AMC 12 Number Theory - Complete Collection

AMC 12 Number Theory - Complete Collection

Welcome to the ultimate AMC 12 Number Theory problem collection! This page contains 41 authentic competition problems spanning 23 years (2000-2023), all focused on number theory concepts that appear repeatedly on the AMC 12.

What You'll Find Here

Study Progress

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Congratulations! You've completed all items!

• Review introductory problems

  Build foundational skills (27 questions)
• Tackle intermediate challenges

  Push your limits (14 questions)
• Master key concepts

  Understand patterns across topics
• Complete full practice set

  Test yourself on all 41 problems

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How to Use This Collection

Study Strategy

For beginners: Start with the Introductory section. These problems (typically #3-#16 on the exam) build core skills in divisibility, modular arithmetic, and basic number properties.

For advanced students: Jump straight to Intermediate problems (typically #17-#25). These require deeper insight and often combine multiple concepts.

For everyone: After solving, analyze why certain approaches work. Number theory rewards pattern recognition!

What Makes Number Theory Special?

Number theory problems on the AMC 12 test your ability to:

• See patterns in integers and their properties
• Apply modular arithmetic to simplify complex calculations
• Use divisibility rules creatively
• Factor and manipulate expressions strategically

The beauty? Once you recognize the patterns, many “hard” problems become straightforward!

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Topics Covered

This collection spans the full breadth of AMC 12 number theory:

Topic	Key Concepts	Question Count
Divisibility & Factorization	Prime factorization, factor counting, GCD/LCM	8+ problems
Modular Arithmetic	Remainders, congruences, Chinese Remainder Theorem	7+ problems
Prime Numbers	Prime testing, prime factorization, prime properties	6+ problems
Number Bases	Base conversion, digit properties	3+ problems
Sequences	Recurrence relations, Fibonacci-like sequences	4+ problems
Diophantine Equations	Integer solutions, linear equations	5+ problems
Repeating Decimals	Period length, fraction conversion	3+ problems
Factorials	Trailing zeros, factorial properties	3+ problems
Combinatorial Number Theory	Frobenius numbers, representability	2+ problems
Special Functions	Digit sums, divisor function, multiplicative functions	4+ problems

Note

Many problems combine multiple topics! For example, a modular arithmetic problem might also involve prime factorization or sequences.

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Quick Reference: Essential Concepts

Before diving into problems, here are the tools you’ll need most:

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Must-Know Formulas

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Problem Collection

Organization

Problems are grouped by difficulty:

• Intermediate (14 problems): Challenges requiring deeper insight
• Introductory (27 problems): Foundation builders

Within each section, problems are sorted by year (newest first) so you can track how the AMC 12 has evolved.

Problem-Solving Tip

Before looking at the answer, ask yourself:

1. What number theory concept is being tested?
2. Can I simplify using modular arithmetic?
3. Is there a pattern or formula I should recognize?
4. Can I test small cases to find a pattern?

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Intermediate Number Theory

These 14 problems represent the toughest number theory challenges on the AMC 12. Expect to spend 3-5 minutes per problem and use multiple concepts.

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Problem 1: AMC 12A 2023 #17

Flora’s Frog Jumps

Flora the frog starts at 0 on the number line and makes a sequence of jumps to the right. In any one jump, independent of previous jumps, Flora leaps a positive integer distance with probability .

What is the probability that Flora will eventually land at 10?

Answer

Key Insight

This problem combines probability with partition theory. The probability of reaching position satisfies a beautiful recursive formula. It turns out that for all positive integers !

Why? Consider that reaching requires either:

• Jumping directly (probability ), or
• Reaching some first, then jumping

The recursion simplifies beautifully to .

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Problem 2: AMC 12A 2023 #22

Möbius Function Challenge

Let be the unique function defined on the positive integers such that for all positive integers . What is ?

Answer

Key Insight

This is the Möbius inversion formula in disguise! The given condition defines as related to the Möbius function .

Since (note the prime factorization), you can use Möbius inversion to compute .

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Problem 3: AMC 12A 2022 #23

Harmonic Series and LCM

Let and be the unique relatively prime positive integers such that Let denote the least common multiple of the numbers . For how many integers with is ?

Answer

(namely )

Key Insight

When you add fractions , the denominator divides .

The question is: when is strictly less than ? This happens when the sum simplifies due to cancellation. Analyzing the prime factorization reveals specific values of where this occurs.

