Page 14 - NCERT Books
P. 14

102   MATHEMATICS


                              Solution: We have, 9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7
                              If we divide 9408 by the factor 3, then
                              9408 ÷ 3 = 3136 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7  which is a perfect square. (Why?)
                              Therefore, the required smallest number is 3.
                              And,                            3136   = 2 × 2 × 2 × 7 = 56.

               2    6, 9, 15  Example 8:  Find the smallest square number which is divisible by each of the numbers
               3    3, 9, 15  6, 9 and 15.
               3    1, 3, 5   Solution: This has to be done in two steps. First find the smallest common multiple and
               5    1, 1, 5   then find the square number needed. The least number divisible by each one of 6, 9 and
                    1, 1, 1   15 is their LCM. The LCM of 6, 9 and 15 is 2 × 3 × 3 × 5 = 90.
                              Prime factorisation of 90 is 90 = 2 × 3 × 3 × 5.
                                 We see that prime factors 2 and 5 are not in pairs. Therefore 90 is not a perfect
                              square.
                                 In order to get a perfect square, each factor of 90 must be paired. So we need to
                              make pairs of 2 and 5. Therefore, 90 should be multiplied by 2 × 5, i.e., 10.
                              Hence, the required square number is 90 × 10 = 900.

                                             EXERCISE 6.3

                               1. What could be the possible ‘one’s’ digits of the square root of each of the following
                                   numbers?
                                    (i) 9801         (ii) 99856          (iii) 998001        (iv) 657666025
                               2. Without doing any calculation, find the numbers which are surely not perfect squares.
                                    (i) 153            (ii) 257            (iii) 408           (iv)  441
                               3. Find the square roots of 100 and 169 by the method of repeated subtraction.
                               4. Find the square roots of the following numbers by the Prime Factorisation Method.
                                    (i) 729            (ii) 400            (iii) 1764          (iv) 4096
                                   (v) 7744            (vi) 9604          (vii) 5929          (viii) 9216
                                   (ix) 529            (x) 8100
                               5. For each of the following numbers, find the smallest whole number by which it should
                                   be multiplied so as to get a perfect square number. Also find the square root of the
                                   square number so obtained.
                                    (i) 252            (ii) 180            (iii) 1008          (iv) 2028
                                   (v) 1458            (vi) 768
                               6. For each of the following numbers, find the smallest whole number by which it should
                                   be divided so as to get a perfect square. Also find the square root of the square
                                   number so obtained.
                                    (i) 252            (ii) 2925           (iii) 396           (iv) 2645
                                   (v) 2800            (vi) 1620
                               7. The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’s
                                   National Relief Fund. Each student donated as many rupees as the number of students
                                   in the class. Find the number of students in the class.
   9   10   11   12   13   14   15   16   17   18   19