2024 AMC amc10a 真题 答案 详解

---

中英双语真题

2024 AMC 10A 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 a trail is 4.2 miles long and ascends 4000 feet?
(A) 240 (B) 246 (C) 252 (D) 258 (E) 264

Problem 3
What is the sum of the digits of the smallest prime that can be written as a sum of 5 distinct primes?
(A) 5 (B) 7 (C) 8 (D) 10 (E) 13

Problem 4
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 5
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 6
What is the minimum number of successive swaps of adjacent letters in the string ABCDEF that are needed to change the string to FEDCBA? (For example, 3 swaps are required to change ABC to CBA; one such sequence of swaps is ABC → BAC → BCA → CBA.)
(A) 6 (B) 10 (C) 12 (D) 15 (E) 24

Problem 7
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 8
Amy, Bomani, Charlie, and Daria work in a chocolate factory. On Monday, Amy, Bomani, and Charlie started working at 1:00 PM and were able to pack 4, 3, and 3 packages, respectively, every 3 minutes. At some later time, Daria joined the group, and Daria was able to pack 5 packages every 4 minutes. Together, they finished packing 450 packages at exactly 2:45 PM. At what time did Daria join the group?
(A) 1:25 PM (B) 1:35 PM (C) 1:45 PM (D) 1:55 PM (E) 2:05 PM

Problem 9
In how many ways can 6 juniors and 6 seniors form 3 disjoint teams of 4 people so that each team has 2 juniors and 2 seniors?
(A) 720 (B) 1350 (C) 2700 (D) 3280 (E) 8100

Problem 10
Consider the following operation. Given a positive integer $n$, if $n$ is a multiple of 3, then you replace $n$ by $\frac{n}{3}$. If $n$ is not a multiple of 3, then you replace $n$ by $n+10$. For example, beginning with $n=4$, this procedure gives $4→14→24→8→18→6→2→12→\cdots$. Suppose you start with $n=100$. What value results if you perform this operation exactly 100 times?

英文真题

2024 AMC 10A Problems Problem 1 What is the value of $9901\cdot101-99\cdot10101$? (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 What is the sum of the digits of the smallest prime that can be written as a sum of 5 distinct primes? (A) 5 (B) 7 (C) 8 (D) 10 (E) 13 Problem 4 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 5 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 6 What is the minimum number of successive swaps of adjacent letters in the string ABCDEF that are needed to change the string to FEDCBA? (For example, 3 swaps are required to change ABC to CBA; one such sequence of swaps is ABC→ BAC→BCA→CBA.) (A) 6 (B) 10 (C) 12 (D) 15 (E) 24 Problem 7 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 8 Amy, Bomani, Charlie, and Daria work in a chocolate factory. On Monday Amy, Bomani, and Charlie started working at 1:00PM and were able to pack 4, 3, and 3 packages, respectively, every 3 minutes. At some later time, Daria joined the group, and Daria was able to pack 5 packages every 4 minutes. Together, they finished packing 450 packages at exactly 2:45PM. At what time did Daria join the group? (A) 1:25 PM (B) 1:35 PM (C) 1:45 PM (D) 1:55 PM (E) 2:05 PM Problem 9 In how many ways can 6 juniors and 6 seniors form 3 disjoint teams of 4 people so that each team has 2 juniors and 2 seniors? (A) 720 (B) 1350 (C) 2700 (D) 3280 (E) 8100 Problem 10 Consider the following operation. Given a positive integer $n$, if $n$ is a multiple of 3, then you replace $n$ by $\frac{n}{3}$ if $n$ is not a multiple of 3, then you replace $n$ by $n+10$. For example, beginning with $n=4$ this procedure gives $4 \rightarrow 14 \rightarrow 24 \rightarrow 8 \rightarrow 18 \rightarrow 6 \rightarrow 2 \rightarrow 12 \rightarrow \cdots$. Suppose you start with $n=100$. What value results if you perform this operation exactly 100 times? (A) 10 (B) 20 (C) 30 (D) 40 (E) 50 Problem 11 How many ordered pairs of integers $(m,n)$ satisfy $\sqrt{m^2 - 49} = m$? (A) 1 (B) 2 (C) 3 (D) 4 (E) infinitely many Problem 12 Zelda played the Adventures of Math game on August 1 and scored 1,700 points. She continued to play daily over the next 5 days. The bar chart below shows the daily

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
$$ 9901\cdot101 - 99\cdot10101 = (10000-99)\cdot101 - 99\cdot(10000+101)\ = 10000\cdot101 - 99\cdot101 - 99\cdot10000 - 99\cdot101 \ = (10000\cdot101 - 99\cdot10000) - 2\cdot(99\cdot101)\ = 2\cdot10000 - 2\cdot9999\ =\boxed{\text{(A)}\ 2}. $$

