Page 22 - classs 6 a_Neat

EXAMPLE :30 We know that 13 is a factor of each one of the numbers 65 and 117.
So, 13 must be a factor of(l 17 - 65) = 52. PRACTICE EXERCISE 2 .3
Clearly. 13 divides 52 exactly.
1. Test the divisibility of the following numbers by 2.
TO FIND PRIME NUMBERS BETWEEN 100 AND 200 2. Test the divisibility of the following numbers by 3.
a. 417 b. 8,146 c. 5,190 d. 98,117 e. 1,34,712

We know that 15 × 15 > 200. 3. Test the divisibility of the following numbers by 4.
So, we adopt the following rule: a. 31,672 b.560 c. 3,807 d. 9,865 e. 8,73,248
RULE Examine whether the given number ts divisible by any prime number less than 15. If yes then it is not 4. Test the divisibility of the following numbers by 5.
prime: otherwise it is prime a. 3,140 b. 1,560 c. 12,345 d. 24,685 e. 55,557
5. Test the divisibility of the following numbers by 6.
. a. 18,198 b. 13,467 c. 34,542 d. 3,21,003 e. 3,00,630
EXAMPLE :30 Which of the following are prime numbers? 6. Test the divisibility of the following numbers by 7.
(i) 117 (ii) 139 (iii) 193 a. 364 b. 1,505 c. 3,199 d. 7,021 e. 4,507
7. Test the divisibility of the following numbers by 8.
SOLUTION (i) Test the divisibility of 117 by each one of the prime numbers 2, 3, 5, 7, 11, 13, a. 1,728 b. 47,780 c. 63,904 d. 1,36,979 e.17,90,184
taking one by one. We find that 117 is divisible by 13 . So, 117 is no t a prime number. 8. Test the divisibility of the following numbers by 9.
a. 3,159 b.7,218 c. 7,878 d.1,34,755 e. 4,57,893
(ii)Test the divisibility of 139 by each one of the prime numbers 2, 3, 5, 7, 11, 13. We find that 9. Test the divisibility of the following numbers by 10.
139 is divisible by none of them. So, 139 is a prime number. a. 10,010 b. 1,00,005 c. 21,420 d. 89,760 e. 64,64,645
Test the divisibility of 193 by each one of the prime numbers 2, 3, 5, 7, 11, 13 . 10. Test the divisibility of the following numbers by 11.
We find that 193 is divisible by none of them. a. 7,69,483 b. 3,581 c. 17,864 d. 20,916 e. 4,598
So, 193 is a prime number. f. 492 g. 683 h. 7,876 i.1,120 j. 2,467

TO FIND PRIME NUMBERS BETWEEN 100 AND 400
PRIME FACTORIZATION
We know that 20 × 20 = 400. PRIME FACTOR A factor of a given number is called a prime factor if this factor is a prime number.
Rule Examine whether the given number is divisible by any prime number less than 20. if yes then it is not
prime; otherwise it is prime. EXAMPLE 1: 2 and 3 are prime factors of 12.
EXAMPLE :31 Which oj the jollowing is a prime number?
(i) 263 (ii) 323 (iii) 361 PRIME FACTORIZATION To express a given number as a product of prime factors is called prime factoriza-
SOLUTION (i) Testthe divisibilityof263 byeach one of theprimenumbers2,3,5, 7, 11, 13, 17, 19 . tion or complete factorization of the given number.
We find that 263 is no t divisible by any of these numbers.
So, 263 is a prime number. EXAMPLE 2: Let us factorize 36 in three different ways as given below:
36 = 2 × 18 36 = 3 × 12 36 = 9 × 4
(ii)Test the divisibilityof 323 byeach one of thenumbers2,3,5, 7, 11, 13, 17, 19. 2 × 9 2 × 6 3 × 3 2 × 2
We find that 323 is divisible by 1 7. 3 × 3 2 × 3
∴ 323 is no t a prime number. Thus, 36 = 2 × 2 × 3 × 3: 36 = 3 × 2 × 2 × 3; 36 = 3 × 3 × 2 × 2.

(iii) Testthe divisibilityof361 byeach one oftheprimenumbers2,3,5, 7, 11, 13, 17, 19. We notice here that in each of the prime factorizations, the factors may be arranged differently but, in fact, they
We find that 361 is divisible by 19 . are the same.
Hence, 361 is no t a prime number. Thus, we generalise this result as under.
JUST TRY:1 121, 2,442, 9a,ns, 2a,4s7. 61,9oa Every composite number can be factorized into primes in only one way, except for the order of primes.
This property is known as unique factorization property.