Problem 1 · Divisibility & Remainders
AMC 8
Number Theory
★★★★☆☆☆
📐 Key Concept
When a number $n$ is divided by $d$, write $n = dq + r$ where $0 \le r < d$. Simultaneous congruences $x \equiv a \pmod{m}$ and $x \equiv b \pmod{n}$ (with $\gcd(m,n)=1$) have a unique solution mod $mn$ — this is the Chinese Remainder Theorem.✏️ Worked Example
Find the smallest positive integer leaving remainder 2 when divided by 3, and remainder 3 when divided by 5.Solution: Write $x = 3k+2$ and test: $k=0 \Rightarrow x=2$, $k=1 \Rightarrow x=5$, $k=2 \Rightarrow x=8$. Check: $8 \bmod 5 = 3$ ✓
$x = 8$
Problem: What is the smallest positive integer that leaves a remainder of 3 when divided by 7, and a remainder of 5 when divided by 11?
💡 Full Solution
We need $x \equiv 3 \pmod{7}$ and $x \equiv 5 \pmod{11}$.
Write $x = 7k + 3$ and substitute into the second congruence: $$7k + 3 \equiv 5 \pmod{11} \implies 7k \equiv 2 \pmod{11}$$ Modular inverse of 7 mod 11: since $7 \times 8 = 56 = 5 \times 11 + 1$, we have $7^{-1} \equiv 8 \pmod{11}$.
So $k \equiv 8 \times 2 = 16 \equiv 5 \pmod{11}$, giving $x = 7(5) + 3 = \boxed{38}$.
Write $x = 7k + 3$ and substitute into the second congruence: $$7k + 3 \equiv 5 \pmod{11} \implies 7k \equiv 2 \pmod{11}$$ Modular inverse of 7 mod 11: since $7 \times 8 = 56 = 5 \times 11 + 1$, we have $7^{-1} \equiv 8 \pmod{11}$.
So $k \equiv 8 \times 2 = 16 \equiv 5 \pmod{11}$, giving $x = 7(5) + 3 = \boxed{38}$.
Problem 2 · Inscribed Circle
AMC 10
Geometry
★★★★★☆☆
📐 Key Concept
For a right triangle with legs $a, b$ and hypotenuse $c$, the inradius is $$r = \frac{a + b - c}{2}$$ The area satisfies $K = r \cdot s$ where $s$ is the semi-perimeter.✏️ Worked Example
A right triangle has legs 3 and 4. Find the inradius.Solution: $c = 5$. Then $r = \dfrac{3+4-5}{2} = 1$.
$r = 1$
Problem: A right triangle has legs of length 5 and 12. A circle is inscribed in the triangle. What is the area of the inscribed circle?
💡 Full Solution
Hypotenuse: $c = \sqrt{25 + 144} = \sqrt{169} = 13$.
Inradius: $r = \dfrac{5 + 12 - 13}{2} = \dfrac{4}{2} = 2$.
Area of inscribed circle: $\pi r^2 = \pi (2)^2 = \boxed{4\pi}$.
Inradius: $r = \dfrac{5 + 12 - 13}{2} = \dfrac{4}{2} = 2$.
Area of inscribed circle: $\pi r^2 = \pi (2)^2 = \boxed{4\pi}$.
Problem 3 · Vieta's Formulas
AMC 12
Algebra
★★★★★★☆
📐 Key Concept
For $x^3 + bx^2 + cx + d = 0$ with roots $r_1, r_2, r_3$ (Vieta's Formulas): $$r_1+r_2+r_3 = -b,\quad r_1r_2+r_1r_3+r_2r_3 = c,\quad r_1r_2r_3 = -d$$ Also: $r_1^2+r_2^2+r_3^2 = (r_1+r_2+r_3)^2 - 2(r_1r_2+r_1r_3+r_2r_3)$.✏️ Worked Example
If $x^3 - 6x^2 + 11x - 6 = 0$, find $r_1^2+r_2^2+r_3^2$.Solution: $\sum r_i = 6$, $\sum r_ir_j = 11$. So $\sum r_i^2 = 36 - 22 = 14$.
