SOLUTION IN 131 TO 140

131.   Clearly, 325325 is divisible by a ll7 , 11 and 13.

132.   Sum of digits = 35 and so it is not divisible by 3.

               

                   (Sum of digits at odd places) – (sum of digits at even places) = (19 – 16) = 3, not                                        divisible by 11

                   So, the given number is neither divisible by 3 nor by 11.

133.   Since 111111 is divisible by each one of 7, 11 and 13, so each one of given type of           numbers is divisible by each one of 7, 11 and 13, so each one of given type of numbers is  divisible by each one of 7, 11, 13, as we may write, 222222 = 2 x 111111, 333333 = 3 x  11111, etc.

134.   Smallest 3-digit prime number is 101. Clearly, 2525 = 25 x 101; 3232 = 32 x 101, etc each    such number is divisible by 101.

135.   256256 = 2456 x 1001; 678678 = 678 x 101, etc

        So, any number of this form is divisible by 1001.

136.   Required number = 1 x 2 x 3 x 4 = 24.

137   Required number = ( 2 x 4 x 6 x ) = 48.

139.   Let the three consecutive odd numbers be (2 X + 1), (2 X + 3) and (2 X+ 5).

        Their sum = (6 X +9) ( 2 X + 3) , which is always divisible by 3.

140.   Let the two consecutive odd integers be (2 X + 1) and ( 2 X + 3)

        Then (2 X + 3 )² -  (2 X +3 + 2 X + 1 ) *( 2 X + 3 – 2 X -1) = ( 4 X + 4) x 

        = 8 ( X + 1), which is always divisible by 8/