Formula sheet

A factorial is the product of all positive integers less than or equal to a given positive integer. In other words,

The number of ways of arranging $n$ distinct objects is just

The number of ways to order $k$ objects out of $n$ total objects is

The number of ways of arranging $n$ objects in a circle, where rotations of the same arrangement aren't considered distinct, is

If reflections (mirror images) of the same arrangement aren't considered distinct either, it is

A combination is a collection of items where the order of the items does not matter.

The number of ways to choose $k$ objects out of a total of $n$ objects is

The binomial identity states that

The number of ways to order a word is

where $n$ is the number of letters and $d_1, d_2, d_3, \ldots$ are the number of times each of the duplicate letters that appear in the word.

This works when every outcome is equally likely, like the faces of a fair die.

Often, when a problem asks for the probability of an event, you may have to do casework to find the number of successful outcomes. Then divide by the total number of outcomes, which is usually easy to find.

Casework is solving counting or probability problems by considering the different cases and adding them together. Many counting or probability problems can be solved by dividing a problem into several cases and calculating arrangements and probabilities for each case before summing them together.

You should try casework when there are no other obvious approaches, as it is one of the most common techniques on the AMC 8.

Good cases are disjoint (no outcome is in two cases) and complete (every outcome is in some case). Then the case sizes add up to the answer.

$|A \cup B|$ is the union of elements in both $A$ and $B$ (i.e. all elements that are in set $A$ or set $B$, but duplicates are only written once).

$|A \cap B|$ is the intersection of elements in both $A$ and $B$ (i.e. only those elements which are in both sets).

Given two sets, $|A_1|$ and $|A_2|$,

Basically, we count the number of possibilities in 2 "things" and subtract the duplicates.

Given three sets, $|A_1|, |A_2|, |A_3|$,

In this formula, we count the number of possibilities in 3 "things", subtract the possibilities that are duplicates in all 3 pairs of sets, and add back the number of duplicates that are in all 3 sets.

The same principle applies when counting more than 3 events. For any $n$ events, to find the total number of ways such that any of the events occur:

1. Find the number of ways each individual event occurs in.
2. For every pair of 2 events, subtract the number of ways where both of them occur.
3. For every triplet of 3 events, add back the number of ways where all 3 of them occur.
4. For every quadruplet of 4 events, subtract the number of ways where all 4 occur …
5. For all $n$ events, add (if $n$ is odd) or subtract (if $n$ is even) the number of ways where all of them occur.

Alternate between adding the number of ways and subtracting the number of ways until you reach the case where all $n$ events occur. Whether you add or subtract the number of ways where all $n$ events occur depends on whether $n$ is odd, since we alternate between adding and subtracting.

Stated more formally, if $(A_i)_{1 \le i \le n}$ are finite sets, then:

The number of ways to place $n$ indistinguishable objects into $k$ distinguishable bins is

Stars and bars is extremely useful and can often be adapted to the situation. For example, if each bin has to have at least 1 object in it, give each bin 1 object to start with. Then apply the formula with $n-k$ objects and $k$ distinguishable bins:

The number of squares in a $n \times n$ grid of squares is

The general formula for the number of rectangles of all sizes in a rectangular grid of size $m \times n$ is

The number of ways to get from $(0,0)$ to a point $(m,n)$ moving only up and right is

To solve a problem with recursion, follow these steps:

1. Solve the problem for a few smaller base cases by hand.
2. In general, you need as many base cases as the number of terms on the right-hand side of your recursion.
3. For a given $f(n)$, find a recursive equation in terms of earlier values of the function. Think about the different ways to get to that point.
4. Calculate the values of $f(n)$ one by one until you reach the number you want.

For probability recursions, it can be a good idea to simply make a table and find the probability of being at each location after every time step. Remember: the total probability after each hop must always be $1$.

Equivalently,

If one person can do something in $a$ amount of time, and someone else can do it in $b$ amount of time, together they can do it in

time.

If we have a quadratic of the form $x^2 + bx + c$, it can be factored into $(x + p)(x + q)$ such that $p + q = b$ and $pq = c$. We can generally do this by guess and check.

Then, we just solve the linear equations of the form $x + r = 0$.

If we have a quadratic of the form $x^2 + bx + c$ with 2 roots $r$ and $s$, then $r + s = -b$ and $rs = c$. Sometimes, you may not need to calculate the roots and just find expression in terms of the roots.

The solutions to the quadratic equation

are

Once you've found the system of equations in your word problem and/or the problem itself gives you a system of equation, you should consider one or more of the following methods for solving them.

