Seed content: Number Properties & Exponents (+6 questions, +2 lessons)

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 and Roots"
+tags: [exponents, roots, powers, rules-of-exponents]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Exponent questions on the GMAT rarely test heavy computation. They test whether you can
+**rewrite** expressions into a common base and apply a small set of rules cleanly. Get the base
+consistent, and most problems collapse in one or two steps.
+
+## Core concepts
+
+For any nonzero base \(a\) and integers \(m, n\):
+
+- **Product rule** — same base, add exponents: \(a^m \cdot a^n = a^{m+n}\).
+- **Quotient rule** — same base, subtract exponents: \(\dfrac{a^m}{a^n} = a^{m-n}\).
+- **Power of a power** — multiply exponents: \((a^m)^n = a^{mn}\).
+- **Zero exponent** — \(a^0 = 1\) for any \(a \neq 0\).
+- **Negative exponent** — a reciprocal: \(a^{-n} = \dfrac{1}{a^n}\).
+- **Fractional exponent** — a root: \(a^{1/n} = \sqrt[n]{a}\), and \(a^{m/n} = \sqrt[n]{a^m}\).
+
+The single most useful move is **matching bases**. Since \(4 = 2^2\), \(8 = 2^3\), \(9 = 3^2\),
+\(27 = 3^3\), you can turn a messy fraction of different bases into one base and just subtract
+exponents.
+
+Two rules that do **not** exist (the classic traps):
+
+- \(a^m + a^n \neq a^{m+n}\). Addition of powers is **not** a rule — you must factor instead.
+- \((a + b)^2 \neq a^2 + b^2\). Expand it: \((a+b)^2 = a^2 + 2ab + b^2\).
+
+## Worked examples
+
+**Example 1 — match the base.** Simplify \(\dfrac{4^{10}}{2^{15}}\).
+
+Rewrite \(4^{10} = (2^2)^{10} = 2^{20}\). Then \(\dfrac{2^{20}}{2^{15}} = 2^{20-15} = 2^5 = 32\).
+
+**Example 2 — factor a sum of equal powers.** If \(3^n + 3^n + 3^n = 3^7\), find \(n\).
+
+You cannot add the exponents. Instead, three copies of \(3^n\) is \(3 \cdot 3^n = 3^{1+n}\).
+So \(3^{n+1} = 3^7\), which gives \(n + 1 = 7\), so \(n = 6\).
+
+**Example 3 — negative and fractional exponents.** Evaluate \(27^{-2/3}\).
+
+Work outward: \(27^{1/3} = 3\), then \(3^2 = 9\), and the negative sign takes the reciprocal:
+\(27^{-2/3} = \dfrac{1}{9}\).
+
+## Common traps
+
+- **Adding powers as if it were a rule.** When you see \(2^x + 2^x\), factor: it equals
+  \(2 \cdot 2^x = 2^{x+1}\), not \(2^{2x}\) or \(4^x\)... well, \(2^{x+1}\) *is* correct; the
+  point is to factor rather than invent an addition rule.
+- **Forgetting to convert to a common base** before applying the quotient rule. \(\tfrac{4^{10}}{2^{15}}\)
+  is not \(2^{10-15}\) — the numerator's base must become 2 first.
+- **Mishandling the negative sign vs. the exponent.** \(-2^4 = -16\) (the exponent binds before
+  the sign), but \((-2)^4 = 16\).
+- **Answering the wrong quantity.** If they ask for \(2^{x+3}\) and you find \(x\), remember to
+  finish the substitution.
+
+## Key takeaways
+
+- Convert everything to a **common base**, then add/subtract/multiply exponents by the rules.
+- **Sums of powers are factored, not combined** — there is no addition rule for exponents.
+- A fractional exponent is a root; a negative exponent is a reciprocal. Apply them one at a time.
+- Memorize small powers of 2 and 3 (\(2^1\ldots2^{10}\), \(3^1\ldots3^4\)) to move quickly.