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Problem 4: AMC 12A 2022 #25

Circle Tangents and Pythagorean Triples

A circle with integer radius is centered at . Distinct line segments of length connect points to for and are tangent to the circle, where , , and are all positive integers and . What is the ratio for the least possible value of ?

Answer

Key Insight

This problem connects geometry with Pythagorean triples! The tangency condition creates a Diophantine equation. You need to find the smallest such that there are at least 14 distinct integer solutions.

The answer involves systematically finding Pythagorean triples and checking the tangency condition.

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Problem 5: Fall AMC 12B 2021 #25

Remainder Functions

For a positive integer, let be the sum of the remainders when is divided by , , , , , , , , and . For example, . How many two-digit positive integers satisfy ?

Answer

(namely and )

Problem-Solving Approach

This requires careful modular arithmetic! Notice that:

This difference is usually non-zero, but at special values of , cancellations occur. Check systematically for two-digit numbers.

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Problem 6: AMC 12A 2021 #25

Divisor Function Optimization

Let denote the number of positive integers that divide , including and . Let There is a unique positive integer such that for all positive integers . What is the sum of the digits of ?

Answer

(where )

Key Insight

This is an optimization problem! You want to maximize .

Intuitively, you want to have many divisors relative to its size. Numbers with many small prime factors work well. The answer is highly composite.

The sum of digits: .

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Problem 7: AMC 12B 2018 #17

Rational Approximation (Farey Sequence)

Let and be positive integers such that and is as small as possible. What is ?

Answer

Key Insight

This uses the mediant property from Farey sequences! The fraction with smallest denominator between and is the mediant .

Here: . Check: ✓

So , , and .

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Problem 8: AMC 12B 2017 #19

Concatenated Number Modulo

Let be the -digit number that is formed by writing the integers from to in order, one after the other. What is the remainder when is divided by ?

Answer

Problem-Solving Approach

Since , use the Chinese Remainder Theorem! Find and separately.

Mod 9: A number has the same remainder mod 9 as the sum of its digits. Mod 5: Look at the last digit pattern.

Then combine using CRT.

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Problem 9: AMC 12B 2017 #21

Integer Averages

Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?

Answer

Key Insight

Work backwards using modular arithmetic! If the average after tests is an integer, then the sum of the first scores is divisible by .

Let be the sum after tests. Then:

This means .

Testing scores between 91-100 with these constraints gives .

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Problem 10: AMC 12B 2016 #22

Repeating Decimal Periods

For a certain positive integer less than , the decimal equivalent of is , a repeating decimal of period , and the decimal equivalent of is , a repeating decimal of period . In which interval does lie?

Answer

(where )

Key Insight

The period of is related to the multiplicative order of 10 modulo .

If the period is 6, then (and 6 is the smallest such exponent). If the period is 4 for , then .

This constrains the possible values of . Systematic checking gives .

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Problem 11: AMC 12B 2014 #23

Binomial Sums and Primes

The number is prime. Let . What is the remainder when is divided by ?

Answer

Key Insight

This combines Lucas’ Theorem with properties of binomial coefficients!

Note that , and we’re summing roughly half the binomial coefficients. The answer is no coincidence—it comes from analyzing the sum using properties of binomial expansions modulo a prime.

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Problem 12: AMC 12A 2010 #23

Factorial Trailing Digits

The number obtained from the last two nonzero digits of is equal to . What is ?

Answer

Problem-Solving Approach

This requires careful analysis of the prime factorization of .

First, count the trailing zeros (from factors of 10 = 2 × 5). Then work modulo 100 to find the last two nonzero digits by systematically computing modulo 100, ignoring factors of 2 and 5.

The calculation is intricate but follows a systematic pattern.

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Problem 13: AMC 12A 2009 #18

Powers of Two Factorization

For , let , where there are zeros between the and the . Let be the number of factors of in the prime factorization of . What is the maximum value of ?

Answer

Key Insight

Write as .

Factor out powers of 2:

The maximum occurs when the expression in parentheses is even but not divisible by higher powers of 2. Testing values of gives maximum .

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Problem 14: AMC 12B 2004 #25

First Digits of Powers

Given that is a -digit number whose first digit is , how many elements of the set have a first digit of ?

Answer

or

Key Insight

This uses Benford’s Law and logarithms! The first digit of is if:

where denotes the fractional part of .

For first digit 4: to .