14
Problem: The polynomial $x^3 - 9x^2 + 26x - 24 = 0$ has three positive integer roots. What is the sum of the squares of the roots?
💡 Full Solution
By Vieta's: $r_1+r_2+r_3 = 9$, $\;r_1r_2+r_1r_3+r_2r_3 = 26$, $\;r_1r_2r_3 = 24$.
Test integers summing to 9 with product 24: try $\mathbf{2, 3, 4}$.
Check: $2+3+4=9$ ✓ and $6+8+12=26$ ✓
Sum of squares: $2^2 + 3^2 + 4^2 = 4 + 9 + 36 = \boxed{49}$.
Test integers summing to 9 with product 24: try $\mathbf{2, 3, 4}$.
Check: $2+3+4=9$ ✓ and $6+8+12=26$ ✓
Sum of squares: $2^2 + 3^2 + 4^2 = 4 + 9 + 36 = \boxed{49}$.
Problem 4 · Complementary Counting
AMC 8
Probability
★★★★☆☆☆
📐 Key Concept
Complementary Counting: Count what you do NOT want, then subtract from the total. $$P(A) = 1 - P(A^c), \qquad |A| = |\text{Total}| - |A^c|$$✏️ Worked Example
How many 3-digit numbers do NOT contain digit 5?Solution: Hundreds: 8 choices; tens, units: 9 each. Total: $8 \times 9 \times 9 = 648$.
648
Problem: Two fair dice are rolled. What is the probability that the product of the two numbers is even?
💡 Full Solution
Product is even unless both dice show odd numbers.
$$P(\text{both odd}) = \frac{3}{6} \times \frac{3}{6} = \frac{1}{4}$$ $$P(\text{product even}) = 1 - \frac{1}{4} = \boxed{\frac{3}{4}}$$
$$P(\text{both odd}) = \frac{3}{6} \times \frac{3}{6} = \frac{1}{4}$$ $$P(\text{product even}) = 1 - \frac{1}{4} = \boxed{\frac{3}{4}}$$
Problem 5 · GCD, LCM & Primes
AMC 10
Number Theory
★★★★★☆☆
📐 Key Concept
For any two positive integers: $\text{lcm}(a,b) \cdot \gcd(a,b) = a \cdot b$.Use prime factorization: $\gcd$ takes minimum exponents; $\text{lcm}$ takes maximum.
✏️ Worked Example
Find $\gcd(120,180)$ and $\text{lcm}(120,180)$.Solution: $120=2^3\!\cdot\!3\!\cdot\!5$; $180=2^2\!\cdot\!3^2\!\cdot\!5$. $\gcd=60$, $\text{lcm}=360$.
$\gcd=60,\; \text{lcm}=360$
Problem: The LCM of two numbers is 180 and their GCD is 6. If one of the numbers is 36, what is the other?
💡 Full Solution
Use $a \cdot b = \text{lcm}(a,b) \cdot \gcd(a,b)$:
$$36 \times b = 180 \times 6 = 1080 \implies b = \frac{1080}{36} = \boxed{30}$$
Problem 6 · Law of Cosines & Area
AMC 12
Trigonometry
★★★★★★☆
📐 Key Concept
- Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$
- Triangle area: $K = \dfrac{1}{2}ab\sin C$
✏️ Worked Example
In $\triangle ABC$, $a=5, b=7, C=60°$. Find $c$.Solution: $c^2 = 25+49-2(5)(7)(\frac{1}{2})=39$. So $c=\sqrt{39}$.
$c = \sqrt{39}$
Problem: In triangle $ABC$, $a = 8$, $b = 10$, and $C = 60°$. What is the area?
