Tamilnadu Board Class 10 Maths Solutions Chapter 2 Numbers and Sequences Ex 2.3

Tamilnadu State Board Class 10 Maths Solutions Chapter 2 Numbers and Sequences Ex 2.3

Question 1.
Find the least positive value of x such that
(i) 71 = x (mod 8)
(ii) 78 + x = 3 (mod 5)
(iii) 89 = (x + 3) (mod 4)
(iv) 96 = \(\frac { x }{ 7 } \) (mod 5)
(v) 5x = 4 (mod 6)
Solution:
To find the least value of x such that
(i) 71 = x (mod 8)
71 = 7 (mod 8)
∴ x = 7.[ ∵ 71 – 7 = 64 which is divisible by 8]

(ii) 78 + x = 3 (mod 5)
⇒ 78 + x – 3 = 5M for some integer n.
75 + x = 5 n 75 + x is a multiple of 5.
75 + 5 = 80. 80 is a multiple of 5.
Therefore, the least value of x must be 5.

(iii) 89 = (x + 3) (mod 4)
89 – (x + 3) = 4n for some integer n.
86 – x = 4 n
86 – x is a multiple of 4.
∴ The least value of x must be 2 then
86 – 2 = 84.
84 is a multiple of 4.
∴ x value must be 2.

(iv) 96 ≡ \(\frac { x }{ 7 } \) (mod 5)
96 – \(\frac { x }{ 7 } \) = 5n for some integer n.
\(\frac { 672-x }{ 7 } \) = 5n
672 – x = 35n.
672 – x is a multiple of 35.
∴ The least value of x must be 7 i.e. 665 is a multiple of 35.

(v) 5x = 4 (mod 6)
5x – 4 = 6M for some integer n.
5x = 6n + 4
x = \(\frac { 6n+4 }{ 5 } \)
When we put 1, 6, 11, … as n values in x = \(\frac { 6n+4 }{ 5 } \) which is divisible by 5.
When n = 1, x = \(\frac { 10 }{ 5 } \) = 2
When n = 6, x = \(\frac { 36+4 }{ 5 } \) = \(\frac { 40 }{ 5 } \) = 8 and so on.
∴ The solutions are 2,8,14…..
∴ Least value is 2.

Question 2.
If x is congruent to 13 modulo 17 then 7x -3 is congruent to which number modulo 17?
Solution:
x ≡ 13 (mod 17)
Let p be the required number …(1)
7x – 3 = p (mod 17) …(2)
From (1),
x – 13 = 17n for some integer M.
x – 13 is a multiple of 17.
x must be 30.
∴ 30 – 13 = 17
which is a multiple of 17.
From (2),
7 × 30 – 3 ≡ p (mod 17)
210 – 3 ≡ p (mod 17)
207 ≡ p (mod 17)
207 ≡ 3 (mod 17)
∴ P ≡ 3

Question 3.
Solve 5x ≡ 4 (mod 6)
Solution:
5x ≡ 4 (mod 6)
5x – 4 = 6M for some integer n.
5x = 6n + 4
x = \(\frac { 6n+4 }{ 5 } \) where n = 1,6,11,…..
∴ x = 2, 8, 14,…

Question 4.
Solve 3x – 2 = 0 (mod 11)
Solution:
3x – 2 = 0 (mod 11)
3x – 2 = 11 n for some integer n.
3x = 11n + 2

Question 5.
What is the time 100 hours after 7 a.m.?
Solution:
100 ≡ x (mod 12) (∵7 comes in every 12 hrs)
100 ≡ 4 (mod 12) (∵ Least value of x is 4)
∴ The time 100 hrs after 7 O’ clock is 7 + 4 = 11 O’ clock i.e. 11 a.m

Question 6.
What is the time 15 hours before 11 p.m.?
Solution:
15 ≡ x (mod 12)
15 – x = 12n
15 – x is a multiple of 12 x must be 3.
∴ The time 15 hrs before 11 O’clock is 11 – 3 = 8 O’ clock i.e. 8 p.m

Question 7.
Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming?
Solution:
No. of days in a week = 7 days.
45 ≡ x (mod 7)
45 – x = In
45 – x is a multiple of 7.
∴ Value of x must be 3.
∴ Three days after Tuesday is Friday. Uncle will come on Friday.

Question 8.
Prove that 2ⁿ + 6 × 9ⁿ is always divisible by 7 for any positive integer n.
Solution:
2¹ + 6 × 9¹ = 2 + 54 = 56 is divisible by 7
When n = k,
2^k + 6 × 9^k = 7 m [where m is a scalar]
⇒ 6 × 9^k = 7 m – 2^k …(1)
Let us prove for n = k + 1
Consider 2^k+1 + 6 × 9^k+1 = 2^k+1 + 6 × 9^k × 9
= 2^k+1 + (7m – 2^k)9 (using (1))
= 2^k+1 + 63m – 9.2^k = 63m + 2^k.2¹ – 9.2^k
= 63m – 2^k (9 – 2) = 63m – 7.2^k
= 7 (9m – 2^k) which is divisible by 7
∴ 2ⁿ + 6 × 9ⁿ is divisible by 7 for any positive integer n

Question 9.
Find the remainder when 2⁸¹ is divided by 17.
Solution:
2⁸¹ = x (mod 17)
2⁴⁰ × 2⁴⁰ × 2⁴¹ = x (mod 17)
(24)¹⁰ × (24)¹⁰ × 2¹ = x (mod 17)
(16)¹⁰ × (16)¹⁰ × 2 = x(mod 17)
(16⁵)² × (16⁵)² × 2
(16⁵) = 16 (mod 17)
(16⁵)² = 16² (mod 17)
(16⁵)² = 256 (mod 17)
= 1 (mod 17)
[∵ 255 is divisible by 17]
(16⁵)² × (16⁵)² × 2 ≡ 1 × 1 × 2 (mod 17)
∴ 2⁸¹ = 2(mod 17)
∴ x = 2

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.
Solution:
The duration of flight from Chennai to London is 11 hours.
Starting time at Chennai is 23.30 hrs. = 11.30 p.m.
Travelling time = 11.00 hrs. / = 22.30 hrs = 10.30 a.m.
Chennai is 4 \(\frac { 1 }{ 2 } \) hrs ahead to London.
= 10.30 – 4.30 = 6.00
∴ At 6 a.m. on Monday the flight will reach at London Airport.

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