Since , approximately powers have first digit 4.

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Introductory Number Theory

These 27 problems build foundational skills. Most appear as problems #3-#16 on the AMC 12, requiring 2-4 minutes each.

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Problem 15: AMC 12B 2023 #16

Frobenius Coin Problem

In the state of Coinland, coins have values and cents. Suppose is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of ?

Answer

(where )

Key Concept

This is a Frobenius number problem! However, the classic formula only works for two coprime numbers.

With three coins (6, 10, 15), you need to check systematically. Since , eventually all sufficiently large numbers can be represented.

Testing shows 29 cannot be made, but all numbers can. Sum of digits: .

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Problem 16: AMC 12B 2018 #15

Divisibility with Digit Constraints

How many odd positive -digit integers are divisible by but do not contain the digit ?

Answer

Counting Strategy

Break it down:

1. 3-digit: where (8 choices, no 3)
2. Odd: (4 choices, no 3)
3. Divisible by 3:

For each choice of and , determine valid values of using and .

Count systematically by cases to get 96.

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Problem 17: AMC 12B 2016 #16

Consecutive Integer Sums

In how many ways can be written as the sum of an increasing sequence of two or more consecutive positive integers?

Answer

Key Insight

If consecutive integers start at , their sum is:

So .

For each divisor of 690 with , check if is a positive integer.

Testing all divisors gives 7 valid representations.

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Problem 18: AMC 12B 2013 #9

Perfect Square in Factorial

What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides ?

Answer

Solution Method

First, find the prime factorization of :

• Powers of 2:
• Powers of 3:
• Powers of 5:
• Powers of 7:
• Powers of 11:

So .

Largest perfect square dividing : (take even powers).

Its square root: . Sum of exponents: .

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Problem 19: AMC 12B 2013 #14

Fibonacci-like Sequences

Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is . What is the smallest possible value of ?

Answer

Key Insight

These are generalized Fibonacci sequences. If the first two terms are and , the sequence is:

The seventh term is .

To minimize with two different sequences, try:

• Sequence 1:
• Sequence 2:

Both give with different first terms!

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Problem 20: AMC 12B 2013 #15

Factorial Ratios

The number is expressed in the form

where and are positive integers and is as small as possible. What is ?

Answer

Solution Strategy

Factor .

To express this as a ratio of factorials with small :

• Try to build 61 from factorials
• (too large)
• Better: Use and

Working through the algebra, the optimal representation has and , giving .

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Problem 21: AMC 12B 2012 #11

Number Bases

In the equation below, and are consecutive positive integers, and , , and represent number bases:

What is ?

Answer

Solution Method

Convert to base 10:

• where

Substitute :

So (taking positive), , and .

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Problem 22: AMC 12B 2011 #15

Factors of

How many positive two-digit integers are factors of ?

Answer

Key Insight

Factor using difference of squares repeatedly:

Further: .

So .

Count two-digit divisors systematically: gives 12 factors.

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Problem 23: AMC 12A 2010 #11

Exponential Equations

The solution of the equation can be expressed in the form . What is ?

Answer

Solution Method

Take logarithms of both sides:

So .

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Problem 24: AMC 12A 2007 #11

Cyclic Digit Sequences

A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let be the sum of all the terms in the sequence. What is the largest prime factor that always divides ?

Answer

Key Insight

Let the sequence have terms. Each digit appears exactly once in each position (hundreds, tens, units) as we cycle through.

If we denote digits as , the sum is:

Since , we have always. The largest prime factor is 37.

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Problem 25: AMC 12A 2007 #12

Parity of Products

Integers and , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that is even?

Answer

Parity Analysis

is even when and have the same parity.

Both are even when at least one factor in each pair is even. Both are odd when all four numbers are odd.

Among 0-2007: 1004 even, 1004 odd (probability each).

Working through: answer is .

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Problem 26: AMC 12A 2007 #22

Digit Sum Equations

For each positive integer , let denote the sum of the digits of For how many values of is

Answer

(solutions: 1977, 1980, 1983, 2001)

Solution Strategy

Since , we have (at most 4 digits).

So .

This means:

Check each value in this range systematically. Testing gives exactly 4 solutions.

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Problem 27: AMC 12A 2006 #4

Digital Clock Maximum

A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?

Answer

Solution

The time format is HH:MM (where HH goes from 01 to 12, MM from 00 to 59).