💡 Full Solution
$$\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}(8)(10)\sin 60° = 40 \cdot \frac{\sqrt{3}}{2} = \boxed{20\sqrt{3}}$$
Problem 7 · Work & Rate Problems
AMC 8
Rates
★★★☆☆☆☆
📐 Key Concept
If A takes $a$ hours and B takes $b$ hours alone, working together their rate is $\dfrac{1}{a} + \dfrac{1}{b}$ jobs/hour, so time together $= \dfrac{ab}{a+b}$ hours.✏️ Worked Example
Alice: 4 hr, Bob: 6 hr. Together?Solution: Rate $=\frac{1}{4}+\frac{1}{6}=\frac{5}{12}$. Time $=\frac{12}{5}=2.4$ hr.
2.4 hours
Problem: Pipe A fills a tank in 3 hr; Pipe B fills it in 6 hr; Pipe C drains it in 9 hr. All open simultaneously — how long to fill?
💡 Full Solution
Net rate $= \dfrac{1}{3} + \dfrac{1}{6} - \dfrac{1}{9}$.
Common denominator 18:
$$\frac{6}{18} + \frac{3}{18} - \frac{2}{18} = \frac{7}{18} \text{ tanks/hr}$$
Time to fill $= \dfrac{18}{7} = \boxed{\dfrac{18}{7}}$ hours.
Problem 8 · Chord-Chord Angle
AMC 10
Geometry
★★★★★☆☆
📐 Key Concept
- Inscribed angle $= \frac{1}{2} \times$ intercepted arc
- Chord-chord angle $= \frac{1}{2} \times$ (sum of intercepted arcs)
✏️ Worked Example
Inscribed angle intercepts arc $100°$.Solution: Angle $= \frac{100°}{2} = 50°$.
$50°$
Problem: Chords $AB$ and $CD$ intersect inside a circle at $P$. Arc $AC = 80°$, arc $BD = 120°$. Find $\angle APD$.
💡 Full Solution
When two chords intersect inside a circle, the angle equals half the sum of intercepted arcs:
$$\angle APD = \frac{\text{arc }AC + \text{arc }BD}{2} = \frac{80° + 120°}{2} = \boxed{100°}$$
Problem 9 · Arithmetic Series
AMC 12
Sequences
★★★★★★☆
📐 Key Concept
- Arithmetic series: $S_n = \dfrac{n}{2}(a_1 + a_n)$
- General term: $a_n = S_n - S_{n-1}$ for $n \ge 2$
- Infinite geometric ($|r|<1$): $S = \dfrac{a}{1-r}$
✏️ Worked Example
If $S_n = n^2 + 2n$, find $a_5$.Solution: $a_5 = S_5 - S_4 = 35 - 24 = 11$.
$a_5 = 11$
Problem: The sum of the first $n$ terms of an arithmetic sequence is $S_n = 3n^2 + 5n$. Find $a_{10}$.
💡 Full Solution
$a_n = S_n - S_{n-1} = (3n^2+5n) - [3(n-1)^2+5(n-1)]$
$$= 3n^2+5n - 3n^2+6n-3-5n+5 = 6n+2$$
Therefore $a_{10} = 6(10)+2 = \boxed{62}$.
Problem 10 · Combinations: At Least
AMC 10
Combinatorics
★★★★★☆☆
📐 Key Concept
$\dbinom{n}{k} = \dfrac{n!}{k!(n-k)!}$ counts $k$-element subsets from $n$ elements.For "at least $k$" problems: sum over each valid case.
✏️ Worked Example
Ways to choose 3 ordered officers from 8 people?Solution: $P(8,3) = 8 \times 7 \times 6 = 336$.
336 ways
Problem: A committee of 4 is chosen from 6 men and 5 women. How many committees include at least 2 women?
💡 Full Solution
Sum over each case (exactly 2W, 3W, 4W):
$$\binom{5}{2}\binom{6}{2} + \binom{5}{3}\binom{6}{1} + \binom{5}{4}\binom{6}{0}$$
$$= 10{\cdot}15 + 10{\cdot}6 + 5{\cdot}1 = 150 + 60 + 5 = \boxed{155}$$
Problem 11 · De Moivre's Theorem
AMC 12
Complex Numbers
★★★★★★☆
📐 Key Concept
For $z = r(\cos\theta + i\sin\theta)$, De Moivre's Theorem gives: $$z^n = r^n(\cos n\theta + i\sin n\theta)$$✏️ Worked Example
Find $(1+i)^8$.Solution: $|1+i|=\sqrt{2}$, $\theta=45°$. $(\sqrt{2})^8(\cos360°+i\sin360°) = 16 \cdot 1 = 16$.
