac1n (M + C)\text{。}\]

Angela 的一杯由总牛奶的 \(\tfrac14\) 与总咖啡的 \(\tfrac16\) 组成，所以 \(\tfrac1n\) 严格介于 \(\tfrac16\) 和 \(\tfrac14\) 之间。这迫使 \(4 \lt n \lt 6\)，即 \(n = 5\)。

所以正确答案是 C。

English

Let there be \(n\) people, drinking \(8n\) ounces total, split into milk \(M\) and coffee \(C.\) Angela drank one cup, so \[\tfrac14 M + \tfrac16 C = \tfrac1n (M + C).\]

The left side is a weighted average of \(\tfrac14\) and \(\tfrac16,\) so \(\tfrac1n\) lies strictly between \(\tfrac16\) and \(\tfrac14.\) That forces \(4 \lt n \lt 6,\) so \(n = 5.\)

Thus, the correct answer is C.

23 答案 E

众数总是 \(2\)，平均数是 \(\mu = \dfrac{25+x}{7}\)。

若 \(x \leq 2\)，中位数也是 \(2\)。两个相等的数不可能属于一个非常数三项等差数列，所以这种情况没有解。

若 \(2 < x < 4\)，三个数按 \(2 < x < \mu\) 排列。它们恰好在 \(x-2 = \mu-x\) 时构成等差数列。代入 \(\mu\) 得 \[ x-2 = \dfrac{25+x}{7}-x\text{，}\]解得 \(x=3\)。

若 \(x \geq 4\)，中位数是 \(4\)，并且 \(\mu>4\)。因此条件是 \[ 4-2 = \mu-4\text{，}\]所以 \(\mu=6\)，且 \(x=17\)。

两个可能值是 \(3\) 和 \(17\)，其和为 \(20\)。

所以正确答案是 E。

English

The mode is always \(2,\) and the mean is \(\mu = \dfrac{25+x}{7}.\)

If \(x \leq 2,\) the median is also \(2.\) Two equal values cannot belong to a non-constant three-term arithmetic progression, so this case gives no solutions.

If \(2 < x < 4,\) the values occur in the order \(2 < x < \mu.\) They form an arithmetic progression exactly when \(x-2 = \mu-x.\) Substituting for \(\mu\) gives \[ x-2 = \dfrac{25+x}{7}-x, \] whose solution is \(x=3.\)

If \(x \geq 4,\) the median is \(4\) and \(\mu>4.\) Thus the condition is \[ 4-2 = \mu-4, \] so \(\mu=6\) and \(x=17.\)

The two possible values are therefore \(3\) and \(17,\) whose sum is \(20.\)

Thus, the correct answer is E.

24 答案 B

为了计算 \(f(3z)\)，令 \(\dfrac{x}{3} = 3z\)，所以 \(x = 9z\)。于是 \[ \begin{aligned} f(3z) &= (9z)^2 + 9z + 1 \\ &= 81z^2 + 9z + 1 \end{aligned}\text{。}\]

令它等于 \(7\)，得到 \(81z^2 + 9z - 6 = 0\)。

由根和公式，所有 \(z\) 的和为 \(-\dfrac{9}{81} = -\dfrac19\)。

所以正确答案是 B。

English

To evaluate \(f(3z),\) set \(\dfrac{x}{3} = 3z,\) so \(x = 9z.\) Then \[ \begin{aligned} f(3z) &= (9z)^2 + 9z + 1 \\ &= 81z^2 + 9z + 1. \end{aligned} \]

Setting this equal to \(7\) gives \(81z^2 + 9z - 6 = 0.\)

By the sum-of-roots formula, the sum of the values of \(z\) is \(-\dfrac{9}{81} = -\dfrac19.\)

Thus, the correct answer is B.

25 答案 A

从第 \(300\) 天（\(N\) 年）到第 \(200\) 天（\(N + 1\) 年）相隔 \((L - 300) + 200\) 天，其中 \(L\) 是 \(N\) 年的天数。若 \(N\) 不是闰年，相隔 \(265 \equiv 6 \pmod 7\) 天，对应星期一而不是星期二。所以 \(N\) 是闰年，相隔 \(266 = 7 \cdot 38\) 天，正好仍是星期二。

所以 \(N - 1\) 年和 \(N + 1\) 年都不是闰年。

第 \(100\) 天（\(N - 1\) 年）比星期二（第 \(300\) 天，\(N\) 年）早 \((365 - 100) + 300 = 565\) 天。由于 \(565 = 7 \cdot 80 + 5\)，它比星期二早 \(5\) 天，也就是星期四。

所以正确答案是 A。

English

From day \(300\) of year \(N\) to day \(200\) of year \(N + 1\) is \((L - 300) + 200\) days, where \(L\) is the length of year \(N.\) If \(N\) were not a leap year, this is \(265 \equiv 6 \pmod 7,\) giving a Monday, not a Tuesday. So year \(N\) is a leap year, and the count is \(266 = 7 \cdot 38,\) consistent with Tuesday.

Then years \(N - 1\) and \(N + 1\) are not leap years.

The \(100\)th day of year \(N - 1\) precedes the Tuesday (day \(300\) of year \(N\)) by \((365 - 100) + 300 = 565\) days. Since \(565 = 7 \cdot 80 + 5,\) that day is \(5\) days earlier in the week than Tuesday, which is a Thursday.

Thus, the correct answer is A.