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, cyclicity]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+A remainder question is really a question about **structure**: what an integer looks like when
+you write it in terms of the divisor. Once you write \(n = dq + r\), most remainder problems
+become simple substitution or a short search over cases.
+
+## 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 at least 0 and strictly less than the divisor. "Divided by 5" means the
+remainder is one of \(0, 1, 2, 3, 4\).
+
+**Remainders travel through addition and multiplication.** If you only care about the remainder
+of a sum or product, you can replace each number by its own remainder first:
+
+- remainder of \((a + b)\) = remainder of (remainder of \(a\) + remainder of \(b\));
+- remainder of \((a \cdot b)\) = remainder of (remainder of \(a\) \(\times\) remainder of \(b\)).
+
+Then reduce again if the result is still \(\ge d\). This lets you handle huge numbers without
+ever computing them.
+
+**Units digits cycle.** The units digit of a power repeats on a short cycle, which is a
+remainder-mod-10 fact. For example, powers of 3 cycle \(3, 9, 7, 1, 3, 9, 7, 1, \ldots\) with
+period 4.
+
+## Worked examples
+
+**Example 1 — substitute the form.** When \(n\) is divided by 5 the remainder is 3. What is the
+remainder when \(3n\) is divided by 5?
+
+Write \(n = 5k + 3\). Then \(3n = 15k + 9\). Since \(15k\) is a multiple of 5, only the 9
+matters, and \(9 = 5 + 4\), so the remainder is **4**. (Shortcut with the multiplication rule:
+remainder of \(3 \times 3 = 9\), reduce mod 5 to get 4.)
+
+**Example 2 — combine two conditions.** A number leaves remainder 2 when divided by 3 and
+remainder 2 when divided by 4. What is the smallest such positive integer greater than 2?
+
+Numbers that are 2 more than a multiple of both 3 and 4 are 2 more than a multiple of 12:
+\(12m + 2\). The smallest above 2 is \(14\).
+
+**Example 3 — units digit via cyclicity.** What is the units digit of \(3^{23}\)?
+
+The cycle of the units digit of powers of 3 has length 4: \(3, 9, 7, 1\). Divide the exponent by
+4: \(23 = 4 \cdot 5 + 3\), so we are at position 3 in the cycle, which is **7**.
+
+## Common traps
+
+- **Remainder of at least the divisor.** A remainder when dividing by 5 can never be 5 or more.
+  If your arithmetic gives 9, reduce it (\(9 \bmod 5 = 4\)).
+- **Assuming the remainder is unchanged when you scale.** Multiplying \(n\) by 3 does *not* keep
+  the remainder at 3 — you must recompute (Example 1).
+- **Zero is a valid remainder.** "Divisible by \(d\)" is exactly "remainder 0."
+- **Off-by-one on cyclicity.** A remainder of 0 in the exponent lands on the *last* term of the
+  cycle, not the first.
+
+## Key takeaways
+
+- Write \(n = dq + r\) with \(0 \le r < d\); substitute and simplify.
+- Remainders pass cleanly through addition and multiplication — reduce early, reduce often.
+- For units digits of powers, find the cycle length and take the exponent mod that length.
+- Always sanity-check that your final remainder is less than the divisor.

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, powers-of-two, substitution]
+choices:
+  A: "35"
+  B: "64"
+  C: "128"
+  D: "256"
+  E: "512"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(2^x = 32\), what is the value of \(2^{x+3}\)?
+
+## Explanation
+
+You don't even need to solve for \(x\). Use the product rule to split the exponent:
+\[2^{x+3} = 2^x \cdot 2^3.\]
+
+You're told \(2^x = 32\), and \(2^3 = 8\), so
+\[2^{x+3} = 32 \cdot 8 = 256.\]
+
+(If you prefer to find \(x\): \(2^x = 32 = 2^5\), so \(x = 5\), and \(2^{5+3} = 2^8 = 256\).)
+**Correct: (D).**
+
+**The trap:** choice **(A) 35** comes from adding \(32 + 3\) — treating the exponent's "+3" as
+if it were added to the value. Exponents add when powers of the same base are *multiplied*, so
+the "+3" multiplies by \(2^3 = 8\).
+
+## Hints
+
+- \(2^{x+3}\) can be rewritten as \(2^x \cdot 2^3\) — and you already know \(2^x\).
+- "+3" in the exponent is not "+3" in the value; it means one more factor of \(2^3\).

