Math Magic: Modular Arithmetic Adventure

Problem 1: Finding Least Positive Values

1

Find the least positive value of x such that:

71 ≡ x (mod 8)
78 + x ≡ 3 (mod 5)
89 ≡ (x + 3) (mod 4)
96 ÷ 7 ≡ x (mod 5)
5x ≡ 4 (mod 6)

i) 71 ≡ x (mod 8)

71 ÷ 8 = 8 with remainder 7

So, 71 ≡ 7 (mod 8)

Answer: x = 7

ii) 78 + x ≡ 3 (mod 5)

First find 78 mod 5: 78 ÷ 5 = 15 with remainder 3

So: 3 + x ≡ 3 (mod 5) ⇒ x ≡ 0 (mod 5)

Least positive x: x = 5

iii) 89 ≡ (x + 3) (mod 4)

89 ÷ 4 = 22 with remainder 1

So: 1 ≡ x + 3 (mod 4) ⇒ x ≡ -2 ≡ 2 (mod 4)

Answer: x = 2

iv) 96 ÷ 7 ≡ x (mod 5)

First calculate 96 ÷ 7 ≈ 13.714

We take integer part: 13

Now 13 mod 5: 13 ÷ 5 = 2 with remainder 3

Answer: x = 3

v) 5x ≡ 4 (mod 6)

Test values of x from 1 upwards:

x=1: 5 ≡ 5 ≢ 4

x=2: 10 ≡ 4 ≡ 4

Answer: x = 2

Problem 2: Congruence Transformation

2

If x ≡ 13 (mod 17), then 7x - 3 is congruent to which number modulo 17?

Given: x ≡ 13 (mod 17)

Calculate 7x - 3: 7×13 - 3 = 91 - 3 = 88

Now find 88 mod 17: 17×5 = 85, remainder 3

So 88 ≡ 3 (mod 17)

Answer: 7x - 3 ≡ 3 (mod 17)

Problem 3: Solving Linear Congruence

3

Solve 5x ≡ 4 (mod 6)

We need to find all integers x such that when 5x is divided by 6, remainder is 4

Test values of x from 0 to 5 (since modulo 6 repeats every 6 numbers):

x=0: 0 ≡ 0 ≢ 4

x=1: 5 ≡ 5 ≢ 4

x=2: 10 ≡ 4 ≡ 4 ✔️

x=3: 15 ≡ 3 ≢ 4

x=4: 20 ≡ 2 ≢ 4

x=5: 25 ≡ 1 ≢ 4

Solution: x ≡ 2 (mod 6)

General solution: x = 2 + 6k for any integer k

Problem 4: Another Linear Congruence

4

Solve 3x - 2 ≡ 0 (mod 11)

Rewrite the equation: 3x ≡ 2 (mod 11)

We need to find the multiplicative inverse of 3 modulo 11

Find a number y such that 3y ≡ 1 (mod 11)

Testing: 3×4 = 12 ≡ 1 (mod 11), so inverse is 4

Multiply both sides by 4: x ≡ 8 (mod 11)

General solution: x = 8 + 11k for any integer k

Problem 5: Time Calculations

5

What is the time 100 hours after 7 a.m.?

Time repeats every 24 hours, so we find 100 mod 24

100 ÷ 24 = 4 with remainder 4

So 100 hours = 4 full days and 4 extra hours

7 a.m. + 4 hours = 11 a.m.

Answer: 11 a.m. 4 days later

Problem 6: Time Calculations

6

What is the time 15 hours before 11 p.m.?

11 p.m. is 23:00 in 24-hour format

23 - 15 = 8

8:00 is 8 a.m.

Answer: 8 a.m. on the previous day

Problem 7: Day Calculation

7

Today is Tuesday. My uncle will come after 45 days. On which day will my uncle arrive?

Weeks repeat every 7 days, so find 45 mod 7

45 ÷ 7 = 6 with remainder 3

So 45 days = 6 weeks and 3 extra days

Tuesday + 3 days = Friday

Answer: Friday

Problem 8: Divisibility Proof

8

Prove that 2ⁿ + 6×9ⁿ is always divisible by 7 for any positive integer n.

We'll use mathematical induction:

Base case (n=1): 2 + 6×9 = 56, which is divisible by 7 (56÷7=8)

Inductive step: Assume true for n=k, i.e., 2ᵏ + 6×9ᵏ ≡ 0 mod 7

For n=k+1: 2ᵏ⁺¹ + 6×9ᵏ⁺¹ = 2×2ᵏ + 6×9×9ᵏ = 2×2ᵏ + 54×9ᵏ

Rewrite 54 as 54 ≡ 5 mod 7, so: 2×2ᵏ + 5×9ᵏ

From inductive hypothesis: 2ᵏ ≡ -6×9ᵏ mod 7

Substitute: 2×(-6×9ᵏ) + 5×9ᵏ = -12×9ᵏ + 5×9ᵏ = -7×9ᵏ ≡ 0 mod 7

Thus, if true for n=k, then true for n=k+1

By induction, 2ⁿ + 6×9ⁿ is divisible by 7 for all positive integers n

Problem 9: Remainder Calculation

9

Find the remainder when 2⁸¹ is divided by 17.

We'll use Fermat's Little Theorem: for prime p, aᵖ⁻¹ ≡ 1 mod p

For p=17, a=2: 2¹⁶ ≡ 1 mod 17

Express 81 as: 81 = 16×5 + 1

So 2⁸¹ = (2¹⁶)⁵ × 2¹ ≡ 1⁵ × 2 ≡ 2 mod 17

Answer: Remainder is 2

Problem 10: Time Zone Calculation

10

The duration of flight travel from Chennai to London through British Airlines is approximately 11 hours. The airplane begins its journey on Sunday at 23:30 hours. If the time at Chennai is four and half hours ahead to that of London's time, then find the time at London when the flight lands at London Airport.

Departure time in Chennai: Sunday 23:30

Flight duration: 11 hours

Arrival time in Chennai: Monday 10:30 (23:30 + 11 hours)

Chennai is 4.5 hours ahead of London

So London time is 4.5 hours behind Chennai time

10:30 - 4:30 = 6:00

Answer: Monday 6:00 a.m. London time