Seed content: Quant number-properties/remainders lesson + 6 questions

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, cyclicity, number-properties]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Remainders show up all over Focus Quant and Data Insights, and almost every question rests on a
+single identity plus a couple of rules for combining remainders. Learn the identity, learn how
+remainders behave under addition and multiplication, and learn the one trap that catches most
+test-takers: assuming a remainder is determined when it isn't.
+
+## 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 \(r\):
+
+\[n = dq + r, \qquad 0 \le r < d\]
+
+The remainder is always at least \(0\) and strictly less than the divisor. So "remainder 7"
+immediately tells you the divisor is **at least 8**.
+
+**Remainders add and multiply.** To find the remainder of a sum or product, you may replace each
+number by its own remainder first, then reduce again at the end:
+
+\[(a + b) \bmod d = \big((a \bmod d) + (b \bmod d)\big) \bmod d\]
+\[(a \times b) \bmod d = \big((a \bmod d) \times (b \bmod d)\big) \bmod d\]
+
+That is why you never need the actual large number — only its remainder. For powers, this is just
+repeated multiplication: \(n^k \bmod d\) depends only on \((n \bmod d)^k \bmod d\).
+
+**Cyclicity for large powers.** The remainders of \(2^1, 2^2, 2^3, \dots\) modulo \(5\) run
+\(2, 4, 3, 1, 2, 4, 3, 1, \dots\) — a repeating block of length \(4\). For any base and divisor
+the remainders of the powers eventually cycle. Find the cycle length, then use the exponent's
+position within the cycle. To locate \(2^{100}\), note \(100\) is a multiple of the period \(4\),
+so it lands on the **last** value of the block.
+
+## Worked examples
+
+**Combining a sum.** When \(x\) is divided by \(12\) the remainder is \(5\), and when \(y\) is
+divided by \(12\) the remainder is \(11\). The remainder of \(x + y\) is
+
+\[(5 + 11) \bmod 12 = 16 \bmod 12 = 4.\]
+
+Note \(16\) is not a legal remainder (it exceeds \(12\)); you must reduce once more.
+
+**A power modulo a small number.** The remainder of \(7^{100}\) divided by \(5\): since
+\(7 \equiv 2 \pmod 5\), this equals \(2^{100} \bmod 5\). The powers of \(2\) cycle with period
+\(4\), and \(100 = 4 \times 25\), so \(2^{100}\) sits at the end of a block — remainder \(1\).
+
+## Common traps
+
+- **Leaving an illegal remainder.** \(5 + 11 = 16\) is not the answer mod \(12\); a remainder must
+  be less than the divisor, so reduce again to \(4\).
+- **Assuming a smaller divisor is determined.** Knowing \(n \bmod 12 = 5\) does **not** fix
+  \(n \bmod 8\): \(n = 5\) gives remainder \(5\), but \(n = 17\) gives remainder \(1\). Unless the
+  new divisor divides the old one, test a few values of \(n = dq + r\) before concluding.
+- **Testing only one value.** When a question asks what "must be true," plug in at least two
+  numbers of the form \(dq + r\) (e.g. \(q = 0\) and \(q = 1\)); a pattern from a single case is
+  often a coincidence.
+
+## Key takeaways
+
+- Everything follows from \(n = dq + r\) with \(0 \le r < d\); a remainder of \(r\) forces the
+  divisor to be greater than \(r\).
+- Reduce each part first, combine, then reduce again — remainders add and multiply.
+- For big powers, find the repeating cycle of remainders and use the exponent's position in it.
+- A remainder for one divisor rarely fixes the remainder for a different divisor — test values before you commit.

content/questions/quant/quant-number-properties-remainders-0017.md

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+---
+id: quant-number-properties-remainders-0017
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: easy
+tags: [remainders, sum]
+choices:
+  A: "0"
+  B: "1"
+  C: "2"
+  D: "3"
+  E: "6"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+When the positive integer \(n\) is divided by 5, the remainder is 3. What is the remainder when
+\(n + 3\) is divided by 5?
+
+## Explanation
+
+Write \(n\) using the division identity: \(n = 5q + 3\) for some non-negative integer \(q\). Then
+
+\[n + 3 = 5q + 6 = 5q + 5 + 1 = 5(q + 1) + 1.\]
+
+The part \(5(q + 1)\) is a multiple of 5, so the remainder is **1**.
+
+A quick check with a concrete value: take \(n = 8\) (which leaves remainder 3 on division by 5).
+Then \(n + 3 = 11\), and \(11 = 5 \times 2 + 1\) — remainder 1.
+
+**The trap:** answer **E (6)** comes from adding the remainders \(3 + 3 = 6\) and stopping. But 6
+is larger than the divisor 5, so it is not a legal remainder — reduce once more: \(6 = 5 + 1\),
+giving 1. Answer **D (3)** comes from assuming adding 3 leaves the remainder unchanged.
+
+## Hints
+
+- Write \(n\) as \(5q + 3\) and substitute into \(n + 3\).
+- A remainder must be less than the divisor — if you get 6, reduce it modulo 5 again.

