SOLUTION IN 61 TO 70

  61.   6 = 3 x 2. clearly, 5 * 2 is divisible by 2 2. Replace * by 4.

Then, (5 + x + 2) must be divisible boy 3.so, x=2

62.   24. = 3x 8, where 3 and 8 are co –prime

Clearly, 35718 is not divisible by 8, as 718 is not divisible by 8

Similarly, 63810is not divisible by 8and 537804 is not divisible by 9.

 Consider part (d).

 Sum of digits = (3 + 1+ 2+5 + 7 + 3 + 6+) = 27, which is divisible by 3.

Also, 736 is divisible by 8

3125736 is divisible by (3 x 8 ), i.e., 24

63.   132 = 11 x 3 x 4.

Clearly, 968 is not divisible by3

None of 462 and2178 is divisible by 4.

And, 5184 is not divisible by 11.

Each one of the remaining four numbers is divisible by 11.so, there are 4 such numbers.

64.   Let the given numbers be 476 Xy 0.

Then ( 4+ 7 + 6 +X + y + 0) = (17 + X+ y)  must be divisible bu 3.

And, (0 + X + 7) – ( y + 6+ 4 ) = ( X –y- 3) must be either 0 or 11.

X- y – 3 =0         y = X -3

(27 + X + y ) = (17 + X + X – 3)= ( 2X + + 14) X = 2 or X = 8

X = 8 and y + 5.

65.   (4 +5 + 2 ) – (1 + 6 + 3) =1, not divisible by 11.

        (2 + 6 + 4) – (4 + 5 + 2) = 1, not divisible by 11.

        (4 + 6+1) – (2 + 5 + 3) = 1, not divisible by 11.

        (4 + 6 + 1) – (2 5 + 4) = 0, so, 415624 is divisible by 11.

66.   Required numbers are 102, 108, 114, ….., 996

          This is an A.P. in which a = 102, d = 6 and l = 996

           Let the number of terms be a. then,

           A + (n – 1)d = 996         102 + (n -1) x 6 = 996

           6 x (n – 1) 894                 (n- 1) = 149   n = 150

             Required number of terms = 150.

SOLUTION IN 51 TO 60

  51.

Given exp. = (a² + b² + - ab)  = 1  =         1                  = 1

                                                                       (a³ + b³)           (a + b)   (753 +247)    1000

52.   Sum of digits  = (5 +1 + 7 + X + 3+ 2 + 4) = (22 + X), which must be divisible by 3.

X=2.

53.   Sum of digits = (4 + 8 +1 X + 6 + 7 + 3) = (29 + X), which must be divisible by 9.

      X = 7.

54.   Given number = 97215X6

(6 + 5 + 2 + 9) – (X + 1 + 7) = (14 – X), which must be divisible by 11.

   X = 7.

55.   The number 6X2 must be divisible by 8.

(X =3, as 632 is divisible by 8.

56.   45 = 5 x 9, where 5 and 9 are co - primes.

Unit digit must be 0 or 5 and sum of digits must be divisible by 9.

Among given numbers, such number is 202860.

57.   99 = 11 x 9 where 11 and 9 are co-primes.

By hit and trial, we find that 114345 is divisible by 11 as well as 9. So, it is divisible by 99.

58.   72 = 9 x 8, where 9 and 8 are co-prime

The minimum value of x for which 73 x is divisible by 8 is, X = 6.

Sum of digits in 425736 = (4 _ 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 9.

Required value of * is 6.

59.   80 = 2 x 5 x 8.

Since 653 Xy is divisible by 2 and 5 both, so Y= 0.

Now, 653 X0 is divisible by 8, so 3X0 should be divisible by 8. This happens when X =2 .

X + y =(2 +|0) = 2.