Seed content: Quant exponents & roots, remainders, and combinatorics (3 lessons, 21 questions)

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-remainders.md

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+---
+id: quant-number-properties-remainders
+section: quant
+topic: number-properties
+subtopic: remainders
+title: "Remainders"
+tags: [remainders, divisibility, modular-arithmetic, number-properties]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Remainder questions look like they need long division, but almost none of them do. The whole
+topic rests on one identity and a few rules for how remainders behave when you **add**,
+**multiply**, and take **powers**. Master those and you can answer most remainder questions in
+under a minute — often just by picking the smallest number that fits and testing it.
+
+## Core concepts
+
+**The division identity.** When a positive integer \(n\) is divided by a positive divisor \(d\),
+there is a unique quotient \(q\) and remainder \(r\) with
+
+\[n = dq + r, \qquad 0 \le r < d.\]
+
+The remainder is always **less than the divisor** (a remainder of 7 is impossible when dividing
+by 5). Reading a word problem, "quotient 7, remainder 4, divisor 6" becomes
+\(n = 6\cdot 7 + 4 = 46\) — nothing more.
+
+**Remainders add and multiply, then reduce.** If you only care about the remainder mod \(d\), you
+may replace any number by its remainder at every step, as long as you **reduce again at the end**:
+
+\[\text{rem}(a+b) = \text{rem}\big(\text{rem}(a) + \text{rem}(b)\big), \qquad
+\text{rem}(ab) = \text{rem}\big(\text{rem}(a)\cdot \text{rem}(b)\big).\]
+
+So if \(a\) leaves remainder 5 and \(b\) leaves remainder 8 when divided by 12, then \(ab\) leaves
+the remainder of \(5\times 8 = 40\), which is \(40 - 36 = 4\). The single most common mistake is
+stopping at 40 (or at \(5+8=13\)) and forgetting the final reduction.
+
+**Powers cycle.** Remainders of \(a^1, a^2, a^3, \dots\) repeat in a short cycle. Dividing by 10
+is the same as asking for the **units digit**, and units digits cycle with period at most 4:
+
+\[7^1 \to 7,\quad 7^2 \to 9,\quad 7^3 \to 3,\quad 7^4 \to 1,\quad 7^5 \to 7,\ \dots\]
+
+To find \(7^{20} \bmod 10\), note the cycle length is 4 and \(20\) is a multiple of 4, so \(7^{20}\)
+lands on the **last** entry of the cycle: remainder **1**. Match the exponent to its position with
+"exponent mod cycle-length," and remember that a remainder of 0 means the *end* of the cycle, not
+the start.
+
+**Pick the smallest number that fits.** "\(n\) leaves remainder 2 when divided by 5" just means
+\(n\) is one of \(2, 7, 12, 17, \dots\). For two conditions at once, list each and find the first
+overlap; the pattern then repeats every \(\text{lcm}\) of the divisors.
+
+## Worked examples
+
+**Translate and compute.** A positive integer divided by 6 gives quotient 7 and remainder 4.
+The integer is \(6\cdot 7 + 4 = 46\). (Swapping quotient and remainder — \(6\cdot 4 + 7 = 31\) — is
+the classic trap.)
+
+**Combine two conditions.** Find the smallest integer above 10 that leaves remainder 2 mod 5 and
+remainder 1 mod 3. Numbers \(\equiv 2 \pmod 5\): \(2, 7, 12, 17, 22, \dots\); of these, the ones
+\(\equiv 1 \pmod 3\) are \(7, 22, 37, \dots\) (they repeat every \(\text{lcm}(5,3)=15\)). The first
+one above 10 is **22**.
+
+**Reason with squares.** If \(n^2\) leaves remainder 4 when divided by 7, what can \(n\) leave?
+Test each possible remainder \(r = 0,1,\dots,6\): \(r^2 \bmod 7\) gives \(0,1,4,2,2,4,1\). The value
+4 appears at \(r = 2\) and \(r = 5\), so \(n\) leaves remainder **2 or 5** — *not* 4. The remainder of
+\(n\) is generally **not** the same as the remainder of \(n^2\).
+
+## Common traps
+
+- **Forgetting to reduce at the end.** \(5\times 8 = 40\) is not the answer when dividing by 12; the
+  remainder is \(40 - 36 = 4\).
+- **Adding remainders when the numbers are multiplied** (or vice versa). Combine remainders the
+  same way you combine the numbers, then reduce.
+- **Swapping quotient and remainder** in \(n = dq + r\).
+- **A remainder as large as the divisor.** Dividing by \(d\), the remainder must be \(0\) to \(d-1\).
+- **Assuming \(n\) and \(n^2\) (or \(n\) and \(2n\)) share a remainder.** They usually don't — test it.
+
+## Key takeaways
+
+- Everything flows from \(n = dq + r\) with \(0 \le r < d\).
+- To combine numbers, combine their remainders **and reduce again**: works for sums and products.
+- Powers of a number cycle; use "exponent mod cycle-length," and treat a result of 0 as the end of
+  the cycle.
+- For "which could be" and two-condition problems, list the smallest values that fit and look for
+  the overlap; it repeats every lcm of the divisors.

content/questions/quant/quant-arithmetic-exponents-roots-0017.md

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+---
+id: quant-arithmetic-exponents-roots-0017
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: easy
+tags: [exponents, power-rules]
+choices:
+  A: "4"
+  B: "8"
+  C: "16"
+  D: "64"
+  E: "512"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(\dfrac{2^{5} \cdot 2^{3}}{2^{6}}\)?
+
+## Explanation
+
+Keep everything in base 2 and just track exponents. When you multiply like bases you **add**
+exponents; when you divide, you **subtract**:
+
+\[\frac{2^{5} \cdot 2^{3}}{2^{6}} = 2^{\,5 + 3 - 6} = 2^{2} = 4\]
+
+**The trap:** answer **E (512)** comes from *multiplying* the exponents in the numerator
+(\(5 \times 3 = 15\), then \(15 - 6 = 9\), giving \(2^{9} = 512\)). You add exponents when the bases
+are multiplied, not multiply them. Answer **D (64)** is \(2^{6}\), what you get if you drop one of
+the numerator terms.
+
+## Hints
+
+- Rewrite the whole expression as a single power of 2.
+- Multiply like bases → add exponents; divide → subtract exponents.

content/questions/quant/quant-arithmetic-exponents-roots-0018.md

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+---
+id: quant-arithmetic-exponents-roots-0018
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: easy
+tags: [exponents, negative-exponent, zero-exponent]
+choices:
+  A: "0"
+  B: "1/9"
+  C: "10/9"
+  D: "2"
+  E: "9"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(3^{-2} + 3^{0}\)?
+
+## Explanation
+
+Two rules do all the work here. A **negative exponent** means take the reciprocal, and **anything
+nonzero raised to the 0 power is 1**:
+
+\[3^{-2} = \frac{1}{3^{2}} = \frac{1}{9}, \qquad 3^{0} = 1\]
+
+So the sum is
+
+\[\frac{1}{9} + 1 = \frac{1}{9} + \frac{9}{9} = \frac{10}{9}\]
+
+**The traps:** answer **B (1/9)** drops the \(3^{0} = 1\) term (a studier who thinks \(3^0 = 0\)).
+Answer **D (2)** comes from mistakenly treating \(3^{-2}\) as \(1\) as well. A negative exponent is
+never a negative number, so **A (0)** and **E (9)** are out.
+
+## Hints
+
+- A negative exponent means "reciprocal," not "negative": \(3^{-2} = \tfrac{1}{3^2}\).
+- Remember that \(3^{0} = 1\), not \(0\).