Samacheer Kalvi Class 10 Maths Solutions Chapter 2 Numbers and Sequences Exercise 2.3

Question 1. Find the least positive value of x such that
(i) \( 71 = x \pmod{8} \)
(ii) \( 78 + x = 3 \pmod{5} \)
(iii) \( 89 = (x + 3) \pmod{4} \)
(iv) \( 96 = \frac{x}{7} \pmod{5} \)
(v) \( 5x = 4 \pmod{6} \)
Answer:
(i) To solve \( 71 = x \pmod{8} \), we need to find the remainder when 71 is divided by 8. When 71 is divided by 8, the quotient is 8 and the remainder is 7. So, \( 71 \equiv 7 \pmod{8} \). Thus, the least positive value for x is 7.
(ii) We are solving \( 78 + x = 3 \pmod{5} \). This means \( 78 + x - 3 \) must be a multiple of 5. Simplifying, we get \( 75 + x \) must be a multiple of 5. Since 75 is already a multiple of 5, the smallest positive value for x that makes the sum a multiple of 5 is 5 itself.
(iii) We need to solve \( 89 = (x + 3) \pmod{4} \). This means \( 89 - (x + 3) \) must be a multiple of 4. Simplifying the expression, we get \( 86 - x \) must be a multiple of 4. When 86 is divided by 4, the remainder is 2. For \( 86 - x \) to be perfectly divisible by 4, x must be 2. So, the least positive value for x is 2.
(iv) To solve \( 96 = \frac{x}{7} \pmod{5} \), we first write it as \( 96 - \frac{x}{7} = 5n \) for some integer n. Multiplying by 7, we get \( 672 - x = 35n \). This means \( 672 - x \) must be a multiple of 35. When 672 is divided by 35, the remainder is 7. For \( 672 - x \) to be exactly divisible by 35, x must be 7. This is the least positive value.
(v) We need to solve \( 5x = 4 \pmod{6} \). This can be written as \( 5x - 4 = 6n \), where n is any integer. Rearranging, we get \( 5x = 6n + 4 \), so \( x = \frac{6n+4}{5} \). We need x to be a whole number. By trying integer values for n, when \( n=1 \), \( x = \frac{6(1)+4}{5} = \frac{10}{5} = 2 \). This is the smallest positive integer value for x.
In simple words: To find the value of x in these questions, we use modular arithmetic to find the remainder or a number that fits the division rule. We look for the smallest positive whole number that makes the equation true.

🎯 Exam Tip: Always make sure to find the *least positive* value for x, which means the smallest whole number greater than zero that satisfies the condition.

Question 2. If x is congruent to 13 modulo 17 then 7x -3 is congruent to which number modulo 17?
Answer: We are given that \( x \equiv 13 \pmod{17} \). This means x leaves a remainder of 13 when divided by 17. To find what \( 7x - 3 \) is congruent to, we can substitute \( x=13 \) into the expression. So, we calculate \( 7(13) - 3 \), which is \( 91 - 3 = 88 \). Next, we find the remainder when 88 is divided by 17. When 88 is divided by 17, the quotient is 5 and the remainder is 3. Therefore, \( 7x - 3 \) is congruent to \( 3 \pmod{17} \).
In simple words: First, replace x with 13 in the expression \( 7x-3 \) and calculate the result. Then, divide that result by 17 and the remainder you get is the answer.

🎯 Exam Tip: When working with congruences, you can substitute a congruent number into an expression to simplify calculations, as any number in the same congruence class will yield the same result.

Question 3. Solve \( 5x = 4 \pmod{6} \)
Answer: To solve the congruence \( 5x \equiv 4 \pmod{6} \), we can rewrite it as \( 5x - 4 = 6n \), where n is any integer. Rearranging, we get \( 5x = 6n + 4 \), which means \( x = \frac{6n+4}{5} \). We need x to be an integer. By trying different integer values for n, we find the corresponding integer values for x.
When \( n=1 \), \( x = \frac{6(1)+4}{5} = \frac{10}{5} = 2 \).
When \( n=6 \), \( x = \frac{6(6)+4}{5} = \frac{40}{5} = 8 \).
When \( n=11 \), \( x = \frac{6(11)+4}{5} = \frac{70}{5} = 14 \).
When \( n=16 \), \( x = \frac{6(16)+4}{5} = \frac{100}{5} = 20 \).
So, the values of x that solve this congruence are 2, 8, 14, 20, and so on.
In simple words: We are looking for numbers x that, when multiplied by 5, give a remainder of 4 when divided by 6. We found that 2, 8, 14, 20, and other numbers following this pattern are the solutions.

