Seed content: 8 Quant bundles — exponents-roots, remainders, combinatorics, mixtures, interest, sequences-functions, odds-evens-signs, and translation (8 lessons, 56 questions)

content/lessons/quant/quant-algebra-sequences-functions.md

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+---
+id: quant-algebra-sequences-functions
+section: quant
+topic: algebra
+subtopic: sequences-functions
+title: "Sequences & Functions"
+tags: [sequences, arithmetic-sequence, geometric-sequence, recursion, functions, function-notation]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Two closely related ideas show up together on GMAT Focus Quant: **sequences** (ordered lists of
+numbers built by a rule) and **functions** (a rule that turns an input into an output). Both reward
+the same discipline — read the rule exactly, mind the indexing, and don't do more arithmetic than the
+question needs. The traps are almost always **off-by-one** errors and **misreading the rule**, not
+hard computation.
+
+## Core concepts
+
+**Arithmetic sequences** add a fixed **common difference** \(d\) each step. If the first term is
+\(a_1\), the \(n\)th term is
+
+\[a_n = a_1 + (n-1)\,d.\]
+
+The multiplier is \((n-1)\), **not** \(n\): you take \(d\) one fewer time than the term number,
+because the first term takes zero steps. To **sum** the first \(n\) terms, average the first and last
+term and multiply by how many there are:
+
+\[S_n = \frac{n(a_1 + a_n)}{2}.\]
+
+**Geometric sequences** multiply by a fixed **common ratio** \(r\) each step:
+
+\[a_n = a_1 \, r^{\,n-1}.\]
+
+Again the exponent is \(n-1\), not \(n\). Terms grow (or shrink) by a constant *factor*, so they
+change far faster than an arithmetic sequence's constant *steps*.
+
+**Recursive sequences** define each term from the previous one, e.g. \(a_{n+1} = 2a_n - 1\) with a
+given starting value. There is no shortcut unless you spot a pattern — just apply the rule one step at
+a time, and **stop at the right term** (writing out \(a_1, a_2, a_3, \dots\) with labels prevents the
+usual off-by-one slip).
+
+**Functions** are rules written as \(f(x)\). To evaluate \(f(\text{something})\), substitute that
+"something" for **every** \(x\) in the definition. For a **composition** \(f(f(x))\), work
+**inside-out**: compute the inner \(f\) first, then feed its result back into \(f\).
+
+## Worked examples
+
+**Arithmetic term.** First term \(a_1 = 7\), common difference \(d = 4\). The 20th term is
+\(a_{20} = 7 + (20-1)\cdot 4 = 7 + 76 = 83\) — not \(7 + 20\cdot 4 = 87\). The \((n-1)\) is the whole
+game.
+
+**Geometric term.** First term \(3\), ratio \(2\). The 6th term is \(3\cdot 2^{6-1} = 3\cdot 32 = 96\),
+not \(3\cdot 2^6 = 192\).
+
+**Recursion.** With \(a_1 = 2\) and \(a_{n+1} = 2a_n - 1\): \(a_2 = 3,\; a_3 = 5,\; a_4 = 9,\;
+a_5 = 17\). Label each step so you hand back \(a_5\), not \(a_4\) or \(a_6\).
+
+**Composition.** If \(f(x) = 2x - 3\), then \(f(f(4))\): first \(f(4) = 2\cdot 4 - 3 = 5\), then
+\(f(5) = 2\cdot 5 - 3 = 7\). Inner first, then outer — and don't drop the \(-3\) either time.
+
+## Common traps
+
+- **Off-by-one in the term formula.** Using \(a_1 + n\,d\) or \(a_1 r^{\,n}\) instead of the
+  \((n-1)\) version overshoots by exactly one step. This is the single most common error here.
+- **Forgetting the first term.** Computing \((n-1)d\) or \(r^{\,n-1}\) but not adding/multiplying the
+  first term.
+- **Dropping ÷2 in the arithmetic sum,** or using the wrong count of terms (the number of terms from
+  \(a\) to \(b\) with step \(d\) is \(\frac{b-a}{d} + 1\), not \(\frac{b-a}{d}\)).
+- **Answering the wrong quantity.** A problem that gives two terms and asks for the *first term* is
+  easy to finish by reporting \(d\) instead. Reread what's being asked.
+- **Composition outside-in or stopping early.** \(f(f(x))\) is not \(f(x)\), and it is not
+  \([f(x)]^2\); evaluate the inner function first, then the outer.
+
+## Key takeaways
+
+- Arithmetic: \(a_n = a_1 + (n-1)d\); sum \(= \dfrac{n(a_1+a_n)}{2}\).
+- Geometric: \(a_n = a_1 r^{\,n-1}\); terms change by a constant factor, so they move fast.
+- Recursive sequences: apply the rule step by step and **stop at the right index** — label your terms.
+- Functions: substitute for every \(x\); compose **inside-out** and keep the constant term.
+- Nearly every miss here is an **off-by-one** or a **misread rule**, not hard arithmetic — slow down on
+  the indexing.

