**Samacheer Kalvi 10th Maths Book Back Solution:**

Tamil Nadu **10th Maths Book Back Answers Unit 2 – Numbers and Sequences Ex 2.3** are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete** Samacheer Kalvi Books Back Answers/Solutions** are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 2 Numbers and Sequences consists of 5 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Numbers and Sequences Ex 2.3 Answers in English.

**10th Maths Book Back Answers/Solutions:**

TN Samacheer Kalvi 10th Maths Unit 2 Chapter 3 Book Back Exercise has given below. The 10th Maths Book Back Solutions Guide is uploaded below,

**Chapter 2**

** Numbers and Sequences Ex 2.3**

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 = x7 (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 ≡ x7 (mod 5)

96 – x7 = 5n for some integer n.

672−x7 = 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 = 6n+45

When we put 1, 6, 11, … as n values in x = 6n+45 which is divisible by 5.

When n = 1, x = 105 = 2

When n = 6, x = 36+45 = 405 = 8 and so on.

∴ The solutions are 2, 8, 14…..

∴ Least value is 2.

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

3.Solve 5x ≡ 4 (mod 6)

**Solution:**

5x ≡ 4 (mod 6)

5x – 4 = 6M for some integer n.

5x = 6n + 4

x = 6n+45 where n = 1, 6, 11,…..

∴ x = 2, 8, 14,…

4.Solve 3x – 2 ≡ 0 (mod 11)

**Solution:**

3x – 2 ≡ 0 (mod 11)

3x – 2 = 11 n for some integer n.

3x = 11n + 2

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

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

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.

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

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

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 412 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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