$(1+i)^8 = 16$
Problem: What is the value of $(\cos 20° + i\sin 20°)^{18}$?
💡 Full Solution
By De Moivre's Theorem ($r=1$, $\theta=20°$):
$$(\cos20°+i\sin20°)^{18} = \cos(18\times20°)+i\sin(18\times20°)$$
$$= \cos360°+i\sin360° = \boxed{1}$$
Problem 12 · Successive Percent Change
AMC 8
Percents
★★★☆☆☆☆
📐 Key Concept
Successive changes of $+p\%$ then $-q\%$ give net factor: $$\left(1+\frac{p}{100}\right)\!\left(1-\frac{q}{100}\right)$$ Net change $= (\text{factor} - 1)\times 100\%$.✏️ Worked Example
Rise 20%, fall 20%.Solution: $1.2 \times 0.8 = 0.96 \Rightarrow -4\%$ net.
$-4\%$ net decrease
Problem: A store raises prices by 25%, then gives a 20% discount. What is the overall percent change?
💡 Full Solution
Net factor: $1.25 \times 0.80 = 1.00$.
Percent change $= (1.00 - 1) \times 100 = \boxed{0\%}$.
The two changes cancel exactly — price returns to original.
Percent change $= (1.00 - 1) \times 100 = \boxed{0\%}$.
The two changes cancel exactly — price returns to original.
Problem 13 · Inverse Functions
AMC 10
Algebra
★★★★★☆☆
📐 Key Concept
To find $f^{-1}(c)$: solve $f(x)=c$ for $x$.Alternatively, swap $x \leftrightarrow y$ in $y=f(x)$ and solve for $y$.
✏️ Worked Example
$f(x)=2x+3$, find $f^{-1}(x)$.Solution: $y=2x+3 \Rightarrow x=\frac{y-3}{2}$. So $f^{-1}(x)=\frac{x-3}{2}$.
$f^{-1}(x)=\dfrac{x-3}{2}$
Problem: If $f(x) = \dfrac{3x+1}{x-2}$, what is $f^{-1}(5)$?
💡 Full Solution
Solve $f(x) = 5$:
$$\frac{3x+1}{x-2} = 5 \implies 3x+1 = 5x-10 \implies -2x = -11 \implies x = \frac{11}{2}$$
Therefore $f^{-1}(5) = \boxed{\dfrac{11}{2}}$.
Problem 14 · Log Equations & Change of Base
AMC 12
Logarithms
★★★★★★☆
📐 Key Concept
Key log laws: $$\log(ab)=\log a+\log b,\quad \log a^n = n\log a$$ Change of base: $\log_b a = \dfrac{\log a}{\log b}$, so $\log_9 x = \dfrac{\log_3 x}{2}$.✏️ Worked Example
Solve $\log_2 x + \log_2(x-2)=3$.Solution: $x(x-2)=8 \Rightarrow x^2-2x-8=0 \Rightarrow x=4$.
$x = 4$
Problem: If $\log_3 x + \log_9 x = 3$, find $x$.
💡 Full Solution
Convert using change of base: $\log_9 x = \dfrac{\log_3 x}{2}$.
$$\log_3 x + \frac{\log_3 x}{2} = 3 \implies \frac{3}{2}\log_3 x = 3 \implies \log_3 x = 2$$
Therefore $x = 3^2 = \boxed{9}$.
Problem 15 · Volume of Cones
AMC 10
Geometry
★★★★★☆☆
📐 Key Concept
- Cone volume: $V = \dfrac{1}{3}\pi r^2 h$
- Similar solids with scale factor $k$: volumes scale by $k^3$
✏️ Worked Example
Cone with $r=3$, $h=4$. Find volume.Solution: $V = \frac{1}{3}\pi(9)(4) = 12\pi$.