1. Using Symmetry
2. Substituting Variables
3. Elimination
4. Adding/Subtracting/Multiplying Equations
5. Difference of Squares: $x^2 - y^2 = (x - y)(x + y)$
6. Binomial Expansions: $(x + y)^2 = x^2 + 2xy + y^2$ and $(x - y)^2 = x^2 - 2xy + y^2$
7. Other common factorization tricks (see algebraic number theory chapter for more)
8. Squaring equations if you have square roots
9. Introducing new variables and substituting
10. Graphing equation (Recommend graph paper)

This is one of the most common topics on the AMC 8. The trick to solving these types of problems is to just assign variables to the unknowns in the problem and solve them.

Equivalently,

In general, the terms of an arithmetic sequence can be represented as:

where

• $d$ is the common difference between consecutive terms
• $n$ is the number of terms

which basically means the $n$th term of an arithmetic sequence is equal to

We also have that

which means

Essentially,

which essentially means

If the number of terms is even, the average $x$ = the average of the middle 2 terms.

If the number of terms is odd, the average $x$ = the middle term.

which essentially means

We can also substitute $a_n = a_1 + (n-1)d$ to get

In simple terms, the

To intuitively think about it, just take $2$ common from each term:

In simple terms, the

In general, the terms of a geometric sequence can be represented as:

where

• $r$ is the common ratio between consecutive terms
• $n$ is the number of terms

which basically means

A general form for calculating the $n$th term:

which basically means

Essentially,

Number of Terms = $1$ more than the number of times we needed to multiply $r$ by $g_1$ to get $g_n$.

which essentially means

For $-1 < r < 1$,

The strategy for these problems is just to break the sequence into multiple geometric sequences. Another strategy is to let the expression be $S$. Then, divide $S$ by some number such that when you subtract it from $S$, most of the terms cancel. This strategy is faster, but harder to figure out.

After arranging the numbers in increasing or decreasing order:

If the number of terms is odd,

If the number of terms is even,

Expand the first few and last few terms, and cancel out any terms you see.

Primes are numbers that have exactly two factors: $1$ and the number itself.

Example: $2, 3, 5, 7, 11, 13, 17, 19, 23$ are all primes.

Note: $1$ is not a prime (it has only one factor), and $2$ is the only even prime.

The number of factors of a prime $p$ in $n!$ is

Here $\lfloor x \rfloor$ means "round $x$ down to a whole number".

If the prime factorization of the number is expressed as:

then the number of factors of this number is

Basically, in order to find the number of factors of a number:

1. Find the prime factorization of the number
2. Add 1 to all of the exponents
3. Multiply them together

Here is a logical way to think about this formula: for each prime $p_k$, the exponent can be anything from $0$ to $e_k$. Therefore, there are $e_k + 1$ choices. Therefore, the total number of factors is just the product of all such $e_k + 1$ terms.

For example, in $112 = 2^4 \times 7$, in the factor, the power of $2$ can be $2^0$, $2^1$, $2^2$, $2^3$, or $2^4$. The power of $7$ can be $7^0$ or $7^1$.

Therefore, in total there would be $5 \times 2 = 10$ ways just as we calculated in the previous problem. This logic will be useful when other conditions are imposed.

If the prime factorization of the number is expressed as:

then the sum of factors of this number is

Essentially, for a prime $p$ in the prime factorization, first find the sum of $p^k$ for all possible exponents $k$ in the prime factorization. Then, we will multiply all such sums for all of the primes to get the sum of all the factors of the number. This may seem complicated, but it will make more sense with an example.

The product of factors of a number $n$ where it has $f$ factors (this can be calculated using the number of factors formula) is

The Greatest Common Divisor (GCD) of two or more integers (which are not all zero) is the largest positive integer that divides each of the integers.

Note: This is also known as GCF (Greatest Common Factor), and the terms GCF and GCD are often used interchangeably.

The Least Common Multiple (LCM) of two or more integers (which are not all zero) is the smallest positive integer that is divisible by both the numbers.

Greatest common divisor of $m$ and $n$ $= \gcd(m, n)$ can be found by taking the lowest prime exponents from the prime factorizations of $m$ and $n$.

Least common multiple of $m$ and $n$ $= \operatorname{lcm}(m, n)$ can be found by taking the highest prime exponents from the prime factorizations of $m$ and $n$.

The product of GCD and LCM of two numbers is equal to the product of the two numbers:

If a number is divisible by two numbers $a$ and $b$, it will also be divisible by $\operatorname{lcm}(a, b)$.

The Euclidean algorithm states that

where $x > y$ and $k$ is a positive integer.

means the number $n$ leaves the same remainder as $a$ when divided by $b$.

If $a \equiv x \pmod{n}$ and $b \equiv y \pmod{n}$, then

If $a \equiv x \pmod{n}$, then

We can generally combine different mods together. For example, if $x \equiv 1 \pmod{5}$ and $x \equiv 1 \pmod{7}$, then $x \equiv 1 \pmod{35}$.

To calculate large digit(s) of a number $a^b$, a strategy that may work is to just look for a pattern by computing the first few values of $a^b$ and then seeing that the pattern will repeat for large values of $b$.