content/lessons/quant/quant-arithmetic-exponents-roots.md

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+---
+id: quant-arithmetic-exponents-roots
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+title: "Exponents & Roots"
+tags: [exponents, roots, radicals, powers]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Most GMAT Focus exponent questions reward you for **rewriting everything in the same base** and
+then just tracking the exponents. Almost nothing here needs a calculator — the whole game is a
+handful of rules applied cleanly and, above all, **not confusing multiplication of bases with
+addition of exponents**.
+
+## Core concepts
+
+**The power rules.** For any nonzero base \(a\):
+
+\[a^m \cdot a^n = a^{m+n}, \qquad \frac{a^m}{a^n} = a^{m-n}, \qquad (a^m)^n = a^{mn}\]
+
+Read them carefully: you **add** exponents when you *multiply* like bases, and you **multiply**
+exponents when you *raise a power to a power*. Mixing these two up is the single most common
+exponent error on the test.
+
+**Zero and negative exponents.**
+
+\[a^0 = 1 \ (a \neq 0), \qquad a^{-n} = \frac{1}{a^n}\]
+
+A negative exponent means "reciprocal," not "negative number": \(2^{-3} = \tfrac{1}{8}\), not \(-8\).
+
+**Fractional exponents are roots.**
+
+\[a^{1/n} = \sqrt[n]{a}, \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^{m} = \sqrt[n]{a^{m}}\]
+
+So \(27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9\). Take the root first when it keeps the numbers
+small.
+
+**Roots simplify by pulling out perfect squares.** \(\sqrt{ab} = \sqrt{a}\,\sqrt{b}\), so
+
+\[\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}\]
+
+You can only **add or subtract** radicals when the part under the root matches:
+\(5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}\). You may **never** merge \(\sqrt{a} + \sqrt{b}\) into
+\(\sqrt{a+b}\).
+
+**Factoring out a common power** cracks the hardest questions. Terms like \(2^n + 2^n\) are not a
+new base — they collapse:
+
+\[2^n + 2^n = 2 \cdot 2^n = 2^{n+1}, \qquad 3^{k+1} - 3^{k} = 3^{k}(3 - 1) = 2\cdot 3^{k}\]
+
+## Worked examples
+
+**Same-base equation.** Solve \(2^{x+1} = 8^{x-1}\). Rewrite \(8 = 2^3\), so the right side is
+\(2^{3(x-1)} = 2^{3x-3}\). Same base ⟹ equal exponents: \(x + 1 = 3x - 3\), giving \(2x = 4\) and
+\(x = 2\).
+
+**Collapse a sum of equal powers.** If \(2^n + 2^n + 2^n + 2^n = 2^{12}\), the left side is
+\(4 \cdot 2^n = 2^2 \cdot 2^n = 2^{n+2}\). So \(n + 2 = 12\) and \(n = 10\).
+
+**Ratio of powers.** With \(x = 3^{20}\) and \(y = 3^{18}\), \(\dfrac{x}{y} = 3^{20-18} = 3^2 = 9\).
+
+## Common traps
+
+- **Multiplying exponents when you should add them** (and vice versa). \(2^5 \cdot 2^3 = 2^8\), not
+  \(2^{15}\).
+- **Reading a negative exponent as a negative number.** \(3^{-2} = \tfrac{1}{9}\), and \(3^0 = 1\)
+  (not \(0\)).
+- **Merging unlike radicals.** \(\sqrt{50} + \sqrt{18} \neq \sqrt{68}\); simplify each first, then add.
+- **Treating a sum of powers as a new base.** \(2^n + 2^n\) is \(2^{n+1}\), not \(4^n\) or \(2^{2n}\).
+
+## Key takeaways
+
+- Get everything into a **common base**, then just manipulate exponents.
+- Multiply like bases → **add** exponents; raise a power to a power → **multiply** exponents.
+- \(a^0 = 1\), \(a^{-n} = 1/a^n\), and \(a^{m/n} = \sqrt[n]{a^m}\).
+- Simplify radicals by pulling out perfect squares; only combine radicals with the same radicand.
+- Sums of equal powers **factor**: \(k\) copies of \(a^n\) is \(k \cdot a^n\).

