Add Number Properties content: remainders lesson + 4 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 & Modular Thinking"
+tags: [number-properties, remainders, divisibility, units-digit, cyclicity]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Remainder questions are a GMAT Focus favorite because they reward *thinking in terms of the
+divisor* instead of grinding through big arithmetic. The whole topic rests on one identity: when
+a positive integer \(n\) is divided by a positive divisor \(d\), there is a unique quotient \(q\)
+and remainder \(r\) with
+
+\[n = qd + r, \qquad 0 \le r < d.\]
+
+The two constraints on \(r\) do a lot of work: the remainder is never negative, and it is always
+**strictly less than the divisor**. A "remainder of 7 when divided by 5" is impossible.
+
+## Core concepts
+
+**Write the number in the form \(qd + r\).** The fastest move on almost every remainder problem is
+to replace "\(n\) leaves remainder \(r\) when divided by \(d\)" with the algebraic statement
+\(n = dk + r\) for some integer \(k\), then substitute. You rarely need to know \(n\) itself.
+
+**Remainders add and multiply.** When you divide by \(d\), you can replace any number by its
+remainder before adding or multiplying, then take the remainder again at the end:
+
+\[(a + b) \bmod d = \big((a \bmod d) + (b \bmod d)\big) \bmod d,\]
+\[(a \times b) \bmod d = \big((a \bmod d) \times (b \bmod d)\big) \bmod d.\]
+
+This is why you can find the remainder of a huge product without ever computing it.
+
+**Combining two conditions.** "\(n\) leaves remainder 3 mod 5 and remainder 2 mod 3" pins \(n\)
+down modulo \(15\) (the two divisors multiplied, when they share no common factor). List the
+numbers satisfying the harder condition and scan for the first that also satisfies the other.
+
+**Units digit = remainder mod 10.** The units digit of a number *is* its remainder upon division
+by 10. Powers cycle: \(7^1,7^2,7^3,7^4\) end in \(7, 9, 3, 1\) and then repeat with period 4. To
+find the units digit of \(7^{100}\), reduce the exponent mod the cycle length.
+
+## Worked examples
+
+**Example 1 — substitute the form.** When \(n\) is divided by 7 the remainder is 4. What is the
+remainder when \(2n + 5\) is divided by 7?
+
+Write \(n = 7k + 4\). Then \(2n + 5 = 14k + 8 + 5 = 14k + 13 = 7(2k+1) + 6\). The remainder is
+**6**. (Shortcut: work with the remainder only — \(2\times 4 + 5 = 13\), and \(13 \bmod 7 = 6\).)
+
+**Example 2 — combine conditions.** \(n\) leaves remainder 3 when divided by 5 and remainder 2
+when divided by 3. What is \(n \bmod 15\)?
+
+Numbers that are \(3\) more than a multiple of 5: \(3, 8, 13, 18, \dots\). Check each against
+"remainder 2 mod 3": \(3 \to 0\), \(8 \to 2\ \checkmark\). So \(n \equiv 8 \pmod{15}\).
+
+**Example 3 — cyclicity.** Units digit of \(7^{100}\)? The cycle \(7, 9, 3, 1\) has length 4, and
+\(100\) is a multiple of 4, so \(7^{100}\) lands on the **last** entry of the cycle: units digit
+**1**.
+
+## Common traps
+
+- **Remainder \(\ge\) divisor.** The remainder must satisfy \(0 \le r < d\). If your work produces
+  a remainder as large as the divisor, take one more step.
+- **Forgetting to reduce at the end.** \(2\times 4 + 5 = 13\) is *not* the answer when dividing by
+  7 — you still have to take \(13 \bmod 7 = 6\).
+- **Off-by-one in the cycle.** An exponent that is a multiple of the cycle length maps to the
+  *last* term of the cycle, not the first. \(7^4\) ends in 1, and so does \(7^{100}\).
+- **Assuming you must find \(n\).** You almost never need the actual number — the form \(qd + r\)
+  carries all the information the question uses.
+
+## Key takeaways
+
+- Replace "leaves remainder \(r\) mod \(d\)" with \(n = dk + r\) and substitute; the specific value
+  of \(n\) rarely matters.
+- You may reduce each term to its remainder *before* adding or multiplying, then reduce once more.
+- Two coprime divisors pin a number down modulo their product — list one condition and scan.
+- The units digit is the remainder mod 10; powers cycle, so reduce the exponent by the cycle length.

content/questions/quant/quant-number-properties-odds-evens-signs-0019.md

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+---
+id: quant-number-properties-odds-evens-signs-0019
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+type: problem-solving
+difficulty: medium
+tags: [parity, odd-even, must-be]
+choices:
+  A: "xy"
+  B: "x + 2y"
+  C: "x - y"
+  D: "x^2 + y^2"
+  E: "2x + y"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(x\) and \(y\) are integers and \(x + y\) is odd, which of the following must be even?
+
+## Explanation
+
+A sum of two integers is odd exactly when **one is even and the other is odd**. So among \(x\) and
+\(y\), one is even and one is odd — you don't know which, but you know the parities differ.
+
+Test each option with that fact (and remember: even × anything = even; adding an even number
+never changes parity):
+
+- **(A) \(xy\)** — one factor is even, so the product is **even**. This must be even. ✓
+- (B) \(x + 2y\) — \(2y\) is even, so this has the same parity as \(x\), which could be odd or even.
+  Not determined.
+- (C) \(x - y\) — a difference of one even and one odd is **odd**. Never even.
+- (D) \(x^2 + y^2\) — squaring preserves parity, so this is (even) + (odd) = **odd**. Never even.
+- (E) \(2x + y\) — \(2x\) is even, so this has the same parity as \(y\), which could be odd or even.
+  Not determined.
+
+Only (A) is guaranteed to be even, so the answer is **A**.
+
+The key trap is (C) and (D): both are guaranteed to have a *fixed* parity, but that parity is
+**odd**, not even — the opposite of what the question asks.
+
+## Hints
+
+- "\(x + y\) is odd" forces one of \(x, y\) to be even and the other odd.
+- A product with at least one even factor is always even; adding an even number never flips parity.

