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317 Modular Arithmetic 010417.notebook
TBLS IB Math HL Year Two, R3 & R9
January 02, 2017
3‑17
January 4, 2016
AIM: DWBAT use simple rules for modular arithmetic: addition, subtraction, multiplication, and exponentiation.
Warm Up:
HW: p.55 (Ex. 5B) #2abcd(i only), 3‑5
Copy down these modular arithmetic rules: If a ≡ b (mod m) and c ≡ d (mod m) then: ka ≡ kb (mod m) for all k ∈ ZZ a ± c ≡ b ± d (mod m) ac ≡ bd (mod m) an ≡ bn (mod m) for all n ∈ N N Sep 212:36 PM
Forecast Tues, 1/3 ‑ Tues, 1/10: Number congruences Rest of January: (Discrete Mathematics) More Number Theory Mon, 1/16: No School ‑ MLK Day Thurs, 1/19: Last Math Exam for Fall Semester (Number Theory) Tues 1/24 ‑ Fri 1/27: No Classes ‑ NY State Regents Week Sat 1/28: Chinese New Year Mon 1/30: No Classes ‑ Chancellor's Conference Day Tues, 1/31: Classes resume; Spring Semester Begins February: (Discrete Mathematics) Graph Theory March: (Core Curriculum) Statistics & Probability April: Review for IB Exams May: IB Exams
Dec 48:56 PM
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317 Modular Arithmetic 010417.notebook
January 02, 2017
Modular Arithmetic As a general rule, we try to reduce all integers to their least possible positive residue equivalent at all times.
Feb 225:22 PM
Modular Arithmetic Sometimes, it's convenient to use negative remainders if this produces smaller numbers:
To figure out 982 (mod 100), we could say 982 = 9604, and 9604 = 100×96 + 4, so we say that 982 = 9604 ≡ 4 (mod 100) or with less computation: 98 ≡ ‑2 (mod 100), so by squaring both sides, we know 982 ≡ (‑2)2 (mod 100) and we end up with 982 ≡ 4 (mod 100) Feb 225:22 PM
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317 Modular Arithmetic 010417.notebook
January 02, 2017
Modular Arithmetic Find the remainder when 230 is divided by 13.
We are looking for x when 230 ≡ x (mod 13).
2 is a small number, so maybe we can take advantage of small powers of 2:
26 ≡ 64 (mod 13)
26 ≡ 12 (mod 13)
230 ≡ 125 (mod 13)
230 ≡ (‑1)5 (mod 13)
230 ≡ ‑1 (mod 13), so since we use the smallest possible positive number, then 230 gives remainder 12 when divided by 13
Feb 225:22 PM
Modular Arithmetic Find the remainder when 6522 is divided by 7. We are looking for x where 6522 ≡ x (mod 7). Start small and look for ways to take advantage of the (mod 7). What is 65 (mod 7)? 65 ≡ 2 (mod 7)
6522 ≡ 87 × 2 (mod 7)
6522 ≡ 222 (mod 7)
6522 ≡ 17 × 2 (mod 7)
6522 ≡ 23×7 + 1 (mod 7)
6522 ≡ 2 (mod 7)
Feb 225:22 PM
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317 Modular Arithmetic 010417.notebook
January 02, 2017
Modular Arithmetic Prove that 240 ‑ 1 is divisible by 41. We want to show that 240 ‑ 1 ≡ 0 (mod 41), so then we want to show that 240 ≡ 1 (mod 41). 25 ≡ 32 (mod 41) 25 ≡ ‑9 (mod 41) 210 ≡ 81 (mod 41) 210 ≡ ‑1 (mod 41) 240 ≡ 1 (mod 41) as desired. Feb 225:22 PM
Practice
HW: p.55 (Ex. 5B) #2abcd(i only), 3‑5
Find the following remainders: 2a(i) 316 + 7 (mod 5) 2c(i) 532 + 323 (mod 6) 2b(i) 47 ‑ 22 (mod 7) 2d(i) 6×518 (mod 7) 3) Find the last digit of 22223333 (hint: we can use congruence modulo what to focus on the ones digit of any number?)
4) Find the remainder when 5544 + 4455 is divided by 7. 5) Given that x is a whole number such that 2x has last digit 4, find the last digit of 23x + 1. Feb 225:22 PM
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317 Modular Arithmetic 010417.notebook
Answers
January 02, 2017
HW: p.55 (Ex. 5B) #2abcd(i only), 3‑5
2a(i) 3 2b(i) 3 2c(i) 4 2d(i) 6 3) 2 4) 3 5) 5
Feb 225:22 PM
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