content/lessons/quant/quant-counting-probability-combinatorics.md

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+---
+id: quant-counting-probability-combinatorics
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+title: "Combinatorics: Counting Arrangements and Selections"
+tags: [combinatorics, permutations, combinations, factorial, counting]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Combinatorics is the art of counting outcomes **without listing them all**. On GMAT Focus
+Quant the whole game is deciding a single question: **does order matter?** Get that right and
+pick the matching tool, and most counting problems collapse into one line of arithmetic.
+
+## Core concepts
+
+**The factorial.** The number of ways to arrange \(n\) distinct objects in a row is
+\(n!\) (read "\(n\) factorial"):
+
+\[n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1\]
+
+So \(5! = 5\cdot4\cdot3\cdot2\cdot1 = 120\). By convention \(0! = 1\).
+
+**Permutations — order matters.** The number of ways to arrange \(k\) objects chosen from
+\(n\) distinct objects, where the order of the chosen objects matters, is
+
+\[P(n,k) = \frac{n!}{(n-k)!} = n(n-1)\cdots(n-k+1)\]
+
+Use permutations for rankings, seatings, passwords, or codes — anything where "A then B"
+differs from "B then A."
+
+**Combinations — order does NOT matter.** The number of ways to *choose* \(k\) objects from
+\(n\), ignoring order, is
+
+\[C(n,k) = \binom{n}{k} = \frac{n!}{k!\,(n-k)!}\]
+
+Use combinations for committees, teams, handshakes, or any unordered selection. Notice
+\(C(n,k) = \dfrac{P(n,k)}{k!}\): a combination is a permutation with the \(k!\) reorderings of
+the same group divided out.
+
+**Two governing principles.**
+
+- **Multiplication (AND).** If a task is a sequence of independent stages, multiply the counts.
+  Choosing 3 seniors *and* 2 juniors: multiply \(C(5,3)\times C(4,2)\).
+- **Addition (OR).** If outcomes fall into mutually exclusive cases, add the counts.
+
+**Identical items.** When some objects repeat, divide by the factorial of each repeat count.
+Distinct arrangements of a word with letter counts \(n_1, n_2, \dots\) totalling \(n\) letters:
+
+\[\frac{n!}{n_1!\,n_2!\cdots}\]
+
+**Circular arrangements.** Seating \(n\) people around a round table gives \((n-1)!\)
+arrangements, because rotations of the same circle are not distinct — fix one person, then
+arrange the rest.
+
+## Worked examples
+
+**Combination.** A book club picks 3 of its 8 members to lead a discussion. Order does not
+matter, so \(C(8,3) = \dfrac{8\cdot7\cdot6}{3!} = \dfrac{336}{6} = 56\) ways.
+
+**Permutation.** Those same 8 members award a gold, silver, and bronze badge — now order
+matters, so \(P(8,3) = 8\cdot7\cdot6 = 336\) ways. Same people, \(3! = 6\) times as many
+outcomes because the badges are distinguishable.
+
+**Complement ("at least").** From 4 men and 3 women, choose a 3-person team with at least one
+woman. Count the total and subtract the unwanted case: \(C(7,3) - C(4,3) = 35 - 4 = 31\).
+
+## Common traps
+
+- **Order confusion.** The single biggest error. A committee is unordered (combination); a
+  lineup or ranking is ordered (permutation). Using \(P\) where \(C\) belongs overcounts by
+  exactly \(k!\).
+- **Adding instead of multiplying.** "3 seniors AND 2 juniors" multiplies: \(C(5,3)\times C(4,2)\),
+  not \(C(5,3)+C(4,2)\).
+- **Forgetting repeated items.** Arranging the letters of a word with repeats requires dividing
+  \(n!\) by the factorial of *each* repeated letter's count — miss one and you overcount.
+- **"At least one" head-on.** Counting cases directly often double-counts. Use the complement:
+  total minus "none."
+- **Blocks that must stay together.** Glue the block into one unit, arrange the units, then
+  multiply by the internal arrangements of the block.
+
+## Key takeaways
+
+- Ask first: **does order matter?** Yes → permutation \(P(n,k)\); no → combination \(C(n,k)\).
+- \(C(n,k) = \dfrac{P(n,k)}{k!}\) — dividing out the reorderings is the whole difference.
+- Independent stages **multiply**; mutually exclusive cases **add**.
+- Repeated identical items → divide by each repeat's factorial; round tables → \((n-1)!\).
+- For "at least one," count the complement and subtract.

