Mathematical Thinking · Quiz 1 · Jan 2025

**1. Rational and irrational sums — select A and C:** If a/b and c/d are rational with b,d≠0, their sum is (ad+bc)/(bd), also rational. If a rational r plus an irrational x were rational, subtracting r would make x rational, a contradiction. B is false because √2+(−√2)=0 is rational. D is false because the rational set \(\{q\in\mathbb Q:0<q,\ q^2<2\}\) has supremum √2, which is irrational. **2. Compute the three quantities in Question 2:** \[a=\sum_{n=1}^{10}(6-n)=60-\frac{10\cdot11}{2}=5.\] For b, odd n give 1+1/n≤2 with equality at n=1, while even n give 1−1/n<1; hence b=2. An integer divisible by both 4 and 6 is divisible by lcm(4,6)=12. The multiples 12,24,…,96 give c=⌊100/12⌋=8. **3. Check Question 2 statements — select A and C:** a≡3 mod b means 5≡3 mod2, true because 2 divides 5−3. a≡b mod c would require 8 to divide 5−2=3, false. b divides c because 2 divides 8. Finally a+b=7≠8=c. **4. Check Question 3 statements — select A and C:** Any common divisor of 10²⁰⁰+1 and 10²⁰⁰−1 divides their difference 2. Both numbers are odd, so their gcd is 1. The equation 18m+6n=2 is impossible because its left side is divisible by 6. Also gcd(56,98)=14, so all combinations 56m+98n are multiples of 14, and every such multiple occurs: \(56(-2k)+98k=-14k\). Thus the set is 14ℤ, not 28ℤ; for instance −14 is attainable. **5. Set up the language-count problem:** Let P,C,J be the Python,C,Java candidate sets. There are 500 candidates and 50 know none, so |P∪C∪J|=450. The given pair intersections are inclusive of candidates who know all three. If t=|P∩C∩J|, inclusion–exclusion gives \[450=290+190+130-70-70-50+t=420+t,\] hence t=30. **6. Count candidates who know exactly one language:** Python only =290−70−70+30=180. C only =190−70−50+30=100. Java only =130−70−50+30=40. Therefore the requested number is \[180+100+40=320.\] **Answer: A — Q1 A,C; Q2 A,C; Q3 A,C; Q4 320 candidates.** **Convention note:** This exported MCQ wraps four written/MSQ questions under acknowledgement choices. Their substantive answers are Q1: A,C; Q2: A,C with a=5,b=2,c=8; Q3: A,C; Q4: 320. A: **Correct:** The export uses this acknowledgement to record completion; the actual mathematical answers to all four written questions are derived above. B: **Incorrect:** The questions are applicable and have been fully solved; this acknowledgement does not describe the completed work.

**1. Question 5 — finite cases:** A finite union of finite sets is finite and hence countable. If one of A,B is empty, the union is the other countable set. It remains to handle at least one infinite set; finite lists can simply be exhausted while enumerating the other. **2. Question 5 — enumerate the union:** List A as a₁,a₂,… and B as b₁,b₂,… whenever infinite. Read the interleaved list a₁,b₁,a₂,b₂,…, skipping repeated elements and any exhausted finite list. Every element of A∪B appears after finitely many entries. The resulting list is finite or an enumeration indexed by natural numbers. Thus A∪B is countable. **3. Question 6 — expand the seven-digit number:** Writing the number as N=30x0y03 gives \[N=3\cdot10^6+x\cdot10^4+100y+3.\] Modulo 13, 10⁴≡3 and 10⁶≡1 because 10³≡−1. Therefore \(N\equiv3+3x+9y+3=3(x+3y+2)\pmod{13}\). **4. Question 6 — use divisibility and cancellation:** Since 13 divides N and gcd(3,13)=1, the factor 3 is invertible modulo 13. Hence x+3y+2≡0 mod13, or \[x+3y\equiv-2\equiv11\pmod{13}.\] This proves the requested congruence for any allowed digits x,y. **5. Question 7 — find a telescoping identity:** For k≥1, \[\frac1{(k+2)(k+3)}=\frac1{k+2}-\frac1{k+3}.\] This works because subtracting the fractions gives numerator 1 over the product. **6. Question 7 — cancel successive terms:** Summing from k=1 to n cancels every interior reciprocal, leaving \[\sum_{k=1}^n\frac1{(k+2)(k+3)}=\frac13-\frac1{n+3}=\frac{n}{3(n+3)}=\frac{n}{3n+9}.\] The algebra holds for every positive integer n, proving the identity directly. **Answer: A — A∪B is countable; x+3y≡11 mod13; the sum is n/(3n+9).** **Convention note:** The source wraps three proof questions under one acknowledgement MCQ. The proofs, rather than the acknowledgement wording, are the substantive solutions. A: **Correct:** This acknowledgement corresponds to the complete proofs given above for all three written questions. B: **Incorrect:** The requested written questions are applicable and their proofs are provided.