2024 AMC amc12a 真题 答案 详解

---

中英双语真题

2024 AMC 12A Problems

Problem 1
What is the value of $9901 \cdot 101 - 99 \cdot 10101$?
$9901 \cdot 101 - 99 \cdot 10101$ 的值是多少？
(A) 2 (B) 20 (C) 200 (D) 202 (E) 2020

Problem 2
A model used to estimate the time it will take to hike to the top of the mountain on a trail is of the form $T = aL + bG$, where $a$ and $b$ are constants, $T$ is the time in minutes, $L$ is the length of the trail in miles, and $G$ is the altitude gain in feet. The model estimates that it will take 69 minutes to hike to the top if a trail is 1.5 miles long and ascends 800 feet, as well as if a trail is 1.2 miles long and ascends 1100 feet. How many minutes does the model estimate it will take to hike to the top if the trail is 4.2 miles long and ascends 4000 feet?
一个用于估算在一条小径上徒步到山顶所需时间的模型形式为 $T = aL + bG$，其中 $a$ 和 $b$ 是常数，$T$ 是分钟数，$L$ 是小径长度（英里），$G$ 是海拔上升（英尺）。该模型估计如果一条小径长 1.5 英里且上升 800 英尺，以及如果一条小径长 1.2 英里且上升 1100 英尺，则徒步到山顶需要 69 分钟。如果小径长 4.2 英里且上升 4000 英尺，该模型估计徒步到山顶需要多少分钟？
(A) 240 (B) 246 (C) 252 (D) 258 (E) 264

Problem 3
The number 2024 is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum?
该数字 2024 可以表示为若干个两位数的和，其中这些两位数不必互不相同。要表示这个和，至少需要多少个两位数？
(A) 20 (B) 21 (C) 22 (D) 23 (E) 24

Problem 4
What is the least value of $n$ such that $n!$ is a multiple of 2024?
求使得 $n!$ 是 2024 的倍数的最小的 $n$ 值是多少？
(A) 11 (B) 21 (C) 22 (D) 23 (E) 253

Problem 5
A data set containing 20 numbers, some of which are 6, has mean 45. When all the 6s are removed, the data set has mean 66. How many 6s were in the original data set?
一个包含 20 个数字的数据集，其中一些是 6，其平均值为 45。当移除所有 6 后，数据集的平均值为 66。原始数据集中有多少个 6？
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8

Problem 6
The product of three integers is 60. What is the least possible positive sum of the three integers?
三个整数的乘积为 60。这三个整数可能的最小正和是多少？
(A) 2 (B) 3 (C) 5 (D) 6 (E) 13

英文真题

2024 AMC 12A Problems

Problem 1
What is the value of $9901 \cdot 101 - 99 \cdot 10101$?
(A) 2 (B) 20 (C) 200 (D) 202 (E) 2020

Problem 2
A model used to estimate the time it will take to hike to the top of the mountain on a trail is of the form $T = aL + bG$, where $a$ and $b$ are constants, $T$ is the time in minutes, $L$ is the length of the trail in miles, and $G$ is the altitude gain in feet. The model estimates that it will take $69$ minutes to hike to the top if a trail is $1.5$ miles long and ascends $800$ feet, as well as if a trail is $1.2$ miles long and ascends $1100$ feet. How many minutes does the model estimate it will take to hike to the top if the trail is $4.2$ miles long and ascends $4000$ feet?
(A) 240 (B) 246 (C) 252 (D) 258 (E) 264

Problem 3
The number $2024$ is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum?
(A) 20 (B) 21 (C) 22 (D) 23 (E) 24

Problem 4
What is the least value of $n$ such that $n!$ is a multiple of $2024$?
(A) 11 (B) 21 (C) 22 (D) 23 (E) 253

Problem 5
A data set containing $20$ numbers, some of which are $6$, has mean $45$. When all the $6$s are removed, the data set has mean $66$. How many $6$s were in the original data set?
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8

Problem 6
The product of three integers is $60$. What is the least possible positive sum of the three integers?
(A) 2 (B) 3 (C) 5 (D) 6 (E) 13