content/lessons/quant/quant-number-properties-odds-evens-signs.md

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+---
+id: quant-number-properties-odds-evens-signs
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+title: "Odds, Evens & Signs"
+tags: [parity, odd-even, signs, number-properties]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Parity (whether a number is odd or even) and sign rules are the quiet workhorses of GMAT Focus
+number-properties questions. On their own they look trivial, but the exam hides them inside
+"must be true," "could be true," and Data-Insights logic problems where you never learn the actual
+numbers. Knowing the rules cold lets you answer those without any arithmetic at all.
+
+## Core concepts
+
+**Parity under addition and subtraction.** The result is even exactly when the two numbers share
+the same parity:
+
+- even ± even = **even**
+- odd ± odd = **even**
+- even ± odd = **odd**
+
+A useful shortcut: adding or subtracting an **even** number never changes parity; adding or
+subtracting an **odd** number always flips it.
+
+**Parity under multiplication.** A product is even as soon as **one** factor is even:
+
+- even × (anything) = **even**
+- odd × odd = **odd**
+
+So a product of integers is odd only if *every* factor is odd.
+
+**Sums of several odd numbers.** Adding \(k\) odd numbers gives an **odd** total when \(k\) is odd,
+and an **even** total when \(k\) is even. (Each odd number flips the parity once.) Even numbers in
+the sum never affect the parity.
+
+**Consecutive integers.** Among any two consecutive integers \(n\) and \(n+1\), exactly one is
+even, so the product \(n(n+1)\) is **always even**. This is why expressions like \(n^2 + n\) are
+guaranteed even.
+
+**Sign rules for products.** The sign of a product depends only on **how many negative factors**
+there are:
+
+- an **even** number of negative factors → the product is **positive**
+- an **odd** number of negative factors → the product is **negative**
+- any factor of \(0\) → the product is \(0\)
+
+**Sign logic with two variables.** Two facts appear constantly:
+
+- \(xy > 0\) means \(x\) and \(y\) have the **same sign** (both positive or both negative).
+- \(xy < 0\) means they have **opposite signs**.
+
+Combine that with a fact about \(x + y\): if \(xy > 0\) and \(x + y < 0\), the shared sign must be
+negative, so **both are negative**.
+
+## Worked examples
+
+**"Must be odd."** If \(n\) is even, is \(n + 1\) odd? Adding the odd number \(1\) flips even to
+odd, so yes — always. By contrast \(2n\), \(3n\), \(n^2\), and \(n + 2\) are all still even.
+
+**Counting signs.** Evaluate \((-3)(-4)(-2)\). The magnitude is \(3 \cdot 4 \cdot 2 = 24\); there
+are three negative factors, an odd count, so the product is negative: \(-24\).
+
+**Parity from a product.** If \(mn\) is odd, what do you know? A product is odd only when every
+factor is odd, so both \(m\) and \(n\) are odd — and therefore \(m + n\) is even.
+
+## Common traps
+
+- **Confusing "adds up" with "multiplies."** Four copies of an odd number add to an even total
+  (even count of odds), even though each is odd.
+- **Losing track of signs.** With three negative factors the product is negative, not positive —
+  count the negatives, don't just take the magnitude.
+- **Assuming \(n^2\) is even.** \(n^2\) has the same parity as \(n\); it is odd whenever \(n\) is
+  odd.
+- **Forgetting zero.** "\(xy > 0\)" excludes \(x = 0\) or \(y = 0\); "same sign" quietly rules zero
+  out.
+
+## Key takeaways
+
+- Add/subtract: same parity → even, different parity → odd. Adding an even number never changes
+  parity.
+- A product is even if any factor is even; odd only if every factor is odd.
+- Product sign: even number of negatives → positive, odd number → negative, any zero → zero.
+- \(n(n+1)\) is always even — consecutive integers guarantee an even factor.
+- \(xy > 0\) means same sign; pair it with the sign of \(x + y\) to pin down both.