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: medium
+tags: [exponents, factoring, same-base]
+choices:
+  A: "5"
+  B: "6"
+  C: "7"
+  D: "8"
+  E: "21"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(3^n + 3^n + 3^n = 3^7\), what is the value of \(n\)?
+
+## Explanation
+
+There is no rule that lets you add exponents when powers are *added*, so don't touch the 7 yet.
+Instead, notice the left side is three identical copies of \(3^n\):
+\[3^n + 3^n + 3^n = 3 \cdot 3^n.\]
+
+Now use the product rule, since \(3 = 3^1\):
+\[3 \cdot 3^n = 3^{1} \cdot 3^{n} = 3^{\,n+1}.\]
+
+So the equation becomes \(3^{\,n+1} = 3^7\). Same base means equal exponents:
+\[n + 1 = 7 \implies n = 6.\]
+**Correct: (B).**
+
+Why the others fail:
+
+- **(C) 7** treats \(3^n + 3^n + 3^n\) as if it were just \(3^n\), ignoring the coefficient of 3.
+- **(E) 21** comes from multiplying the "3 copies" into the exponent (\(3 \times 7\)) — combining
+  the count of terms with the exponent, which no rule permits.
+- **(A) 5** and **(D) 8** are off-by-one slips on \(n + 1 = 7\).
+
+## Hints
+
+- You can't add the exponents. Factor the left side: how many copies of \(3^n\) are there?
+- Three copies of \(3^n\) is \(3 \cdot 3^n\), and \(3 = 3^1\).

content/questions/quant/quant-number-properties-factors-multiples-0021.md

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+---
+id: quant-number-properties-factors-multiples-0021
+section: quant
+topic: number-properties
+subtopic: factors-multiples
+type: problem-solving
+difficulty: hard
+tags: [factors-multiples, counting, inclusion-exclusion]
+choices:
+  A: "30"
+  B: "33"
+  C: "34"
+  D: "50"
+  E: "66"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+How many positive integers less than 100 are divisible by neither 2 nor 3?
+
+## Explanation
+
+The integers in play are \(1, 2, \ldots, 99\) — that's 99 numbers. It's easier to **count what to
+throw away** (multiples of 2 or 3) and subtract, using inclusion–exclusion.
+
+- Divisible by 2: \(\left\lfloor \tfrac{99}{2} \right\rfloor = 49\).
+- Divisible by 3: \(\left\lfloor \tfrac{99}{3} \right\rfloor = 33\).
+- Divisible by both 2 and 3 (i.e. by 6): \(\left\lfloor \tfrac{99}{6} \right\rfloor = 16\).
+
+By inclusion–exclusion, the count divisible by 2 **or** 3 is
+\[49 + 33 - 16 = 66.\]
+
+So the count divisible by **neither** is
+\[99 - 66 = 33.\]
+**Correct: (B).**
+
+**Traps:**
+
+- **(E) 66** is the number divisible by 2 or 3 — the complement of what's asked.
+- **Forgetting to subtract the overlap (16)** double-counts multiples of 6 and gives \(99 - 82 = 17\),
+  or leads to other wrong totals; inclusion–exclusion prevents this.
+
+## Hints
+
+- Count the "bad" numbers (multiples of 2 or 3) and subtract from 99.
+- Don't double-count: multiples of 6 are in both the "÷2" and "÷3" groups, so subtract them once.

content/questions/quant/quant-number-properties-odds-evens-signs-0020.md

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+---
+id: quant-number-properties-odds-evens-signs-0020
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+type: problem-solving
+difficulty: medium
+tags: [odd-even, parity, consecutive-integers]
+choices:
+  A: "n²"
+  B: "n + 2"
+  C: "3n"
+  D: "n² + n"
+  E: "2n + 1"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(n\) is an odd integer, which of the following must be even?
+
+## Explanation
+
+Track the parity of each expression when \(n\) is odd.
+
+- **(A)** \(n^2\): odd \(\times\) odd \(=\) odd. Not even.
+- **(B)** \(n + 2\): odd \(+\) even \(=\) odd. Not even.
+- **(C)** \(3n\): odd \(\times\) odd \(=\) odd. Not even.
+- **(D)** \(n^2 + n = n(n + 1)\): this is a product of **two consecutive integers**, \(n\) and
+  \(n + 1\). Of any two consecutive integers, exactly one is even, so their product is always
+  even. **Even — correct.**
+- **(E)** \(2n + 1\): even \(+\) odd \(=\) odd. Not even.
+
+So only (D) is guaranteed even. **Correct: (D).**
+
+Note the reasoning for (D) doesn't even require \(n\) to be odd — \(n^2 + n = n(n+1)\) is even for
+*every* integer \(n\), because consecutive integers always include one even number. Recognizing
+\(n^2 + n\) as \(n(n+1)\) is the key move.
+
+## Hints
+
+- Test each choice with a small odd number like \(n = 3\), then confirm it always holds.
+- Factor (D): \(n^2 + n = n(n+1)\). What's special about two consecutive integers?
+