23 Nov 2020 7:53 am

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)

Answer:

71 = 7 (mod 8)

∴ The value of x = 7

(ii) 78 + x = 3 (mod 5)

78 + x – 3 = 5n (n is any integer)

75 + x = 5n

(Let us take x = 5)

75 + 5 = 80 (80 is a multiple of 5)

∴ The least value of x is 5

(iii) 89 = (x + 3) (mod 4)

89 – (x + 3) = 4n

(n may be any integer)

89 – x – 3 = 4n

89 – x = 4n

86 – x is a multiple of 4

(84 is a multiple of 4)

86 – 2 = 4n

84 = 4n

The value of x is 2

(iv) 96 = x7 (mod 5)

96 – x7 = 5n (n may be any integer)

672 – x = 35n (multiple of 35 is 665)

672 – 7 = 665

∴ The value of x = 7

(v) 5x = 4 (mod 6)

5x – 4 = 6n (n may be any integer)

5x = 6n + 4

x = 6n+45

Substitute the value of n as 1, 6, 11, 16 …. as n values in x = 6n+45 which is divisible by 5.

2, 8, 14, 20,…………

The least positive 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)

5x – 4 = 6n (n may be any integer)

5x = 6n + 4

x = 6n+45

The value of n 1, 6, 11, 16 ……..

∴ The value of x is 2, 8, 14, 20 …………..

Question 4.

Solve 3x – 2 = 0 (mod 11)

Answer:

Given 3x – 2 = 0(mod 11)

3x – 2 = 11n (n may be any integer)

3x = 2 + 11n

x = 11n+23

Substitute the value of n = 2, 5, 8, 11 ….

When n ≡ 2 ⇒ x = 22+23 = 243 = 8

When n = 5 ⇒ x = 55+23 = 573 = 19

When n = 8 ⇒ x = 88+23 = 903 = 30

When n = 11 ⇒ x = 121+23 = 1233 = 41

∴ The value of x is 8, 19, 30,41

Question 5.

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

Answer:

100 ≡ x (mod 12) Note: In a clock every 12 hours

100 ≡ 4 (mod 12) the numbers repeats.

The time repeat after 7 am is 7 + 4 = 11 o’ clock (or) 11 am.

Question 6.

What is 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?

Answer:

Number of days in a week = 7

45 ≡ x (mod 7)

45 ≡ 3 (mod 7)

The value of x must be 3.

Three days after tuesday is friday uncle will come on friday.

Question 8.

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

Solution:

2^{1} + 6 × 9^{1} = 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}.2^{1} – 9.2^{k}

= 63m – 2^{k} (9 – 2) = 63m – 7.2^{k}

= 7 (9m – 2^{k}) which is divisible by 7

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

Question 9.

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

Answer:

2^{81} ≡ x(mod 17)

2^{40} × 2^{40} × 2^{1} ≡ x(mod 17)

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

(16)^{10} × (16)^{10} × 2^{1} ≡ x(mod 17)

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

= 1 × 1 × 2 (mod 17)

[(16)^{2} = 256 = 1 (mod 17)]

= 2 (mod 17)

2^{81} = 2(mod 17)

∴ x = 2

The remainder is 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?

Answer:

Duration of the flight time = 11 hours

(Chennai to London)

Starting time on Sunday = 23 : 30 hour

Time difference is 4 12 horns ahead to london

The time to reach London airport = (10.30 – 4.30)

= 6 am

The first reach the london airport next day (monday) at 6 am