Sunday, 22 February 2015

H.C.F. AND L.C.M. OF NUMBERS -(EXERCISE 2 SOLUTIONS 61 TO 70:)

61. Greatest number of 4 digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600.

         On dividing 9999 by, 600, the remainder is 399.
         Required number = (9999 – 399) = 9600.

61. L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37.

        Number to be added = (60 – 37) = 23.

62. The least number divisible by 16, 20, 24

         = L.C.M. of 16, 20, 24 = 240 = 2 x 2 x 2 x 2 x 3 x 5.
         To make it a perfect square, it must be multiplied by 3 x 5.
         Required number =240 x 3 x 5 =360.

63. Required number = (L.C.M. of 12, 16, 18, 21, 28) + 7 = 1008 + 7 =1015.

64. Required number = (L.C.M. of 24, 32, 36, 54) – 5 864 – 5 = 859.

65. Required number = (L.C.M. of 12, 15, 20, 54) + 8 540 + 8 = 548
.
66. Greatest number of 4 digits is 9999. L.C.M.  Of 4, 7 and 13 = 364.

         On dividing 9999 by 363, remainder obtained is 171.
         Greatest number of 4 digits divisible by 4, 7 and 13 = (9999 -171) = 9828.
         Hence, required number = (9828 + 3) 9831.

67. least number of 6 digits is 100000. L.C.M. of 4, 6, 10 and 15= 60

           On dividing 100000 by 60, the remainder obtained is 40.
           Least number of 6 digits divisible by 4, 6, 10 and 15 = 100000 + (60 – 40) = 100020.
           N = (100020) + 2) = 1---22. Sum of digits in N = (1 + 2 + 2) 5.

68. L.C.M.  Of 6, 9, 15 and 18 is 90.

          Let required number be 90 K +4, which is a multiple of 7.
          Least value of k for which (90 x 4 + 4 = 364.

 69. Here (48 -38) = 10, (60 – 50) = 10, (72 – 62) = (108 -98) = 10 & (140 -130) = 10.

        Required number = (L.C.M. of 48,60, 72, 108, 140) - = 15120 – 10 = 15110.

70. Here (18- 7) = 11, (21 – 10) =11 and (24, - 13) =11. L.C.M.  Of 18, 21 and 24 is 504.
        Let required number be 504 k – 11.
        Least value of k for which (504 x 6 – 11 = 3024 – 11 = 3013.

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