🎯 Exam Tip: When "solve" is asked for a congruence, listing a few solutions or indicating the pattern of solutions is usually sufficient, as there can be infinitely many.

Question 4. Solve \( 3x - 2 = 0 \pmod{11} \)
Answer: To solve the congruence \( 3x - 2 \equiv 0 \pmod{11} \), we first add 2 to both sides, which gives \( 3x \equiv 2 \pmod{11} \). This means \( 3x - 2 \) must be a multiple of 11, so we can write \( 3x - 2 = 11n \) for some integer n. Rearranging the equation, we get \( 3x = 11n + 2 \), and thus \( x = \frac{11n+2}{3} \). We need to find integer values for x. By substituting different integer values for n:
When \( n=2 \), \( x = \frac{11(2)+2}{3} = \frac{24}{3} = 8 \).
When \( n=5 \), \( x = \frac{11(5)+2}{3} = \frac{57}{3} = 19 \).
When \( n=8 \), \( x = \frac{11(8)+2}{3} = \frac{90}{3} = 30 \).
When \( n=11 \), \( x = \frac{11(11)+2}{3} = \frac{123}{3} = 41 \).
The solutions for x are 8, 19, 30, 41, and so on.
In simple words: We are looking for numbers x that, when multiplied by 3 and then 2 is subtracted, result in a number that can be divided evenly by 11. The answers form a pattern like 8, 19, 30, 41.

🎯 Exam Tip: For linear congruences, if the coefficient of x and the modulus are coprime, there will be a unique solution modulo the modulus. If not, solutions may not exist or may appear in multiple sets.

Question 5. What is the time 100 hours after 7 a.m.?
Answer: To find the time 100 hours after 7 a.m., we use modular arithmetic because a clock repeats every 12 hours. We need to find the remainder when 100 is divided by 12. `\( 100 \div 12 \)` gives a quotient of 8 and a remainder of 4. So, 100 hours is the same as 4 hours on a 12-hour clock. Adding 4 hours to 7 a.m. gives `\( 7 + 4 = 11 \)` a.m. Therefore, the time will be 11 a.m.
In simple words: Since clocks repeat every 12 hours, we divide 100 hours by 12 to find the leftover hours. The remainder is 4. So, we just add 4 hours to 7 a.m., which makes it 11 a.m.

🎯 Exam Tip: Remember to use modulo 12 for standard clock problems (a.m./p.m.) and modulo 24 for 24-hour clock problems.

Question 6. What is time 15 hours before 11 p.m.?
Answer: To find the time 15 hours before 11 p.m., we first need to figure out how 15 hours fits into a 12-hour clock cycle. We find the remainder when 15 is divided by 12, which is 3. This means going back 15 hours is equivalent to going back just 3 hours on a clock. Subtracting 3 hours from 11 p.m. (`\( 11 - 3 \)` p.m.) gives us 8 p.m. So, the time will be 8 p.m.
In simple words: We want to go back 15 hours from 11 p.m. Since a clock repeats every 12 hours, going back 15 hours is like going back 3 hours. So, 3 hours before 11 p.m. is 8 p.m.

🎯 Exam Tip: When calculating time differences, always find the remainder of the hours (modulo 12 or 24) to simplify the calculation, especially for large numbers of hours.

Question 7. Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming?
Answer: To find the day of the week 45 days after Tuesday, we use modular arithmetic with 7, as there are 7 days in a week. We calculate the remainder when 45 is divided by 7. `\( 45 \div 7 \)` gives a quotient of 6 and a remainder of 3. This means 45 days is equivalent to 3 days in the weekly cycle. Counting forward 3 days from Tuesday (Tuesday + 3 days) brings us to Friday. So, the uncle will come on Friday.
In simple words: There are 7 days in a week. We divide 45 by 7 and find the remainder, which is 3. So, we count 3 days forward from Tuesday. That day is Friday.

🎯 Exam Tip: For problems involving days of the week, the modulus is always 7, representing the 7 days in a cycle.

