Aptitude - Numbers

Let S_n = (2 + 4 + 6 + ... + 30). This is an A.P. in which a = 2, d = 2 and l = 30

Let the number of terms be n. Then,

a + (n - 1)d = 30

     2 + (n - 1) x 2 = 30

    n = 15.

Sn =	n	(a + l)	=	15	x (2 + 30) = (15 x 16) = 240.
	2			2

Then number 6x2 must be divisible by 8.

   x = 3, as 632 is divisible 8.

By hit and trial, we find that

47619 x 7 = 333333.

Let the number be x. Then

60% of	3	of x = 36
	5

	60	x	3	x x = 36
	100		5

x =		36 x	25		= 100
			9

Required number = 100

80 = 2 x 5 x 8

Since 653xy is divisible by 2 and 5 both, so y = 0.

Now, 653x is divisible by 8, so 13x should be divisible by 8.

This happens when x = 6.

x + y = (6 + 0) = 6.

Let the two consecutive odd integers be (2n + 1) and (2n + 3). Then,

(2n + 3)² - (2n + 1)² = (2n + 3 + 2n + 1) (2n + 3 - 2n - 1)

     = (4n + 4) x 2

     = 8(n + 1), which is divisible by 8.

Unit digit in (4137)⁷⁵⁴ = Unit digit in {[(4137)⁴]¹⁸⁸ x (4137)²}

=Unit digit in { 292915317923361 x 17114769 }

= (1 x 9) = 9

 4 | x            z = 6 x 1 + 4  = 10
 -----------
 5 | y -2         y = 5 x z + 3  = 5 x 10 + 3  = 53
 -----------
 6 | z - 3        x = 4 x y + 2  = 4 x 53 + 2  = 214
 -----------
   | 1 - 4
   
Hence, required number = 214.