Problem 7
In $\Delta ABC, \angle ABC=90^{\circ}$ and $BA=BC=\sqrt{2}$. Points $P_1, P_2, \ldots, P_{2024}$ lie on hypotenuse $\overline{AC}$ so that $AP_1=P_1P_2=P_2P_3=\cdots=P_{2023}P_{2024}=P_{2024}C$. What is the length of the vector sum
$\overrightarrow{BP_1}+\overrightarrow{BP_2}+\overrightarrow{BP_3}+\cdots+\overrightarrow{BP_{2024}}?$
(A) 1011 (B) 1012 (C) 2023 (D) 2024 (E) 2025

Problem 8
How many angles $\theta$ with $0 \leq \theta \leq 2\pi$ satisfy $\log(\sin(3\theta)) + \log(\cos(2\theta)) = 0$?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

Problem 9
Let $M$ be the greatest integer such that both $M+1213$ and $M+3773$ are perfect squares. What is the units digit of $M$?
(A) 1 (B) 2 (C) 3 (D) 6 (E) 8

真题文字详解(英文)

Problem 1 The following problem is from both the 2024 AMC 10A #1 and 2024 AMC 12A #1, so both problems redirect to this page. What is the value of $9901\cdot101-99\cdot10101$?
(A) 2 (B) 20 (C) 200 (D) 202 (E) 2020

Solution 1 (Direct Computation) The likely fastest method will be direct computation. $9901\cdot101$ evaluates to $1000001$ and $99\cdot10101$ evaluates to $999999$. The difference is $\boxed{\text{(A)}\ 2}$. Solution by juwushu.

Solution 2 (Distributive Property) We have
$$ \begin{aligned} &9901\cdot101-99\cdot10101\ =&(10000-99)\cdot101-99\cdot(10000+101)\ =&10000\cdot101-99\cdot101-99\cdot10000-99\cdot101\ =&(10000\cdot101-99\cdot10000)-2\cdot(99\cdot101)\ =&2\cdot10000-2\cdot9999\ =&\boxed{\text{(A)}\ 2}. \end{aligned} $$

~MRENTHUSIASM

Solution 3 (Solution 1 but Distributive) Note that $9901\cdot101=9901\cdot100+9901=990100+9901=1000001$ and $99\cdot10101=100\cdot10101-10101=1010100-10101=999999$, therefore the answer is $1000001-999999=\boxed{\text{(A)}\ 2}$.

Solution 4 (Modular Arithmetic) Evaluating the given expression $(\mod 10)$ yields $1-9\equiv2(\mod 10)$, so the answer is either $\text{(A)}$ or $\text{(D)}$. Evaluating $(\mod 101)$ yields $0-99\equiv2(\mod 101)$. Because answer $\text{(D)}$ is $202=2\cdot101$, that cannot be the answer, so we choose choice $\boxed{\text{(A)}\ 2}$.

Solution 5 (Process of Elimination) We simply look at the units digit of the problem we have (or take mod $10$):
$$ 9901\cdot101-99\cdot10101\equiv1\cdot1-9\cdot1\equiv2\ \text{mod}\ 10. $$
Since the only answers with $2$ in the units digit are $\text{(A)}$ and $\text{(D)}$, we can then continue if you are desperate to use guess and check or an actually valid method to find the answer is $\boxed{\text{(A)}\ 2}$.

Solution 6 (Faster Distribution) Observe that $9901=9900+1=99\cdot100+1$ and $10101=10100+1=101\cdot100+1$
$$ \begin{aligned} &\Rightarrow9901\cdot101-99\cdot10101=((9900\cdot101)+(1\cdot101))-((99\cdot10100)+(99\cdot1))\ =&(99\cdot100\cdot101)+101-(99\cdot100\cdot101)-99\ =&101-99\ =&\boxed{\text{(A)}\ 2}. \end{aligned} $$

Solution 7 (Cubes) Let $x=100$. Then, we have
Then, the answer can be rewritten as $(x^3+1)-(x^3-1)=\boxed{\text{(A)}\ 2}$.