Question 8. Prove that \( 2^n + 6 \times 9^n \) is always divisible by 7 for any positive integer \( n \).
Answer: We need to prove that the expression \( 2^n + 6 \times 9^n \) is always divisible by 7 for any positive integer \( n \). We can achieve this using the principle of mathematical induction.
**Step 1: Base Case (n=1)**
For \( n=1 \), the expression becomes \( 2^1 + 6 \times 9^1 = 2 + 54 = 56 \). Since \( 56 = 7 \times 8 \), it is clearly divisible by 7.
**Step 2: Inductive Hypothesis**
Assume that the expression is divisible by 7 for some positive integer \( k \). This means \( 2^k + 6 \times 9^k = 7m \) for some integer \( m \). From this, we can write \( 6 \times 9^k = 7m - 2^k \).
**Step 3: Inductive Step (n=k+1)**
Now, we need to show that the expression is also divisible by 7 for \( n=k+1 \). The expression for \( n=k+1 \) is \( 2^{k+1} + 6 \times 9^{k+1} \).
We can rewrite this as \( 2 \times 2^k + 9 \times (6 \times 9^k) \).
Now, we substitute \( (6 \times 9^k) \) with \( (7m - 2^k) \) from our hypothesis.
\( \implies \) \( 2 \times 2^k + 9 \times (7m - 2^k) \)
\( \implies \) \( 2 \times 2^k + 63m - 9 \times 2^k \)
\( \implies \) \( 63m + 2^k (2 - 9) \)
\( \implies \) \( 63m + 2^k (-7) \)
\( \implies \) \( 7 (9m - 2^k) \)
Since \( 7 (9m - 2^k) \) is a multiple of 7, it is divisible by 7. Therefore, by mathematical induction, the statement is true for all positive integers \( n \).
In simple words: We showed that the formula gives a number divisible by 7 when we test it for the first number, 1. Then, we proved that if it works for any number 'k', it must also work for the next number 'k+1'. This means it will always work for all positive numbers.

🎯 Exam Tip: For divisibility proofs, using modular arithmetic properties (like \( 9 \equiv 2 \pmod{7} \)) can sometimes be a quicker alternative to mathematical induction, but both methods are valid.

Question 9. Find the remainder when \( 2^{81} \) is divided by 17?
Answer: To find the remainder when \( 2^{81} \) is divided by 17, we need to calculate \( 2^{81} \pmod{17} \). We start by finding a pattern of powers of 2 modulo 17.
We know that \( 2^4 = 16 \). Since \( 16 = 17 - 1 \), we can say \( 2^4 \equiv -1 \pmod{17} \).
Now, we can find a higher power that is congruent to 1. Squaring both sides, we get \( (2^4)^2 = 2^8 \equiv (-1)^2 \equiv 1 \pmod{17} \). This means `\( 2^8 \)` leaves a remainder of 1 when divided by 17.
Next, we divide the exponent 81 by 8: \( 81 = 8 \times 10 + 1 \).
So, we can rewrite \( 2^{81} \) as \( (2^8)^{10} \times 2^1 \).
Substituting the congruence, we get \( (1)^{10} \times 2^1 \pmod{17} \).
This simplifies to \( 1 \times 2 \pmod{17} \), which is \( 2 \pmod{17} \).
Therefore, the remainder when \( 2^{81} \) is divided by 17 is 2.
In simple words: We want to find the remainder of \( 2^{81} \) when divided by 17. We found that \( 2^8 \) gives a remainder of 1. So we divide 81 by 8, which gives 10 with a remainder of 1. This means \( 2^{81} \) is like \( (2^8)^{10} \times 2^1 \). Since \( 2^8 \) is 1, the whole thing becomes \( 1^{10} \times 2^1 \), which is 2. So the remainder is 2.

🎯 Exam Tip: When dealing with large exponents in modular arithmetic, try to find a smaller exponent (often \( \phi(n) \) or a divisor of it) that results in a remainder of 1 or -1 to simplify the calculation significantly.

Question 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 will the flight lands at London Airport?
Answer: To find the arrival time in London, we first calculate the arrival time in Chennai's local time. The flight departs at 23:30 hours (which is 11:30 p.m.) on Sunday. The flight duration is 11 hours.
Adding 11 hours to 23:30 gives `\( 23:30 + 11:00 = 34:30 \)` hours.
Since a day has 24 hours, `\( 34:30 \)` is equivalent to `\( 34:30 - 24:00 = 10:30 \)` on the next day. So, the flight lands at 10:30 a.m. on Monday, according to Chennai time.
Next, we account for the time difference: Chennai is 4.5 hours ahead of London. To find London's local time, we subtract 4.5 hours from Chennai's arrival time.
`\( 10:30 - 4:30 = 6:00 \)`.
Therefore, the flight lands at London Airport at 6:00 a.m. on Monday.
In simple words: The flight leaves Chennai on Sunday at 11:30 p.m. and flies for 11 hours. This means it lands in Chennai's time on Monday at 10:30 a.m. Since Chennai is 4.5 hours ahead of London, we subtract 4.5 hours from 10:30 a.m. to get London's time, which is 6:00 a.m. on Monday.

🎯 Exam Tip: When solving time zone problems, always calculate the arrival time in the departure zone first, then adjust for the time difference, being careful with date changes if crossing midnight.