SOLUTION IN 141 TO 150

141   Let the required number be X.

       Then, (11 X + 11) = 11 ( X + 1) is divisible by 13, so, X =12

14     Let the 3-digit number be X y z . then,

      ( 100 X + 10 y + z) – (X + y + z) = 99 X + 9 y = 9 (22 X + y), which is divisible by 9.

143.     Putting X =5 and y + 1, we get (3 X + 7 y) = ( 3 x 5 + 7 x 1 ) = 22, which is divisible           by 11.

4 x + 5 y = (4 x 5 x 5 x 1 ) = 25, which is not divisible by 11.

X + y + = (5 + 1 + 4) = 9, which is not divisible by 11.

9 x + 4 y (9 x 5 x 4 x 1 ) = 49, which is not divisible by 11..

4 X – 9 y = (4 x 5 – 9 x 1) = 11, which is divisible by 11.

144.                   4 a 3

                 9  8  4                                       a + 8 = b     b- a + 8

                 13 b 7

 

Also, 13 b 7 is divisible by 11.

 (7 + 3) – ( b + 1 )  = ( 9 =b) ( 9 –b) = 0  b= 9.

B = 9 and a = 1 (a+ b) =10.

145.                  Required number = (2⁵ - 2 ) =(32 – 2) 30.

146.                  Required number = product of first three multiples of 3 = (3 x 6 x 9 ) = 162

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147.                  On dividing 1000 by 45, we get remainder = 10.

                Required number to be added = ( 45 – 10) = 35.

148.                  On dividing 803642 y 11, we get remainder = 70.

                Required number to be added = (11 – 4 ) = 7.

149.                  On dividing 6709 by 9, we get remainder = 4.

                Required number to be subtracted = 4.

150.                  On dividing 427398 by 15, we get remainder = 3.

                Required number to be subtracted = 3.