Solution 8 (Super Fast) It's not hard to observe and express $9901$ into $99\cdot100+1$, and $10101$ into $101\cdot100+1$. We then simplify the original expression into $(99\cdot100+1)\cdot101-99\cdot(101\cdot100+1)$, which could then be simplified into $99\cdot100\cdot101+101-99\cdot100\cdot101-99$, which we can get the answer of $101-99=\boxed{\text{(A)}\ 2}$.

Problem 2 The following problem is from both the 2024 AMC 12A #2 and 2024 AMC 10A #2, so both problems redirect to this page.

真题文字详解(中英双语)

Problem 1 The following problem is from both the 2024 AMC 10A #1 and 2024 AMC 12A #1, so both problems redirect to this page. What is the value of $9901\cdot101-99\cdot10101$?
(A) 2 (B) 20 (C) 200 (D) 202 (E) 2020

Solution 1 (Direct Computation)
The likely fastest method will be direct computation. $9901\cdot101$ evaluates to $1000001$ and $99\cdot10101$ evaluates to $999999$. The difference is (A) 2.

Solution 2 (Distributive Property)
We have
$$9901\cdot101-99\cdot10101=(10000-99)\cdot101-99\cdot(10000+101)$$
$$=10000\cdot101-99\cdot101-99\cdot10000-99\cdot101$$
$$=(10000\cdot101-99\cdot10000)-2\cdot(99\cdot101)$$
$$=2\cdot10000-2\cdot9999$$
$$=\boxed{\textbf{(A)}\ 2}$$

Solution 3 (Solution 1 but Distributive)
Note that $9901\cdot101=9901\cdot100+9901=990100+9901=1000001$ and $99\cdot10101=100\cdot10101-10101=1010100-10101=999999$, therefore the answer is $1000001-999999=\boxed{\textbf{(A)}\ 2}$.

Solution 4 (Modular Arithmetic)
Evaluating the given expression $(mod\ 10)$ yields $1-9\equiv2\pmod{10}$, so the answer is either (A) or (D). Evaluating $(mod\ 101)$ yields $0-99\equiv2\pmod{101}$. Because answer (D) is $202=2\cdot101$, that cannot be the answer, so we choose choice $\boxed{\textbf{(A)}\ 2}$.

Solution 5 (Process of Elimination)
We simply look at the units digit of the problem we have (or take mod $10$):
$$9901\cdot101-99\cdot10101\equiv1\cdot1-9\cdot1\equiv2\pmod{10}.$$ Since the only answers with $2$ in the units digit are (A) and (D), we can then continue if you are desperate to use guess and check or an actually valid method to find the answer is (A) 2.

Solution 6 (Faster Distribution)
Observe that $9901=9900+1=99\cdot100+1$ and $10101=10100+1=101\cdot100+1$
$$\Rightarrow9901\cdot101-99\cdot10101=((9900\cdot101)+(1\cdot101))-((99\cdot10100)+(99\cdot1))$$
$$=(99\cdot100\cdot101)+101-(99\cdot100\cdot101)-99$$
$$=101-99$$
$$=\boxed{\textbf{(A)}\ 2}$$

Solution 7 (Cubes)
Let $x=100$. Then, we have
Then, the answer can be rewritten as $(x^3+1)-(x^3-1)=\boxed{\textbf{(A)}\ 2}$.

Solution 8 (Super Fast)
It's not hard to observe and express $9901$ into $99\cdot100+1$, and $10101$ into $101\cdot100+1$. We then simplify the original expression into $(99\cdot100+1)\cdot101-99\cdot(101\cdot100+1)$, which could then be simplified into $99\cdot100\cdot101+101-99\cdot100\cdot101-99$, which we can get the answer of $101-99=\boxed{\textbf{(A)}\ 2}$.

Problem 2 The following problem is from both the 2024 AMC 12A #2 and 2024 AMC 10A #2, so both problems redirect to this page.