2019 AMC amc10a 真题 答案 详解

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中英双语真题

2019 AMC 10A
Problem 1
What is the value of
$$2^{\left(0^{\left(19\right)}\right)}+\left(\left(2^{0}\right)^{1}\right)^{9}?$$

下列算式的值是多少
$$2^{\left(0^{\left(19\right)}\right)}+\left(\left(2^{0}\right)^{1}\right)^{9}?$$
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

Problem 2
What is the hundreds digit of $(20! - 15!)$?
$(20! - 15!)$ 的百位数字是多少？
(A) 0 (B) 1 (C) 2 (D) 4 (E) 5

Problem 3
Ana and Bonita are born on the same date in different years, $n$ years apart. Last year Ana was 5 times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n$?
Ana 和 Bonita 出生在不同年份的同一天，相隔 $n$ 年。去年 Ana 的年龄是 Bonita 年龄的 5 倍。今年 Ana 的年龄是 Bonita 年龄的平方。问 $n$ 是多少？
(A) 3 (B) 5 (C) 9 (D) 12 (E) 15

英文真题

2019 AMC 10A Problems Problem 1 What is the value of $$2^{\left(0^{\left(19\right)}\right)}+\left(\left(2^{0}\right)^{1}\right)^{9}?$$ (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 Problem 2 What is the hundreds digit of $(20! - 15!) $? (A) 0 (B) 1 (C) 2 (D) 4 (E) 5 Problem 3 Ana and Bonita are born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n$? (A) 3 (B) 5 (C) 9 (D) 12 (E) 15 Problem 4 A box contains $28$ red balls, $20$ green balls, $19$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least $15$ balls of a single color will be drawn? (A) 75 (B) 76 (C) 79 (D) 84 (E) 91 Problem 5 What is the greatest number of consecutive integers whose sum is $45$? (A) 9 (B) 25 (C) 45 (D) 90 (E) 120 Problem 6 For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral? - a square - a rectangle that is not a square - a rhombus that is not a square - a parallelogram that is not a rectangle or a rhombus - an isosceles trapezoid that is not a parallelogram (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 Problem 7 Two lines with slopes $\frac{1}{2}$ and $2$ intersect at $(2,2)$. What is the area of the triangle enclosed by these two lines and the line $x+y=10$? (A) 4 (B) $4 \sqrt{2}$ (C) 6 (D) 8 (E) $6 \sqrt{2}$ Problem 8 The figure below shows line $\ell$ with a regular, infinite, recurring pattern of squares and line segments.

真题文字详解(英文)

Problem 1 What is the value of $$2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9$$?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

Solution $$2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9=1+1=\boxed{(C)\ 2}.$$

Problem 2 What is the hundreds digit of $(20!-15!)$?
(A) 0 (B) 1 (C) 2 (D) 4 (E) 5

Solution 1 Because we know that $5^3$ is a factor of $15!$ and $20!$, the last three digits of both numbers is a $0$, this means that the difference of the hundreds digits is also $\boxed{(A)\ 0}$.

Solution 2 We can clearly see that $20!\equiv 15!\equiv 0\pmod{1000}$, so $20! - 15!\equiv 0\pmod{100}$ meaning that the last two digits are equal to $00$ and the hundreds digit is $\boxed{(A)\ 0}$.

-abhinav0627

Solution 3 (Brute Force)
$20!=2432902008176640000$
$15!=1307674368000$
Then, we see that the hundreds digit is $0-0=\boxed{(A)\ 0}$.
Please do not do this and only use this solution as a last resort.

Problem 3 Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n$?
(A) 3 (B) 5 (C) 9 (D) 12 (E) 15

Solution Solution 1 Let $A$ be the age of Ana and $B$ be the age of Bonita. Then,

真题文字详解(中英双语)

Problem 1 What is the value of $$2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9$$?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

Solution
$$2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9=1+1=\boxed{2}$$

Problem 2 What is the hundreds digit of $(20!-15!)$?
(A) 0 (B) 1 (C) 2 (D) 4 (E) 5

Solution 1 Because we know that $5^3$ is a factor of $15!$ and $20!$, the last three digits of both numbers is a 0, this means that the difference of the hundreds digits is also $\boxed{0}$.

Solution 2 We can clearly see that $20!\equiv 15!\equiv 0 \pmod {1000}$, so $20! - 15!\equiv 0 \pmod {100}$ meaning that the last two digits are equal to 00 and the hundreds digit is $\boxed{0}$.

-abhinavg0627

Solution 3 (Brute Force)
$20!=2432902008176640000$
$15!=1307674368000$
Then, we see that the hundreds digit is $0-0=\boxed{0}$.
Please do not do this and only use this solution as a last resort.

Problem 3 Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was 5 times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n$?
(A) 3 (B) 5 (C) 9 (D) 12 (E) 15

Solution
Let $A$ be the age of Ana and $B$ be the age of Bonita. Then,