To maximize the digit sum:

• Hours: 9 has digit sum 9 (best single digit hour)
• Better: 12 has digit sum
• Wait: 9:59 gives

But actually 12:59 gives .

The maximum is at 9:59 with sum .

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Problem 28: AMC 12A 2006 #9

Diophantine Price Problem

Oscar buys pencils and erasers for 1.00$. A pencil costs more than an eraser, and both items cost a whole number of cents. What is the total cost, in cents, of one pencil and one eraser?

Answer

Solution Method

Let = pencil cost (cents), = eraser cost (cents).

Solve for :

For to be an integer: .

Since and :

Try : . Check: ✓

Total: .

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Problem 29: AMC 12A 2006 #14

GCD Application (Pigs and Goats)

Two farmers agree that pigs are worth 300$210$390$ debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?

Answer

Key Insight

The smallest debt that can be resolved is .

Any debt that’s a multiple of 30 can be paid using the Euclidean algorithm principle!

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Problem 30: AMC 12A 2005 #18

Prime-Looking Numbers

Call a number prime-looking if it is composite but not divisible by or The three smallest prime-looking numbers are , and . There are prime numbers less than . How many prime-looking numbers are there less than ?

Answer

Counting Method

Step 1: Count integers not divisible by 2, 3, or 5 using inclusion-exclusion.

Result: 266 such integers (including 1).

Step 2: Subtract:

• 1 (not composite)
• 168 primes

Step 3: But we counted 2, 3, 5 in the 168 primes, and they ARE divisible by themselves.

Careful counting gives: prime-looking numbers.

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Problem 31: AMC 12A 2003 #8

Divisor Probability

What is the probability that a randomly drawn positive factor of is less than ?

Answer

Solution

First find all divisors of :

Divisors: (total: 12)

Divisors : (count: 6)

Probability: .

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Problem 32: AMC 12A 2003 #12

Card Stacking with Divisibility

Sally has five red cards numbered through and four blue cards numbered through . She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?

Answer

Solution Strategy

With 5 red and 4 blue cards, the stack alternates starting and ending with red: R-B-R-B-R-B-R-B-R.

Each red must divide its neighboring blue cards.

Working through the constraints:

• Card 1 (red) must divide blue neighbors → very restrictive
• Card 5 (red) similarly constrains

Testing systematically, the only valid arrangement has middle three cards summing to 12.

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Problem 33: AMC 12B 2002 #3

Quadratic Prime Values

For how many positive integers is a prime number?

Answer

Key Insight

Factor the expression:

For this product to be prime, one factor must be 1 and the other prime.

• If , then and ✓ (prime!)
• If , then and (not prime)
• For , both factors are , so product is composite

Answer: Only works.

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Problem 34: AMC 12B 2002 #11

Prime Sum Properties

The positive integers and are all prime numbers. The sum of these four primes is

Answer

Key Insight

Since are all prime and positive, consider parity.

If both and are odd, then and are even. For them to be prime, they must equal 2.

But is impossible if .

So one of is even. The only even prime is 2.

Try : Then must all be prime. The only solution: (giving primes 5, 3, 7).

Sum: (prime!)

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Problem 35: AMC 12B 2002 #12

Perfect Square Ratios

For how many integers is the square of an integer?

Answer

Solution Method

Let for integer .

Then:

For to be an integer, .

Since , we need .

Divisors of 20: .

Check which give :

• → only work

This gives 4 values of .

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Problem 36: AMC 12B 2002 #15

Four-Digit Divisibility

How many four-digit numbers have the property that the three-digit number obtained by removing the leftmost digit is one ninth of ?

Answer

Solution Strategy

Let where is the first digit.

Then: .

Write: .

Solve for :

For to be 4-digit: .

Since is a digit: (can’t be 0 or 9).

But we need to have exactly 4 digits and to have 3 digits.

Testing: all work. Answer: 7.

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Problem 37: AMC 12A 2002 #20

Repeating Decimal Denominators

Suppose that and are digits, not both nine and not both zero, and the repeating decimal is expressed as a fraction in lowest terms. How many different denominators are possible?

Answer

Key Insight

Reduce to lowest terms by dividing by .

Since , possible denominators after reduction are divisors of 99:

But we exclude (both zero) and (both nine).

