EXERCISE 2 SOLUTIONS: IN 1 TO 10:

                                         

1.       Clearly, 252 = 2 x 2 x 3 x 3 7.

2.       99 = 1 x 3 x 3 x 11; 101 = 1 x 101;

176= 1 x 2 x 2 x 2 x 2 x 11; 182 = 1 x 2 x 7 13.

  So, divisors of 99 are 1, 3, 9, 33 and 99;

  Divisors of 101 are 1 and 101;

  Divisors of 176 are 1, 2, 4, 8, 16, 22, 44, 88 and 176;

  Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182,

  Hence, 176 has the most number of divisors.

 3.       N                       divisors excluding n                          sum of divisors

        6                                 1, 2, 3                                               6

        9                                 1, 3                                                   4

      15                                 1, 3, 5                                                9

      21                                 1, 3, 7                                               11

Clearly, 6 is a perfect number

4.      H.C.F. = product of lowest powers of common factors = 2² x 3² x 5

5.       H.C.F. = product of lowest powers of common factors =- 3 x 5 x 7 =105

6.       4 x 27 x 3125 = 2² x 3³x 5⁵; 8 x 9 x 25 x 7 =2³ x 3² x 5² x 7;

1    6 x 81 x 5 x 11 x 49 x = 2⁴ x 3⁴ x 5 x 7² x 11

      H.C.F. = 2² x 3² x 5 = 180

7. 36 =2² X 3²; 84 = 2² X 3 X 7.

          H.C.F. = 2² x 3 = 12

8.       204= 2² x 3 x 17; 1190 = 2 x 5 x 7x 17; 1445 = 5 x 17².

H.C.F. = 17.  

9. H.C.F. of 18 and 25 is 1. So, they are co-primes.

10. L.C.M. =product of highest powers of prime factors = 2⁵ x 3⁴ x 5³ x 7² x 11.