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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.2

Question 1.

For what values of natural number n, 4th can end with the digit 6?

Answer:

4n = (2^{2})^{n} = 2^{2n}

= 2^{n} × 2^{n}

2 is a factor of 4^{n}

∴ 4^{n} is always even.

Question 2.

If m, n are natural numbers, for what values of m, does 2^{n} × 5^{n} ends in 5?

Solution:

2^{n} × 5^{m}

2^{n} is always even for all values of n.

5^{m} is always odd and ends with 5 for all values of m.

But 2^{n} × 5^{m} is always even and ends in 0.

∴ 2^{n} × 5^{m} cannot end with the digit 5 for any values of m. No value of m will satisfy 2^{n} × 5^{m} ends in 5.

Question 3.

Find the H.C.F. of 252525 and 363636.

Answer:

To find the HCF of 252525 and 363636 by using Euclid’s Division algorithm.

363636 = 252525 × 1 + 111111

The remainder 111111 ≠ 0

By division of Euclid’s algorithm

252525 = 111111 × 2 + 30303

The remainder 30303 ≠ 0

Again by division of Euclid’s algorithm

111111 = 30303 × 3 + 20202

The remainder 20202 ≠ 0

Again by division of Euclid’s algorithm.

30303 = 20202 + 10101

The remainder 10101 ≠ 0

Again by division of Euclid’s algorithm.

20202 = 10101 × 2 + 0

The remainder is 0

∴ The H.C.F. is 10101

Question 4.

If 13824 = 2^{a} × 3^{b} then find a and b?

Answer:

Using factor tree method factorise 13824

13824 = 2^{9} × 3^{3}

Given 13824 = 2a × 3b

Compare we get a = 9 and b = 3

Aliter:

13824 = 2^{9} × 3^{3}

Compare with

13824 = 2^{a} × 3^{b}

The value of a = 9 b = 3

Question 5.

If p_{1}^{x1} × p_{2}^{x2} × p_{3}^{x3} × p_{4}^{x4} = 113400 where p_{1} p_{2}, p_{3}, p_{4} are primes in ascending order and x_{1}, x_{2}, x_{3}, x_{4}, are integers, find the value of p_{1},p_{2},p_{3},p_{4} and x_{1},x_{2},x_{3},x_{4}.

Answer:

Given 113400 = p_{1}^{x1} × p_{2}^{x2} × p_{3}^{x3} × p_{4}^{x4}

Using tree method factorize 113400

113400 = 23 × 34 × 52 × 7

compare with

113400 = p_{1}^{x1} × p_{2}^{x2} × p_{3}^{x3} × p_{4}^{x4}

P_{1} = 2, x_{1} = 3

P_{2} = 3, x_{2} = 4

P_{3} = 5, x_{3} = 2

P_{4} = 7, x_{4} = 1

Question 6.

Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of Arithmetic.

Answer:

Factorise 408 and 170 by factor tree method

408 = 2^{3} × 3 × 17

170 = 2 × 5 × 17

To find L.C.M. list all prime factors of 408 and 170 of their greatest exponents.

L.C.M. = 2^{3} × 3 × 5 × 17

= 2040

To find the H.C.F. list all common factors of 408 and 170.

H.C.F. = 2 × 17 = 34

L.C.M. = 2040 ; HCF = 34

Question 7.

Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36?

Answer:

The greatest number of 6 digits is 999999.

The greatest number must be divisible by L.C.M. of 24, 15 and 36

24 = 2^{3} × 3

15 = 3 × 5

36 = 2^{2} × 3^{2}

L.C.M = 2^{3} × 3^{2} × 5

= 360

To find the greatest number 999999 must be subtracted by the remainder when 999999 is divided by 360

The greatest number in 6 digits = 999999 – 279

= 999720

Question 8.

What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?

Solution:

35 = 5 × 7

56 = 2 × 2 × 2 × 7

91 = 7 × 13

LCM of 35, 56, 91 = 5 × 7 × 2 × 2 × 2 × 13 = 3640

∴ Required number = 3647 which leaves remainder 7 in each case.

Question 9.

Find the least number that is divisible by the first ten natural numbers?

Answer:

Find the L.C.M of first 10 natural numbers

The least number is 2520

Modular Arithmetic

Two integers “a” and “b” are congruence modulo n if they differ by an integer multiple of n. That b – a = kn for some integer k. This can be written as a = b (mod n).

Euclid’s Division Lemma and Modular Arithmetic

Let m and n be integers, where m is positive. By Euclid’s division Lemma we can write n = mq + r where 0 < r < m and q is an integer.

This n = r (mod m)