content/questions/quant/quant-number-properties-remainders-0018.md

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+---
+id: quant-number-properties-remainders-0018
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: easy
+tags: [remainders, possible-values]
+choices:
+  A: "20"
+  B: "22"
+  C: "23"
+  D: "24"
+  E: "25"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+When a positive integer \(n\) is divided by 6, the remainder is 5. Which of the following could
+be the value of \(n\)?
+
+## Explanation
+
+By the division identity, \(n = 6q + 5\) for some non-negative integer \(q\), so the possible
+values are \(5, 11, 17, 23, 29, \dots\) — each is 5 more than a multiple of 6. Equivalently,
+\(n\) leaves remainder 5, so \(n - 5\) must be divisible by 6.
+
+Test each choice by dividing by 6:
+
+- \(20 = 6 \times 3 + 2\) — remainder 2.
+- \(22 = 6 \times 3 + 4\) — remainder 4.
+- \(23 = 6 \times 3 + 5\) — **remainder 5.** ✓
+- \(24 = 6 \times 4 + 0\) — remainder 0.
+- \(25 = 6 \times 4 + 1\) — remainder 1.
+
+Only **23** works.
+
+**The trap:** a number ending in 5 (25) *feels* like it should leave remainder 5, but remainder
+depends on division by 6, not on the last digit. Check \(n - 5\) for divisibility by 6 instead:
+\(23 - 5 = 18 = 6 \times 3\). ✓
+
+## Hints
+
+- Every valid \(n\) has the form \(6q + 5\); equivalently \(n - 5\) is a multiple of 6.
+- Subtract 5 from each choice and ask whether the result is divisible by 6.

content/questions/quant/quant-number-properties-remainders-0019.md

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+---
+id: quant-number-properties-remainders-0019
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: medium
+tags: [remainders, squares, product]
+choices:
+  A: "1"
+  B: "2"
+  C: "4"
+  D: "5"
+  E: "7"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+When the positive integer \(n\) is divided by 9, the remainder is 7. What is the remainder when
+\(n^2\) is divided by 9?
+
+## Explanation
+
+Because remainders multiply, the remainder of \(n^2\) modulo 9 depends only on the remainder of
+\(n\):
+
+\[n^2 \bmod 9 = (n \bmod 9)^2 \bmod 9 = 7^2 \bmod 9 = 49 \bmod 9.\]
+
+Now \(49 = 9 \times 5 + 4\), so the remainder is **4**.
+
+To see why the quotient doesn't matter, expand \(n = 9q + 7\):
+
+\[n^2 = (9q + 7)^2 = 81q^2 + 126q + 49.\]
+
+Both \(81q^2\) and \(126q = 9(14q)\) are multiples of 9, so only the \(49\) contributes to the
+remainder, again giving 4.
+
+**The trap:** answer **E (7)** comes from assuming the remainder is unchanged by squaring. Answer
+**A (1)** can come from mishandling \(49 \bmod 9\) (e.g. using \(9 \times 6 = 54\) and getting a
+negative). Reduce carefully: the largest multiple of 9 below 49 is 45, leaving 4.
+
+## Hints
+
+- The remainder of \(n^2\) depends only on the remainder of \(n\): compute \(7^2 \bmod 9\).
+- Find the largest multiple of 9 that is at most 49, then subtract.

content/questions/quant/quant-number-properties-remainders-0020.md

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+---
+id: quant-number-properties-remainders-0020
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: medium
+tags: [remainders, sum, combining]
+choices:
+  A: "0"
+  B: "2"
+  C: "4"
+  D: "6"
+  E: "8"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+When the positive integer \(x\) is divided by 12, the remainder is 5, and when the positive
+integer \(y\) is divided by 12, the remainder is 11. What is the remainder when \(x + y\) is
+divided by 12?
+
+## Explanation
+
+Remainders add, so start by adding the two remainders, then reduce modulo 12:
+
+\[(x + y) \bmod 12 = (5 + 11) \bmod 12 = 16 \bmod 12.\]
+
+Since \(16 = 12 + 4\), the remainder is **4**.
+
+Concretely, take \(x = 5\) and \(y = 11\): \(x + y = 16 = 12 \times 1 + 4\) — remainder 4. Any
+other valid choice gives the same result, e.g. \(x = 17,\ y = 23\): \(x + y = 40 = 12 \times 3 +
+4\).
+
+**The trap:** answer **E (8)** treats 16 as if it were the remainder and subtracts 8 from 16 by
+mistake; answer that leaves **16** un-reduced isn't even offered because a remainder must be less
+than 12. Always reduce the sum of remainders one more time.
+
+## Hints
+
+- Add the two remainders first: \(5 + 11\).
+- The result exceeds 12, so it is not yet a legal remainder — reduce modulo 12 again.