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: medium
+tags: [remainders, modular-arithmetic]
+choices:
+  A: "0"
+  B: "1"
+  C: "3"
+  D: "5"
+  E: "6"
+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 7, the remainder is 4. What is the remainder when
+\(2n + 5\) is divided by 7?
+
+## Explanation
+
+You do not need to know \(n\) — only its remainder. Write \(n\) in the standard division form:
+
+\[n = 7k + 4\]
+
+for some non-negative integer \(k\). Substitute into \(2n + 5\):
+
+\[2n + 5 = 2(7k + 4) + 5 = 14k + 8 + 5 = 14k + 13.\]
+
+Now peel off the multiples of 7: \(14k\) is divisible by 7, and \(13 = 7 + 6\), so
+
+\[14k + 13 = 7(2k + 1) + 6.\]
+
+The remainder is **6**.
+
+Faster still: replace \(n\) by its remainder from the start. \(2(4) + 5 = 13\), and
+\(13 \bmod 7 = 6\). Reducing to the remainder before *and* after the arithmetic gives the same
+answer with smaller numbers.
+
+The trap answer is choosing 13 (or reducing incorrectly) — remember the remainder must be less
+than the divisor 7, so you have to take one more step to get 6.
+
+## Hints
+
+- Write \(n = 7k + 4\) and substitute; the value of \(k\) will drop out.
+- Once you reach \(14k + 13\), the remainder is whatever is left after removing all multiples of 7.

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: hard
+tags: [remainders, simultaneous-conditions, divisibility]
+choices:
+  A: "2"
+  B: "3"
+  C: "5"
+  D: "8"
+  E: "11"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A positive integer \(n\) leaves a remainder of 3 when divided by 5 and a remainder of 2 when
+divided by 3. What is the remainder when \(n\) is divided by 15?
+
+## Explanation
+
+Two conditions on divisors 5 and 3 — and since 5 and 3 share no common factor, together they pin
+\(n\) down to a single remainder modulo \(5 \times 3 = 15\).
+
+Start with the condition that has the larger divisor (fewer cases to scan). Numbers that leave
+remainder 3 when divided by 5 are:
+
+\[3,\ 8,\ 13,\ 18,\ 23,\ \dots\]
+
+Now test each against the second condition, "remainder 2 when divided by 3":
+
+- \(3 \div 3\) leaves remainder 0 — no.
+- \(8 \div 3\) leaves remainder 2 — **yes**.
+
+So the smallest such \(n\) is 8, and every valid \(n\) is 8 more than a multiple of 15
+(\(8, 23, 38, \dots\)). Therefore \(n\) leaves a remainder of **8** when divided by 15.
+
+Common traps: adding the two remainders to get \(3 + 2 = 5\) (C), or picking one of the given
+remainders 3 (B) or 2 (A). None of those respects *both* conditions at once — only 8 does.
+
+## Hints
+
+- Because 5 and 3 have no common factor, the two conditions fix \(n\) modulo \(5 \times 3 = 15\).
+- List the numbers satisfying one condition, then scan for the first that also satisfies the other.

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: hard
+tags: [remainders, units-digit, cyclicity, exponents]
+choices:
+  A: "0"
+  B: "1"
+  C: "3"
+  D: "7"
+  E: "9"
+answer: B
+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 10?
+
+## Explanation
+
+The remainder when any integer is divided by 10 is simply its **units digit**, so the question is
+really: what is the units digit of \(7^{100}\)?
+
+Units digits of powers repeat in a short cycle. List the units digit of successive powers of 7:
+
+\[7^1 = 7,\quad 7^2 = 49,\quad 7^3 = 343,\quad 7^4 = 2401,\quad 7^5 = 16807,\ \dots\]
+
+The units digits are \(7, 9, 3, 1, 7, \dots\) — a cycle of length **4**: \(\{7, 9, 3, 1\}\).
+
+To find where \(7^{100}\) lands, divide the exponent by the cycle length: \(100 = 4 \times 25\), so
+\(100\) is a multiple of 4. An exponent that is a multiple of the cycle length maps to the **last**
+term of the cycle, which is \(1\) (just as \(7^4\) ends in 1).
+
+So the units digit of \(7^{100}\) is 1, and the remainder when \(7^{100}\) is divided by 10 is
+**1**.
+
+Trap answers come from misreading the cycle position: 7 (as if \(7^{100}\) behaved like \(7^1\)),
+9 (\(7^2\)), or 3 (\(7^3\)). The exponent being a multiple of 4 is the signal for the end of the
+cycle, not the start.
+
+## Hints
+
+- The remainder upon division by 10 is just the units digit.
+- Units digits of \(7^n\) cycle as \(7, 9, 3, 1\) with period 4 — reduce the exponent \(100\) modulo 4.