2023 AMC 10A Real Questions and Analysis

In this article, you’ll find:

• Representative real questions from each module with detailed solutions.
• The complete 2023 AMC 10A Answer Key.
• The best resources to prepare effectively for the AMC 10.
• A concise topic distribution chart showing which areas appeared most in the 2023 AMC 10A.
• A module-to-question mapping table highlighting the core concepts tested in each module for the 2023 AMC 10A.

Real Question and Solutions Explained

Algebra Example – Problem 8

Question:

Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at 110 degrees Fahrenheit, which is 0 degrees on the Breadus scale. Bread is baked at 350 degrees Fahrenheit, which is 100 degrees on the Breadus scale. Bread is done when its internal temperature is 200 degrees Fahrenheit. What is this in degrees on the Breadus scale?

(A) 33 (B) 34.5 (C) 36 (D) 37.5 (E) 39

Solution:

Let the conversion be \( B = aF + b \).
Using the two calibration points \((F,B) = (110,0)\) and \((350,100)\):
\[ a = \frac{100 – 0}{350 – 110} = \frac{100}{240} = \frac{5}{12}, \quad b = 0 – \frac{5}{12} \times 110 = -\frac{275}{6} \]
At \( F = 200 \):
\[ B = \frac{5}{12} \times 200 – \frac{275}{6} = \frac{250}{3} – \frac{275}{6} = \frac{225}{6} = 37.5 \]
Therefore, the Breadus temperature is 37.5°.

Answer (D)

Common Mistakes:

• Assuming a direct proportion \( B = kF \) (forgetting the intercept).
• Mixing the two reference points when computing the slope.
• Converting \( 200^{\circ} \) after rounding \( a \) or \( b \).

Number Theory Example – Problem 12

Question:

How many three-digit positive integers \( N \) satisfy the following?

• The number formed by reversing the digits of \( N \) is divisible by 5.

• \( N \) is divisible by 7.

(A) 13 (B) 14 (C) 15 (D) 16 (E) 17

Solution:

A number is divisible by 5 if and only if its last digit is 0 or 5. The reverse of \(N\) ends with the hundreds digit of \(N\), so that hundreds digit must be 0 or 5. For a three-digit \(N\), it cannot start with 0; hence the hundreds digit is 5.

Thus \(N \in \{500, 501, \dots, 599\}\) and \(7 \mid N\).
The smallest multiple of 7 in this range is \(7 \times 72 = 504\), and the largest is \(7 \times 85 = 595\).
Hence the valid multiples are \(7 \times 72, 7 \times 73, \dots, 7 \times 85\), giving
\[ 85 – 72 + 1 = 14 \text{ numbers.} \]

Answer (B)

Common Mistakes:

• Requiring the units digit of \(N\) to be 0 or 5 (it’s the reverse that must be).
• Forgetting that the hundreds digit cannot be 0.
• Off-by-one errors when counting multiples within an interval.

Geometry Example – Problem 22

Question:

Circles \( C_1 \) and \( C_2 \) each have radius 1, and the distance between their centers is \( \frac{1}{2} \). Circle \( C_3 \) is the largest circle internally tangent to both \( C_1 \) and \( C_2 \). Circle \( C_4 \) is internally tangent to both \( C_1 \) and \( C_2 \), and externally tangent to \( C_3 \).
What is the radius of \( C_4 \)?

(A) \( \frac{1}{14} \) (B) \( \frac{1}{12} \) (C) \( \frac{1}{10} \) (D) \( \frac{3}{28} \) (E) \( \frac{1}{9} \)

Solution:

Let the centers of the two unit circles be \((\pm \frac{1}{4}, 0)\), so their distance is \( \frac{1}{2} \).

The center of \( C_3 \) lies on the perpendicular bisector (the y-axis).
For internal tangency to each unit circle, the distance from its center \((0, y)\) to either unit circle center must be \( 1 – r \).
This is minimized when \( y = 0 \), giving
\[ r_3 = 1 – \frac{1}{4} = \frac{3}{4} \]

Let \( C_4 \) have center \((0, x)\) (with \( x > 0 \)) and radius \( r \).
Internal tangency to each unit circle gives
\[ \sqrt{\Big( \frac{1}{4} \Big)^2 + x^2} = 1 – r \tag{1} \]
External tangency to \( C_3 \) (center at the origin) gives
\[ x = r_3 + r = \frac{3}{4} + r \tag{2} \]

Substitute (2) into (1):
\[ \Big( \frac{3}{4} + r \Big)^2 + \frac{1}{16} = (1 – r)^2 \]
Simplify:
\[ \frac{5}{8} + \frac{3}{2}r = 1 – 2r \Rightarrow \frac{7}{2}r = \frac{3}{8} \Rightarrow r = \frac{3}{28} \]

Therefore, the radius of \( C_4 \) is \( \frac{3}{28} \).

Answer (D)

Common Mistakes:

• Placing the small-circle center at the midpoint (implies impossible concentric tangency).
• Using \(x = |r_3 – r|\) instead of \(x = r_3 + r\).
• Forgetting both distances to the unit centers must be equal.

Combinatorics Example – Problem 7

Question:

Janet rolls a standard 6-sided die 4 times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal 3?

