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## Tamilnadu Samacheer Kalvi 10th 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 = 5n 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?

Answer:

Given x ≡ 13 (mod 17) ……(1)

7x – 3 ≡ a (mod 17) ……..(2)

From (1) we get

x- 13 = 17 n (n may be any integer)

x – 13 is a multiple of 17

∴ The least value of x = 30

From (2) we get

7(30) – 3 ≡ a(mod 17)

210 – 3 ≡ a(mod 17)

207 ≡ a (mod 17)

207 ≡ 3(mod 17)

∴ The value of a = 3

Question 3.

Solve 5x ≡ 4 (mod 6)

Solution:

5x ≡ 4 (mod 6)

5x – 4 = 6M for some integer n.

5x = 6n + 4

x = \(\frac{6 n+4}{5}\) 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 time 15 hours before 11 p.m.?

Answer:

15 ≡ x (mod 12)

15 ≡ 3 (mod 12)

The value of x must be 3.

The time 15 hours before 11 o’clock is (11 – 3) 8 pm

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 = 7n

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 2^{n} + 6 × 9n is always divisible by 7 for any positive integer n.

Answer:

9 = 2 (mod 7)

9^{n} = 2^{n} (mod 7) and 2^{n} = 2^{n} (mod 7)

2^{n} + 6 × 9^{n} = 2^{n} (mod 7) + 6 [2^{n} (mod 7)]

= 2^{n} (mod 7) + 6 × 2^{n} (mod 7)

7 × 2^{n} (mod 7)

It is always divisible for any positive integer n

Question 9.

Find the remainder when 2^{81} is divided by 17.

Solution:

2^{81} ≡ x (mod 17)

2^{40} × 2^{40} × 2^{41} ≡ x (mod 17)

(2^{4})^{10} × (2^{4})^{10} × 2^{1} ≡ x (mod 17)

(16)^{10} × (16)^{10} × 2 ≡ x(mod 17)

(16^{5})^{2} × (16^{5})^{2} × 2

(16^{5}) ≡ 16 (mod 17)

(16^{5})^{2} ≡ 16^{2} (mod 17)

(16^{5})^{2} ≡ 256 (mod 17)

≡ 1 (mod 17) [∵ 255 is divisible by 17]

(16^{5})^{2} × (16^{5})^{2} × 2 ≡ 1 × 1 × 2 (mod 17)

∴ 2^{81} ≡ 2(mod 17)

∴ x = 2

Question 10.

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

Solution:

The duration of the 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.