$V = 12\pi$
Problem: A cone (radius 6, height 8) has a similar smaller cone (scale $\frac{1}{2}$) removed from its top. What is the remaining volume?
💡 Full Solution
Large cone: $V_{\text{big}} = \dfrac{1}{3}\pi(36)(8) = 96\pi$.
Small cone ($r=3, h=4$): $V_{\text{small}} = \dfrac{1}{3}\pi(9)(4) = 12\pi$.
Remaining: $96\pi - 12\pi = \boxed{84\pi}$.
Small cone ($r=3, h=4$): $V_{\text{small}} = \dfrac{1}{3}\pi(9)(4) = 12\pi$.
Remaining: $96\pi - 12\pi = \boxed{84\pi}$.
Problem 16 · Perfect Square Divisors
AMC 8
Number Theory
★★★★☆☆☆
📐 Key Concept
A divisor of $n = p_1^{a_1}p_2^{a_2}\cdots$ is a perfect square iff all its prime exponents are even. For each $p_i^{a_i}$: even exponents $0,2,\ldots$ give $\lfloor a_i/2 \rfloor + 1$ choices.✏️ Worked Example
Perfect square divisors of $360 = 2^3\cdot3^2\cdot5$:Solution: $2^3\!$: choices $\{0,2\}$=2; $3^2\!$: $\{0,2\}$=2; $5^1\!$: $\{0\}$=1. Total: $2\times2\times1=4$.
4 perfect square divisors
Problem: How many positive perfect squares divide $5^4 \cdot 7^3 \cdot 11^2$?
💡 Full Solution
Each prime exponent in the divisor must be even:
$5^4$: even exponents $\{0,2,4\}$ → 3 choices
$7^3$: even exponents $\{0,2\}$ → 2 choices
$11^2$: even exponents $\{0,2\}$ → 2 choices
Total: $3 \times 2 \times 2 = \boxed{12}$.
$5^4$: even exponents $\{0,2,4\}$ → 3 choices
$7^3$: even exponents $\{0,2\}$ → 2 choices
$11^2$: even exponents $\{0,2\}$ → 2 choices
Total: $3 \times 2 \times 2 = \boxed{12}$.
Problem 17 · Bayes' Theorem
AMC 12
Probability
★★★★★★☆
📐 Key Concept
$$P(A \mid B) = \frac{P(B \mid A)\,P(A)}{P(B)}$$ where $P(B) = P(B\mid A)P(A) + P(B\mid A^c)P(A^c)$ (Law of Total Probability).✏️ Worked Example
Bag: 3 red, 2 blue. Draw 2 without replacement. Given 1st is red, find $P(\text{2nd red})$.Solution: $\dfrac{2}{4} = \dfrac{1}{2}$.
$\dfrac{1}{2}$
Problem: Machine A makes 60% of items (3% defect rate); Machine B makes 40% (5% defect rate). Given an item is defective, what is the probability it came from Machine B?
💡 Full Solution
$P(D) = 0.03(0.6) + 0.05(0.4) = 0.018 + 0.020 = 0.038$.
$$P(B \mid D) = \frac{P(D \mid B)\,P(B)}{P(D)} = \frac{0.05 \times 0.4}{0.038} = \frac{0.020}{0.038} = \boxed{\dfrac{10}{19}}$$
$$P(B \mid D) = \frac{P(D \mid B)\,P(B)}{P(D)} = \frac{0.05 \times 0.4}{0.038} = \frac{0.020}{0.038} = \boxed{\dfrac{10}{19}}$$
Problem 18 · Quadratic Discriminant
AMC 10
Algebra
★★★★★☆☆
📐 Key Concept
Discriminant $\Delta = b^2 - 4ac$ for $ax^2+bx+c=0$:$\Delta > 0$: two real roots; $\Delta = 0$: one; $\Delta < 0$: no real roots.