~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 $1-9\equiv2\pmod{10}$, so the answer is either $\text{(A)}$ or $\text{(D)}$. Evaluating $(mod\ 101)$ yields $0-99\equiv2\pmod{101}$. 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 $10$):
$$ 9901\cdot101 - 99\cdot10101\equiv1\cdot1-9\cdot1\equiv2\pmod{10}. $$
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 $10101=10100+1=101\cdot100+1$
$$ \Rightarrow 9901\cdot101 - 99\cdot10101=((9900\cdot101)+(1\cdot101))-((99\cdot10100)+(99\cdot1))\ =(99\cdot100\cdot101)+101-(99\cdot100\cdot101)-99\ =101-99\ =\boxed{\text{(A)}\ 2}. $$

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 10A #2 and 2024 AMC 12A #2, so both problems redirect to this page. A model used to estimate the time it will take to hike to the top of the mountain on a trail of the form $T=a

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 $9901 \cdot 101 - 99 \cdot 10101$? Solution 1 (Direct Computation) 解法 1 (直接计算) The likely fastest method will be direct computation. $9901\cdot 101$ evaluates to $1000001$ and $99\cdot 10101$ evaluates to $999999$. The difference is $\boxed{\text{(A)}\ 2}$. 最可能最快的方法是直接计算。$9901\cdot 101$ 的值是 $1000001$，$99\cdot 10101$ 的值是 $999999$。它们的差是 $\boxed{\text{(A)}\ 2}$。 Solution 2 (Distributive Property) 解法 2 (分配律) We have $9901\cdot 101-99\cdot 10101=(10000-99)\cdot 101-99\cdot(10000+101)$ $=10000\cdot 101-99\cdot 101-99\cdot 10000-99\cdot 101$ $=(10000\cdot 101-99\cdot 10000)-2\cdot(99\cdot 101)$ $=2\cdot 10000-2\cdot 9999$ $=\boxed{\text{(A)}\ 2}$ ~MRENTHUSIASM 我们有 ~MRENTHUSIASM Solution 3 (Solution 1 but Distributive) 解法 3 (解法 1 但使用分配律) Note that $9901\cdot 101=9901\cdot 100+9901=990100+9901=1000001$ and $99\cdot 10101=100\cdot 10101-10101=1010100-10101=999999$, therefore the answer is $1000001-999999=\boxed{\text{(A)}\ 2}$. 注意 $9901\cdot 101=9901\cdot 100+9901=990100+9901=1000001$ 和 $99\cdot 10101=100\cdot 10101-10101=1010100-10101=999999$，因此答案是 $1000001-999999=\boxed{\text{(A)}\ 2}$。 Solution 4 (Modular Arithmetic) 解法 4 (模运算) Evaluating the given expression $(\mod 10)$ yields $1-9\equiv 2(\mod 10)$, so the answer is either $\text{(A)}$ or $\text{(D)}$. Evaluating $(\mod 101)$ yields $0-99\equiv 2(\mod 101)$. Because answer $\text{(D)}$ is $202=2\cdot 101$, that cannot be the answer, so we choose choice $\boxed{\text{(A)}\ 2}$. 计算给定的表达式 $(\mod 10)$ 得到 $1-9\equiv 2(\mod 10)$，所以答案是 $\text{(A)}$ 或 $\text{(D)}$。计算 $(\mod 101)$ 得到 $0-99\equiv 2(\mod 101)$。因为答案 $\text{(D)}$ 是 $202=2\cdot 101$，所以那不能是答案，我们选择选项 $\boxed{\text{(A)}\ 2}$。 Solution 5 (Process of Elimination) 解法 5 (排除法) We simply look at the units digit of the problem we have (or take $\mod 10$). $9901\cdot 101-99\cdot 10101\equiv 1\cdot 1-9\cdot 1\equiv 2\mod 10$. 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 $\boxed{\text{(A)}\ 2}$. 我们只需看问题的个位数（或取模 $10$）。$9901\cdot 101-99\cdot 10101\equiv 1\cdot 1-9\cdot 1\equiv 2\mod 10$。由于只有 $\text{(A)}$ 和 $\text{(D)}$ 的个位数字是 $2$，因此我们可以继续，如果你非常想用猜测和检查，或者一个真正有效的方法找到答案是 $\boxed{\text{(A)}\ 2}$。 Solution 6 (Faster Distribution) 解法 6 (