Modular arithmetic revision card

Modular arithmetic revision card

• 1.
  Modular Arithmetic Modulo isall about finding the remainder when one number divides another
• 2.
  WASSCE May/June 2018 Copyand complete the tables for the addition ⊕ and ⨂ in modulo 5 Use the tables to find; a) 4 ⨂ 2 ⊕ 3 ⨂ 4 b) 𝑚 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑚 ⨂ 𝑚 = 𝑚 ⊕ 𝑚 c) 𝑛 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 3 ⊕ 𝑛 = 2⨂𝑛 Solution 𝑎) 4 ⨂ 2 ⨁ 3 ⨂ 4 = 3 ⨁ 2 = 0 𝑏) 𝑚 ⊗ 𝑚 = 𝑚 ⨁𝑚 𝑐) 2 ⊗ 2 = 2 ⨁ 2 ∴ 𝑚 = 2 ⊕ 1 2 3 4 1 2 3 4 0 2 3 3 4 2 4 0 ⨂ 1 2 3 4 1 1 3 4 0 2 2 3 2 4 1 ⨂ 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 ⊕ 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 3
• 3.
  SSSCE Nov 2003 Drawa table for multiplication,⨂, modulo 7 on the set 𝑃 = {2,3,4,5,6} Use your table to find on the set P, the truth set of 𝑛⨂ 𝑛 ⨂ 6 = 3 Solution 𝑏) From the table 𝑛⨂ 𝑛 ⨂ 6 = 3 2⨂ 2 ⨂ 6 = 2 ⨂ 5 = 3 5⨂ 5 ⨂ 6 = 5 ⨂ 2 = 3 𝒏 = {𝟐, 𝟓} ⨂ 𝟐 𝟑 𝟒 𝟓 𝟔 𝟐 4 6 1 3 5 𝟑 6 2 5 1 4 𝟒 1 5 2 6 3 𝟓 3 1 6 4 2 𝟔 5 4 3 2 1 a
• 4.
  WASSCE Nov 2007 Copyand complete the multiplication table modulo 5 on the set {1,2,3,4} From the table solve the expression 2𝑛 ∗ 4 = 3 Solution 𝐺𝑖𝑣𝑒𝑛 2𝑛 ∗ 4 = 3 From the table 𝟐 ∗ 𝟒 = 𝟑 → 𝟐𝒏 = 𝟐, 𝒏 = 𝟏 ∗ 1 2 3 4 1 1 3 2 4 1 3 3 2 4 3 1 ∗ 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1
• 5.
  WASSCE May/June 2006 Drawa table of multiplication ⨂, in 𝑀𝑜𝑑𝑢𝑙𝑜 8 on the set {2,3,5,7} Solution ⨂ 2 3 5 7 2 4 6 2 6 3 6 1 7 5 5 2 7 1 3 7 6 5 3 1
• 6.
  1. Simplify 9𝑚𝑜𝑑6 A.3 B. 2 C. 1 D. 0 E. 5 F. The correct answer is 3 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 9𝑚𝑜𝑑6 6 goes into 9 once remainder 3
• 7.
  2. Evaluate 2⨂4in modulo 5 A. 3 B. 4 C. 2 D. 6 E. 7 The correct answer is 3 Solution 2 ⨂ 4 = 8 8𝑚𝑜𝑑5 = 3
• 8.
  3. Given that𝑛 ⊕ 5 = 0𝑚𝑜𝑑6. Find the value of 𝑛, such that 0 < 𝑛 < 15 . A. 𝑛 = {1,7} B. 𝑛 = {1,7,12} C. 𝑛 = {0,6,12} D. 𝑛 = {1,7,13} E. 𝑛 = {1,13} The correct answer is 𝑛 = {1,7,13} Solution 𝑛 ⊕ 5 = 0𝑚𝑜𝑑6 1 ⊕ 5 = 0𝑚𝑜𝑑6 7 ⊕ 5 = 0𝑚𝑜𝑑6 13 ⊕ 5 = 0𝑚𝑜𝑑6 19 ⊕ 5 = 0𝑚𝑜𝑑6 Etc, For 0 < 𝑛 < 15 𝑛 = {1,7,13}
• 9.
  4. Which ofthe following satisfies the equation 𝑦 𝑚𝑜𝑑 7 = 1 in the interval 0 < 𝑦 < 10? A. {1,9} B. {1,5} C. {1} D. {8} E. {2,8} F. {1,8} The correct answer is 1 2 𝑝𝑞 Solution 1 𝑚𝑜𝑑 7 = 1 8 𝑚𝑜𝑑 7 = 1 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑎𝑟𝑒 {1,8}
• 10.
  5. Solve 𝑥+ 1 𝑚𝑜𝑑 5 = 3 in the interval 0 < 𝑥 < 10. A. 𝑥 = {7,8} B. 𝑥 = {2,8} C. 𝑥 = {7} D. 𝑥 = {2} E. 𝑥 = {2,7} The correct answer is𝑥 = {2,7} Solution 3 𝑚𝑜𝑑 5 = 3 𝑥 + 1 = 3 𝑥 = 2 8 𝑚𝑜𝑑 5 = 3 𝑥 + 1 = 8 𝑥 = 7 13 𝑚𝑜𝑑 5 = 3 𝑥 + 1 = 13 𝑥 = 12 For the given interval 0 < 𝑥 < 10 𝑥 = {2,7}
• 11.
  6. Find theremainder when 1001 is divided by 3 A. 2 B. 3 C. 4 D. 6 E. 5 The correct answer is 2 Solution 1001 = 3 × 999 + 2
• 12.
  7. Evaluate 2018𝑚𝑜𝑑 11 A. 1 B. 2 C. 3 D. 4 E. 5 The correct answer is 5 Solution 2018 = 11 × 183 + 5 2018 𝑚𝑜𝑑 11 = 5
• 13.
  8. Find theremainder when −12 is divided by 5. A. −3 B. 3 C. −7 D. −2 E. 5 The correct answer is 3 Solution −12 = −5 × 3 + 3 −12 𝑚𝑜𝑑 5 = 3
• 14.
  9. Find theremainder when −12 is divided by 3 A. −2 B. −1 C. 0 D. 1 E. 2 The correct answer is 0 Solution −12 = −4 × 3 + 0 −12 𝑚𝑜𝑑 3 = 0
• 15.
  10. Simplify 1+ 8𝑚𝑜𝑑6 A. 3 B. 2 C. 1 D. 0 E. 4 The correct answer is 3 Solution 1 + 8𝑚𝑜𝑑6 = 1 + 2𝑚𝑜𝑑6 = 3𝑚𝑜𝑑6
• 16.
  SUMMARY Modulo is justabout remainder Note that remainder is not negative