With algebraic equations, if you have $n$ variables to solve for, you need $n$ equations. But for Diophantine equations this isn't true, since we have one added condition: every variable is an integer. This condition is the key that lets us solve Diophantine equations.

A palindrome is a number that reads the same forward and backward.

For example, $7$, $44$, $383$ and $12321$ are palindromes, but $123$ is not.

The maximum value that cannot be expressed as the sum of non-negative multiples of $a$ and $b$ is

if $a$ and $b$ are relatively prime.

For relatively prime positive integers $a, b$, there are exactly

positive integers which cannot be expressed in the form $ma + nb$ where $m$ and $n$ are non-negative integers.

• Sum of Angles in Triangle is $180^\circ$
• A triangle with 2 angles equal will have their corresponding sides equal and a triangle with 2 sides equal will have their corresponding angles equal (isosceles triangle)
• Opposite angles in intersecting lines are equal
• Corresponding angles in parallel lines are equal
• The angle made by the arc at the center of the circle is double the angle made by the arc at the boundary of the circle

Complementary angles are a pair of angles with the sum of 90 degrees:

Supplementary angles are a pair of angles with the sum of 180 degrees:

When two lines intersect, the vertical angles are equal. Vertical angles are each of the pairs of opposite angles made by two intersecting lines. "Vertical" in this case means they share the same Vertex (corner point), not the usual meaning of up-down.

Corresponding angles equal: when a line crosses two parallel lines, the angles in matching positions at the two crossings are the same.

The angle formed by an arc in the center or the arc angle is double of the angle formed on the edge.

Every angle on the edge that looks at the same arc is equal.

Inscribed triangle with diameter as one side is always a right triangle. The diameter cuts off an arc of $180^\circ$, so the angle opposite it is $\frac{180^\circ}{2} = 90^\circ$.

The perpendicular bisector of any chord passes through the center. In the figure, the perpendicular bisectors of $AB$ and $CD$ intersect at the center $O$.

• Congruent chords are equidistant from the center of a circle.
• If two chords in a circle are congruent, then their intercepted arcs are congruent.
• If two chords in a circle are congruent, then they determine two central angles that are congruent.

The angle marked in the diagram is half of the difference of the 2 red arcs.

If two chords $AB$ and $CD$ intersect at $P$, then the $\angle BPC$ and $\angle APD$ are equal to the average of the two arcs.

If a tangent $R$ intersects the circle at $Q$, and a chord $QP$ is drawn, then the $\angle RQP$ is equal to half the arc angle.

Equal chords mark out equal arcs.

This basically means that if you have 2 chords of the same length, the sector of the circle they mark out will be equal.

If you connect the center of a circle to the point where the circle and a line are tangent, they will form a right angle.

A triangle with base $b$ and height $h$ has an area of

The height is the perpendicular from the top vertex down to the line of the base.

A triangle with sides $a, b, c$ and semiperimeter $s$ has an area of

The semiperimeter is half the perimeter: $s = \frac{a+b+c}{2}$.

A triangle with inradius $r$ (the radius of the circle that can be inscribed in a triangle) and semiperimeter $s$ has an area of:

Inradius $r$ of a right triangle:

where $a$ and $b$ are the legs of the triangle, and $c$ is the hypotenuse.

A triangle with circumradius $R$ (the radius of the circle that the triangle can be inscribed in) and sides $a, b, c$ has an area of

If the side length of an equilateral triangle is $a$:

This follows directly from the $30-60-90$ triangle.

If the side length of a 45-45-90 triangle is $a$:

If the short leg length of a 30-60-90 triangle is $a$:

If the three sides of a triangle are 13, 14, and 15, it can be divided into two right triangles with side lengths: $5, 12, 13$ and $9, 12, 15$.

Area of this triangle $= \frac{1}{2}\times 14\times 12 = 84$

A right triangle with legs $a$ and $b$ and hypotenuse $c$ satisfies the following relation:

Important Pythagorean Triples

$3, 4, 5$ · $5, 12, 13$ · $7, 24, 25$ · $8, 15, 17$ · $9, 40, 41$ · $20, 21, 29$

If all numbers in a Pythagorean triple are multiplied by a constant, the resulting numbers still form a Pythagorean triple. For example, these are all Pythagorean triples: $3, 4, 5$ · $6, 8, 10$ · $9, 12, 15$ · $12, 16, 20$ · $15, 20, 25$.

In a right triangle $ABC$ where $B$ is the right angle, let $D$ be the foot of the perpendicular from $B$ to $AC$. The following triangles are similar:

Length of the perpendicular to the hypotenuse: $BD = \frac{AB\cdot BC}{AC}$.

Also note that:

The centroid of a triangle is on the median and it is $\frac{2}{3}$ of the way from one of the vertices to the midpoint of the opposite side.