Check which denominators are achievable:

• Denominator 1: → impossible
• Denominator 3: ✓
• Denominator 9: ✓
• Denominator 11: ✓
• Denominator 33: Various ✓
• Denominator 99: Requires ✓

Testing systematically: 5 different denominators are possible.

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Problem 38: AMC 12 2000 #6

Prime Product minus Sum

Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

Answer

Solution Method

Primes between 4 and 18: .

For primes and : .

Wait, better: .

Test the given options by trying pairs:

• : ✓

Answer: (C) 119

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Problem 39: AMC 12 2000 #13

Coffee and Milk Ratios

One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

Answer

Solution Strategy

Let = total milk, = total coffee.

Angela drank: .

Total drinks: people.

From Angela’s equation: , so .

Total ounces: .

Substitute :

For : .

Also need , so .

Thus .

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Problem 40: AMC 12 2000 #16

Checkerboard Renumbering

A checkerboard of rows and columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered , the second row , and so on down the board. If the board is renumbered so that the left column, top to bottom, is the second column and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).

Answer

Key Insight

In system 1 (row-major), square (row , column , both 0-indexed) has number:

In system 2 (column-major):

For :

So . For integer : must be divisible by 3.

Valid: (5 values).

Corresponding and values:

Sum: .

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Problem 41: AMC 12 2000 #18

Calendar Day Calculations

In year , the day of the year is a Tuesday. In year , the day is also a Tuesday. On what day of the week did the th day of year occur?

Answer

Solution Strategy

Between day 300 of year and day 200 of year :

• Remaining days in year : (or 66 if leap year)
• Days in year : 200

If is not a leap year: days. If is a leap year: days.

Since both are Tuesdays, the number of days must be divisible by 7.

• (not divisible)
• (divisible!)

So is a leap year (366 days).

From day 100 of year to day 300 of year :

• Days remaining in : (assuming non-leap)
• Days in year up to day 300: 300
• Total: days

So 5 days before Tuesday is Thursday.

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Summary & Statistics

Problem Collection Overview

Distribution by Year

Recent (2021-2023)	Mid (2010-2018)	Classic (2000-2009)
7 questions	14 questions	20 questions

By Competition Type

AMC 12A	AMC 12B	AMC 12	Fall AMC 12B
18	21	4	1

Problem Number Range

• Easiest: #3 (typically basic concepts)
• Hardest: #25 (competition-level challenges)
• Most Common: #11-#18 (medium difficulty sweet spot)

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Study Recommendations

How to Master AMC 12 Number Theory

Week 1-2: Work through all Introductory problems. Master divisibility, modular arithmetic, and basic prime properties.

Week 3-4: Tackle Intermediate problems. Focus on recognizing when to use special techniques (Chinese Remainder Theorem, Möbius inversion, etc.).

Week 5+: Time yourself. AMC 12 gives 75 minutes for 25 questions (~3 min/problem). Number theory problems in positions #15-#20 should take 3-4 minutes each.

Throughout: Maintain a “pattern notebook.” When you see a technique work, write down the trigger: “When I see _____, I should try _____.”

Number Theory Mastery Checklist

Progress 0%

Congratulations! You've completed all items!

• Divisibility & Factors

  Count divisors, use prime factorization fluently
• Modular Arithmetic

  Solve congruences, apply CRT
• Prime Numbers

  Test primality, factor quickly
• Number Bases

  Convert between bases smoothly
• Sequences

  Recognize Fibonacci-like patterns
• Diophantine Equations

  Find integer solutions systematically
• Repeating Decimals

  Understand period and conversion
• Factorials

  Count trailing zeros, compute mod values
• Frobenius Numbers

  Apply the Chicken McNugget Theorem
• Number-Theoretic Functions

  Use divisor, totient, Möbius functions

Reset Progress

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Final Thoughts

Number theory is one of the most pattern-rich topics on the AMC 12. Unlike geometry or algebra problems that can vary wildly, number theory problems often recycle the same core ideas:

• Modular arithmetic simplifies seemingly complex calculations
• Prime factorization unlocks divisibility questions
• GCD/LCM connections appear everywhere

The 41 problems in this collection represent decades of AMC 12 evolution, but the underlying concepts remain remarkably consistent. Master these patterns, and you’ll find yourself recognizing “Oh, this is just like problem #X!” more and more often.

Happy problem-solving! 🎓

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Note

This collection was compiled from the AMC 12 official problem archives (2000-2023). All problems are authentic competition questions. Solutions and insights are educational interpretations.