Aptitude - Numbers

Let 2³² = x. Then, (2³² + 1) = (x + 1).

Let (x + 1) be completely divisible by the natural number N. Then,

(2⁹⁶ + 1) = [(2³²)³ + 1] = (x³ + 1) = (x + 1)(x² - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.

 23) 1056 (45
      92
      ---
      136
      115
      ---
       21
      ---
     
 Required number = (23 - 21)    
                 = 2.   

132 = 4 x 3 x 11

So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.

264 11,3,4 (/)

396 11,3,4 (/)

462 11,3 (X)

792 11,3,4 (/)

968 11,4 (X)

2178 11,3 (X)

5184 3,4 (X)

6336 11,3,4 (/)

Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.

Required number of number = 4.

935421 x 625 = 935421 x 54 = 935421 x		10		4
		2

=	935421 x 104	=	9354210000
	24		16

= 584638125

Largest 4-digit number = 9999

 88) 9999 (113
     88
     ----
     119
      88
     ----
      319
      264
      ---
       55
      ---
       
 Required number = (9999 - 55)
                 = 9944.        

Unit digit in (6374)¹⁷⁹³ = Unit digit in (4)¹⁷⁹³

    = Unit digit in [(4²)⁸⁹⁶ x 4]

    = Unit digit in (6 x 4) = 4

Unit digit in (625)³¹⁷ = Unit digit in (5)³¹⁷ = 5

Unit digit in (341)⁴⁹¹ = Unit digit in (1)⁴⁹¹ = 1

Required digit = Unit digit in (4 x 5 x 1) = 0.