In a right triangle $ABC$, let the median from point $B$ intersect $AC$ at a point $P$. Then

Basically, in a right triangle $AC$ is the diameter of the circumcircle, and $PC$, $PA$, and $PB$ are radii of the circumcircle.

If the line $AD$ bisects angle $A$, then

Any square with side length $s$ has an area of

and a perimeter of

Any rectangle with base $b$ and height $h$ has an area of

and a perimeter of

A rhombus with diagonals $d_1$ and $d_2$ has an area of

and a perimeter of

A parallelogram with base $b$ and height $h$ has an area of

A trapezoid with 2 bases $b_1$ and $b_2$ and a height $h$ has an area of

A circle with radius $r$ has

An arc of a circle with radius $r$ and angle $a^\circ$:

that is, $\pi \times \text{radius}^2 \times$ (fraction of circle in sector).

that is, $2\pi \times \text{radius} \times$ (fraction of circle in sector).

For congruent triangles:

1. All the angles of the triangles are same
2. All corresponding sides are equal
3. Area, perimeter, inradius, circumradius, etc. are equal

Two triangles are congruent if the three angles and sides in the triangle are the same. In other words, the triangles are the same (not necessarily same orientation). In general, triangles are congruent if:

• AAS congruence: Two angles of the triangles are same and the 1 side next to the 2 angles are equal
• SAS Congruence (Side Angle Side): Two sides are equal and the angle between the sides are equal
• SSS congruence (Side Side Side): All three sides are congruent
• HL congruence (Hypotenuse Leg): In a right triangle, hypotenuse and leg are equal
• LL congruence (LL Leg): In a right triangle, the two legs are equal

Warning: SSA is not a valid triangle congruence: if 2 corresponding sides are congruent and the angle which is not between the 2 congruent sides is equal, then the triangles are not necessarily congruent

Two triangles are similar if the three angles in the triangle are the same. In other words, the triangles are the same shape multiplied by a scale factor.

In general, triangles are similar if:

• AA similarity: Two angles of the triangles are same, which basically means that the third angle will be equal)
• SAS similarity (Side Angle Side): Two sides are proportional and the angle between the sides is equal
• SSS similarity (Side Side Side): All three sides are proportional
• HL similarity (Hypotenuse Leg): In a right triangle, the hypotenuse and leg are proportional
• LL similarity (LL Leg): In a right triangle, the two legs are proportional

Warning: SSA does not mean triangles are similar

An easy way to detect similar triangles is if bases of triangles are parallel and the sides of the triangles are collinear (see the hourglass figure below).

For similar triangles:

1. All the angles of the triangles are same
2. All corresponding sides have same ratio
3. Area ratio is the square of side length ratio

Sum of interior angle of a regular hexagon $= (6-2)\cdot 180 = 720$

Interior angle of a regular hexagon $= \dfrac{(6-2)}{6}\cdot 180 = 120$

Exterior angle of a regular hexagon $= \dfrac{360}{6} = 60$

Area of a regular hexagon $= 6\cdot\dfrac{\sqrt3}{4}s^2$

Length of the diagonal of a regular hexagon $= 2s$

Sum of interior angle of a regular octagon $= (8-2)\cdot 180 = 1080$

Interior angle of a regular octagon $= \dfrac{(8-2)}{8}\cdot 180 = 135$

Exterior angle of a regular octagon $= \dfrac{360}{8} = 45$

Area of a regular octagon $= 2(1+\sqrt2)s^2$

• Divide the shape into “nicer” areas which are easier to calculate
• Extend Lines
  • You generally want to extend lines when they form nicer shapes/areas to work with, such as triangles
• Break up areas
  • A common way to do so is to drop altitudes as doing so generally allows you to form right triangles

• Having equal angles means equal lengths and vice versa
• Be on the lookout for 90 degree angles, as you can use Pythagorean theorem
• Split the length into multiple components by using some of these techniques
  • Drawing extra lines
  • Dropping Altitudes
  • Extending lines to create similar triangles, special triangles, etc. and then subtracting the extra length

Along the way, a face diagonal has length $\sqrt{2}a$.

Unfold the prism: the two bases, plus one long strip whose width is the base perimeter.

The height $h$ is measured straight down from the apex to the base.

When the pyramid has a square base, and all the sides are equal

The two circles give $2\pi r^2$. The side unrolls into a rectangle $2\pi r$ wide and $h$ tall.

which basically means

where $s$ is the lateral or slant height, which can also be written as

The sides of a cone can be unfolded to form a sector a circle as well. The circumference of the base of the cone is equal to the length of the arc of the sector and the slant height of the cone is equal to the radius of the sector.

So the sector's angle is $\dfrac{2\pi r}{2\pi s} \times 360^\circ = \dfrac{r}{s} \times 360^\circ$.