Ex. no. 2.3 class 10 Samacheer kalvi

Exercise 2.3 - Modular Arithmetic Solutions

Exercise 2.3: Modular Arithmetic

A complete guide with questions and step-by-step solutions.

Questions

  1. Find the least positive value of $x$ such that
    1. $71 \equiv x \pmod{8}$
    2. $78 + x \equiv 3 \pmod{5}$
    3. $89 \equiv (x+3) \pmod{4}$
    4. $96 \equiv \frac{x}{7} \pmod{5}$
    5. $5x \equiv 4 \pmod{6}$
  2. If $x$ is congruent to 13 modulo 17, then $7x-3$ is congruent to which number modulo 17?
  3. Solve $5x \equiv 4 \pmod{6}$
  4. Solve $3x - 2 \equiv 0 \pmod{11}$
  5. What is the time 100 hours after 7 a.m.?
  6. What is the time 15 hours before 11 p.m.?
  7. Today is Tuesday. My uncle will come after 45 days. In which day will my uncle be coming?
  8. Prove that $2^n + 6 \times 9^n$ is always divisible by 7 for any positive integer $n$.
  9. Find the remainder when $2^{81}$ is divided by 17.
  10. The duration of flight travel from Chennai to London through British Airways 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 of London's time, then find the time at London, when the flight lands at London Airport.

Questions with Detailed Solutions

  1. Find the least positive value of $x$ such that...

    (i) $71 \equiv x \pmod{8}$

    To find $x$, we find the remainder when 71 is divided by 8.

    $71 = 8 \times 8 + 7$. The remainder is 7.

    The least positive value of $x$ is 7.

    (ii) $78 + x \equiv 3 \pmod{5}$

    First, reduce 78 modulo 5: $78 = 15 \times 5 + 3$, so $78 \equiv 3 \pmod{5}$.

    Substitute this into the congruence: $3 + x \equiv 3 \pmod{5}$.

    Subtract 3 from both sides: $x \equiv 0 \pmod{5}$.

    The congruence means $x$ is a multiple of 5. The least positive value is 5.

    The least positive value of $x$ is 5.

    (iii) $89 \equiv (x+3) \pmod{4}$

    First, reduce 89 modulo 4: $89 = 22 \times 4 + 1$, so $89 \equiv 1 \pmod{4}$.

    Substitute this: $1 \equiv x+3 \pmod{4}$.

    Subtract 3 from both sides: $1 - 3 \equiv x \pmod{4}$, which gives $-2 \equiv x \pmod{4}$.

    To find the least positive value, we add the modulus: $x \equiv -2 + 4 \equiv 2 \pmod{4}$.

    The least positive value of $x$ is 2.

    (iv) $96 \equiv \frac{x}{7} \pmod{5}$

    First, reduce 96 modulo 5: $96 = 19 \times 5 + 1$, so $96 \equiv 1 \pmod{5}$.

    Substitute this: $1 \equiv \frac{x}{7} \pmod{5}$.

    Multiply both sides by 7: $7 \times 1 \equiv x \pmod{5}$, which gives $7 \equiv x \pmod{5}$.

    Reduce 7 modulo 5: $7 = 1 \times 5 + 2$, so $x \equiv 2 \pmod{5}$.

    The least positive value of $x$ is 2.

    (v) $5x \equiv 4 \pmod{6}$

    We can rewrite $5 \pmod{6}$ as $5 \equiv -1 \pmod{6}$.

    Substitute this: $-1 \cdot x \equiv 4 \pmod{6}$, which is $-x \equiv 4 \pmod{6}$.

    Multiply by -1: $x \equiv -4 \pmod{6}$.

    To find the least positive value, add the modulus: $x \equiv -4 + 6 \equiv 2 \pmod{6}$.

    The least positive value of $x$ is 2.
  2. If $x$ is congruent to 13 modulo 17, then $7x-3$ is congruent to which number modulo 17?

    We are given $x \equiv 13 \pmod{17}$.

    We want to find the value of $7x - 3 \pmod{17}$.

    Substitute $x=13$ into the expression: $7(13) - 3 = 91 - 3 = 88$.

    Now, we find the remainder when 88 is divided by 17.

    $88 = 17 \times 5 + 3$. The remainder is 3.

    $7x - 3$ is congruent to 3 modulo 17.
  3. Solve $5x \equiv 4 \pmod{6}$

    This is the same problem as 1(v). We are looking for the general solution.

    We have $5x \equiv 4 \pmod{6}$.

    Since $5 \equiv -1 \pmod{6}$, we can write $-x \equiv 4 \pmod{6}$.

    Multiplying by -1, we get $x \equiv -4 \pmod{6}$.

    To make the result positive, we add the modulus: $x \equiv -4 + 6 \pmod{6}$, which simplifies to $x \equiv 2 \pmod{6}$.

    The solution is $x \equiv 2 \pmod{6}$.
  4. Solve $3x - 2 \equiv 0 \pmod{11}$

    First, rewrite the congruence: $3x \equiv 2 \pmod{11}$.