(A) \( \frac{2}{9} \) (B) \( \frac{49}{216} \) (C) \( \frac{25}{108} \) (D) \( \frac{17}{72} \) (E) \( \frac{13}{54} \)

Solution:

Total outcomes: \(6^4=1296\). Count the complement = sequences whose partial sums never equal 3.

• First roll = 3 → already hits 3 (not in complement).
• First roll \(\ge 4\): already exceeds 3 and can never come back → \(3\cdot6^3=648\) sequences in the complement.
• First roll = 1: the second roll cannot be 2.
• If second = 1, sum = 2; third cannot be 1, so 5 choices for the third and 6 for the fourth → \(5\cdot6=30\).
• If second \in\{3,4,5,6\}, sum \(\ge 4\); remaining two rolls free → \(4\cdot6^2=144\).
  Total for first = 1: \(174\).
• First roll = 2: the second roll cannot be 1.
• If second = 2, sum = 4; remaining free → \(6^2=36\).
• If second \in\{3,4,5,6\}, sum \(\ge 5\); remaining free → \(4\cdot6^2=144\).
  Total for first = 2: \(180\).

Complement total: \(648+174+180=1002\). Hence desired count
\[1296-1002=294\]
and probability
\[\frac{294}{1296} = \frac{49}{216}.\]

Answer (B)

Common Mistakes:

• Forgetting that a first roll of 3 already satisfies the condition (should not be part of the complement).
• Overcounting by continuing to restrict later rolls after the running total has already exceeded 3.
• Using \(6^3\) instead of \(6^4\) for the total number of possible outcomes.

2023 AMC 10A Answer Key

Question	Answer
1	E
2	A
3	A
4	D
5	E
6	D
7	B
8	D
9	E
10	D
11	C
12	B
13	C
14	B
15	E
16	B
17	A
18	D
19	E
20	D
21	D
22	D
23	C
24	C
25	A

Free Download: 2023 AMC 10A Problems & Solution

Download AMC 10 Mock Exam & Practice Pack

Download AMC 10 mock exams with full solutions, key AR/HR knowledge reviews, and daily practice sets — perfect for focused and efficient prep.

• Mock Exams: Timed practice with full explanations to build test readiness and review mistakes.
• Knowledge Points: Covers must-know Algebra, Geometry, and Number Theory concepts for AMC 10.
• Daily Practice: Sharpen accuracy and speed with high-frequency AMC-style problems.

*Please enter a valid email address to access the materials and receive future prep updates.

Last 10 Years AMC 10 Real Questions and Analysis

Think Academy provides in-depth breakdowns of the past decade of AMC 10 exams. Click below to explore:

• Year-by-year topic trend insights and concept distributions

• Real AMC 10 exams from the last 10 years

• Official answer keys

• Representative questions, detailed solutions, and common mistakes

AMC 10A	AMC 10B
2024 AMC 10A	2024 AMC10B
2023 AMC 10A	2023 AMC10B
2022 AMC 10A	2022 AMC10B
2021 AMC 10A	2021 AMC10B
2020 AMC 10A	2020 AMC10B
2019 AMC 10A	2019 AMC10B
2018 AMC 10A	2018 AMC10B
2017 AMC 10A	2017 AMC10B
2016 AMC 10A	2016 AMC10B

2023 AMC 10A Topic Distribution

The 2023 AMC 10A featured 25 questions to be completed in 75 minutes, emphasizing advanced problem-solving and proof-based reasoning skills.

Learn more about AMC 10 Format and Scoring Here: AMC 10 FAQ and Resources: Your Ultimate Guide

Learn more about 2023 AMC 10A Strategies Here: 2023 AMC 10A Exam Analysis: Difficulty, Structure, and Strategies

Detailed Module Analysis

Module	Question Numbers	What It Tests (Brief)
Algebra (+ arithmetic reasoning)	1, 2, 6, 8, 10, 16, 21	Ratios, fractions, means, linear equations, functions, and polynomial transformations
Number Theory	3, 5, 12, 23	Primes, factors, divisibility, and digit properties
Geometry	4, 11, 13, 15, 17, 18, 19, 22, 24	Perimeter, area, triangles, circles, 3D solids, rotations, tangents, and similarity
Combinatorics / Counting & Probability	7, 9, 14, 20, 25	Counting, casework, logic, and probability on solids

About Think Academy

At Think Academy, a subsidiary of TAL Education Group, we specialize in advanced coaching and rigorous preparation for math competitions, including AMC 10 and AIME. Guided by experienced educators and supported by partnerships with renowned institutions, Think Academy empowers high-school students to reach their full competitive potential in mathematics. Notably, in 2024, our students achieved exceptional recognition, including selections for prestigious programs such as USA(J)MO (17 students), the Mathematical Olympiad Program (3 students), and the International Mathematical Olympiad teams for the U.S. and South Africa. Explore our AMC 10 training programs to help your child excel.

Check our math programs:

• Early Childhood — PreK
• Lower Elementary — Grades K–2
• Upper Elementary — Grades 3–5
• Middle School — Grade 6-10

Want more insights on math learning and parenting? Fill out the form on this page and subscribe to our newsletter for weekly tips and the latest resources.