content/questions/quant/quant-number-properties-remainders-0021.md

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+---
+id: quant-number-properties-remainders-0021
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: hard
+tags: [remainders, cyclicity, powers]
+choices:
+  A: "1"
+  B: "2"
+  C: "3"
+  D: "4"
+  E: "0"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the remainder when \(7^{100}\) is divided by 5?
+
+## Explanation
+
+You never compute \(7^{100}\). Since remainders multiply, replace the base by its remainder:
+\(7 \equiv 2 \pmod 5\), so
+
+\[7^{100} \bmod 5 = 2^{100} \bmod 5.\]
+
+Now list the remainders of the powers of 2 modulo 5 until they repeat:
+
+\[2^1 \equiv 2,\quad 2^2 \equiv 4,\quad 2^3 \equiv 3,\quad 2^4 \equiv 1,\quad 2^5 \equiv 2,\ \dots\]
+
+The block \(2, 4, 3, 1\) repeats with period 4. The exponent 100 is a multiple of 4
+(\(100 = 4 \times 25\)), so \(2^{100}\) lands on the **last** value of the block, which is **1**.
+
+Check the idea in miniature: \(2^4 = 16 = 5 \times 3 + 1\) leaves remainder 1, and every exponent
+that is a multiple of 4 behaves the same way.
+
+**The trap:** answer **B (2)** comes from using the first term of the cycle instead of the term at
+position \(100 \bmod 4\). When the exponent is an exact multiple of the period, use the *last*
+value of the block, not the first.
+
+## Hints
+
+- Reduce the base first: \(7 \equiv 2 \pmod 5\), so you need \(2^{100} \bmod 5\).
+- List remainders of \(2^1, 2^2, 2^3, \dots\) modulo 5 until they cycle, then find where exponent 100 falls in the cycle.

content/questions/quant/quant-number-properties-remainders-0022.md

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+---
+id: quant-number-properties-remainders-0022
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: hard
+tags: [remainders, cannot-be-determined, divisors]
+choices:
+  A: "1"
+  B: "3"
+  C: "5"
+  D: "7"
+  E: "It cannot be determined from the information given."
+answer: E
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+When the positive integer \(n\) is divided by 12, the remainder is 5. What is the remainder when
+\(n\) is divided by 8?
+
+## Explanation
+
+Knowing the remainder for one divisor does **not** fix the remainder for an unrelated divisor. The
+valid values of \(n\) are \(n = 12q + 5\): \(5, 17, 29, 41, \dots\) Divide each by 8:
+
+- \(n = 5\): \(5 = 8 \times 0 + 5\) — remainder **5**.
+- \(n = 17\): \(17 = 8 \times 2 + 1\) — remainder **1**.
+- \(n = 29\): \(29 = 8 \times 3 + 5\) — remainder **5**.
+- \(n = 41\): \(41 = 8 \times 5 + 1\) — remainder **1**.
+
+The remainder alternates between 5 and 1, so it is not determined — the answer is **E**.
+
+This works because 8 does not divide 12. The remainder modulo a new divisor \(d'\) is fixed by the
+remainder modulo \(d\) only when \(d'\) divides \(d\) (for example, \(n \bmod 12 = 5\) *does* fix
+\(n \bmod 6 = 5\), since 6 divides 12). Here \(8 \nmid 12\), so the extra multiples of 12 shift the
+remainder modulo 8.
+
+**The trap:** answer **C (5)** comes from testing only \(n = 5\) and assuming the remainder carries
+over. Always test at least two values of the form \(12q + 5\); the second one (17) already breaks
+the pattern.
+
+## Hints
+
+- List several values \(n = 12q + 5\) and divide each by 8 — do they all give the same remainder?
+- A remainder modulo 12 fixes the remainder modulo \(d'\) only when \(d'\) divides 12; 8 does not.

curriculum.md

@@ -70,7 +70,7 @@ The taxonomy is ordered so you can study a section top to bottom. A balanced pla
 Early days — the bank is growing. Rough coverage so far (a contributor could automate this
 count from the front-matter):
 
-- **Quant** — arithmetic, number properties, algebra (linear, inequalities), word problems (rates, work, sets), counting & probability, statistics (mean, median, weighted average, standard deviation).
+- **Quant** — arithmetic, number properties (factors, remainders), algebra (linear, quadratics, inequalities), word problems (rates, work, sets), counting & probability, statistics (mean, median, weighted average, standard deviation).
 - **Verbal** — Critical Reasoning (assumptions, strengthen/weaken, inference, flaw) and Reading Comprehension (main idea, detail/inference).
 - **Data Insights** — Data Sufficiency across fundamentals, divisibility, number properties, algebra, statistics, word problems, and overlapping sets.