✏️ Worked Example
Solve $x^2 - 5x + 6 < 0$.Solution: Roots $x=2,3$. Parabola opens up, so solution is $2 < x < 3$.
$2 < x < 3$
Problem: For how many positive integer values of $k$ does $x^2 + kx + 9 = 0$ have no real solutions?
💡 Full Solution
No real solutions when $\Delta = k^2 - 36 < 0$, i.e.\ $k^2 < 36$, i.e.\ $-6 < k < 6$.
Positive integer values: $k \in \{1, 2, 3, 4, 5\}$ — that is $\boxed{5}$ values.
Positive integer values: $k \in \{1, 2, 3, 4, 5\}$ — that is $\boxed{5}$ values.
Problem 19 · Linear Recurrence
AMC 12
Sequences
★★★★★★★
📐 Key Concept
For $a_n = pa_{n-1} + qa_{n-2}$, solve the characteristic equation $r^2 - pr - q = 0$.If there is a repeated root $r_0$: general solution is $a_n = (A + Bn)\,r_0^{\,n}$.
✏️ Worked Example
$a_1=1, a_2=5, a_n=5a_{n-1}-6a_{n-2}$. Char. eq: $r^2-5r+6=0 \Rightarrow r=2,3$.General solution: $a_n = A\cdot2^n + B\cdot3^n$.
$a_n = A\cdot2^n + B\cdot3^n$
Problem: A sequence has $a_1=1$, $a_2=3$, and $a_n = 4a_{n-1} - 4a_{n-2}$ for $n \ge 3$. Find $a_5$.
💡 Full Solution
Characteristic equation: $r^2-4r+4=0 \Rightarrow (r-2)^2=0$. Repeated root $r_0=2$.
General solution: $a_n = (A+Bn)\cdot 2^n$.
Initial conditions: $a_1=(A+B)\cdot2=1$ and $a_2=(A+2B)\cdot4=3$.
Solving: $A+B=\frac{1}{2}$, $A+2B=\frac{3}{4}$, so $B=\frac{1}{4}$, $A=\frac{1}{4}$.
$$a_5 = \left(\tfrac{1}{4}+\tfrac{5}{4}\right)\cdot 2^5 = \tfrac{6}{4}\cdot32 = \tfrac{3}{2}\cdot32 = \boxed{48}$$
General solution: $a_n = (A+Bn)\cdot 2^n$.
Initial conditions: $a_1=(A+B)\cdot2=1$ and $a_2=(A+2B)\cdot4=3$.
Solving: $A+B=\frac{1}{2}$, $A+2B=\frac{3}{4}$, so $B=\frac{1}{4}$, $A=\frac{1}{4}$.
$$a_5 = \left(\tfrac{1}{4}+\tfrac{5}{4}\right)\cdot 2^5 = \tfrac{6}{4}\cdot32 = \tfrac{3}{2}\cdot32 = \boxed{48}$$
Problem 20 · Shared Foci: Ellipse & Hyperbola
AMC 12
Conic Sections
★★★★★★★
📐 Key Concept
- Ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ($a>b>0$): $c^2 = a^2 - b^2$, foci $(\pm c,0)$
- Hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$: $c^2 = a^2 + b^2$, foci $(\pm c,0)$
✏️ Worked Example
Foci of $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$?Solution: $c^2=25-16=9$, $c=3$. Foci: $(\pm3,0)$.
Foci at $(\pm 3,\, 0)$
Problem: The ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$ and the hyperbola $\dfrac{x^2}{k}-\dfrac{y^2}{9}=1$ share the same foci. Find $k$.
💡 Full Solution
Ellipse: $c^2 = 25-9 = 16 \Rightarrow c=4$. Foci: $(\pm4,0)$.
Hyperbola must share these foci, so $c^2 = k + 9 = 16$: $$k = 16 - 9 = \boxed{7}$$
Hyperbola must share these foci, so $c^2 = k + 9 = 16$: $$k = 16 - 9 = \boxed{7}$$