    To solve for $x$, we need to find the multiplicative inverse of 3 modulo 11. We are looking for a number $k$ such that $3k \equiv 1 \pmod{11}$.

    By testing values: $3 \times 4 = 12$, and $12 \equiv 1 \pmod{11}$. So, the inverse of 3 is 4.

    Multiply both sides of the congruence by 4:

    $4 \times (3x) \equiv 4 \times 2 \pmod{11}$

    $12x \equiv 8 \pmod{11}$

    Since $12 \equiv 1 \pmod{11}$, this simplifies to $1 \cdot x \equiv 8 \pmod{11}$.

    The solution is $x \equiv 8 \pmod{11}$.
  5. What is the time 100 hours after 7 a.m.?

    We are working with a 24-hour cycle, so we use modulo 24.

    First, find the remainder of 100 hours when divided by 24.

    $100 = 4 \times 24 + 4$.

    This means 100 hours is equivalent to 4 full days and 4 extra hours.

    The new time will be 4 hours after the starting time of 7 a.m.

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

    The time will be 11 a.m.
  6. What is the time 15 hours before 11 p.m.?

    It's helpful to use a 24-hour clock format. 11 p.m. is 23:00.

    We need to calculate the time 15 hours before 23:00.

    Time = 23:00 - 15 hours = 8:00.

    8:00 in 24-hour format is 8 a.m.

    The time will be 8 a.m.
  7. Today is Tuesday. My uncle will come after 45 days. In which day will my uncle be coming?

    The days of the week repeat in a cycle of 7, so we use modulo 7.

    Let's assign numbers to the days: Sunday=0, Monday=1, Tuesday=2, ..., Saturday=6.

    Today is Tuesday, which is day 2.

    We need to find out what day it will be in 45 days. We calculate $45 \pmod{7}$.

    $45 = 6 \times 7 + 3$. So, $45 \equiv 3 \pmod{7}$.

    This means 45 days is equivalent to 3 days past the start of a week.

    The final day will be (Current Day + 45 days) $\pmod{7}$.

    Day = $(2 + 3) \pmod{7} = 5 \pmod{7}$.

    Day 5 corresponds to Friday.

    The uncle will be coming on a Friday.
  8. Prove that $2^n + 6 \times 9^n$ is always divisible by 7 for any positive integer $n$.

    To prove divisibility by 7, we need to show that $2^n + 6 \times 9^n \equiv 0 \pmod{7}$.

    Let's reduce the bases (6 and 9) modulo 7.

    $6 \equiv -1 \pmod{7}$

    $9 \equiv 2 \pmod{7}$

    Now, substitute these into the expression:

    $2^n + 6 \times 9^n \equiv 2^n + (-1) \times (2^n) \pmod{7}$

    This simplifies to $2^n - 2^n \pmod{7}$, which is $0 \pmod{7}$.

    Since the expression is congruent to 0 modulo 7, it is always divisible by 7.

    The statement is proven.
  9. Find the remainder when $2^{81}$ is divided by 17.

    We need to compute $2^{81} \pmod{17}$.

    We can use Fermat's Little Theorem, which states that if $p$ is a prime number, then for any integer $a$ not divisible by $p$, we have $a^{p-1} \equiv 1 \pmod{p}$.

    Here, $p=17$ (a prime) and $a=2$ (not divisible by 17). Therefore, $2^{17-1} \equiv 1 \pmod{17}$, which means $2^{16} \equiv 1 \pmod{17}$.

    Now, we express the exponent 81 in terms of 16: $81 = 16 \times 5 + 1$.

    So, $2^{81} = 2^{16 \times 5 + 1} = (2^{16})^5 \times 2^1$.

    Now we apply the congruence:

    $2^{81} \equiv (1)^5 \times 2^1 \pmod{17}$

    $2^{81} \equiv 1 \times 2 \pmod{17}$

    $2^{81} \equiv 2 \pmod{17}$

    The remainder is 2.
  10. The duration of flight travel... find the time at London, when the flight lands...

    Let's break down the problem into steps:

    1. Calculate Arrival Time in Chennai's Time Zone:

    Departure Time (Chennai): Sunday, 23:30

    Flight Duration: 11 hours

    Arrival Time = 23:30 Sunday + 11 hours

    23:30 + 0:30 = 00:00 Monday (10 hours 30 mins remaining)

    00:00 Monday + 10 hours 30 mins = 10:30 Monday.

    So, the flight lands at a time corresponding to 10:30 on Monday in Chennai.

    2. Convert Arrival Time to London's Time Zone:

    Time Difference: Chennai is 4.5 hours AHEAD of London.

    This means: London Time = Chennai Time - 4 hours 30 minutes.

    London Landing Time = 10:30 Monday - 4 hours 30 minutes

    10:30 - 4:30 = 6:00.

    The flight will land at 06:00 on Monday at London Airport.