Numbers - General Questions (6)

Then, t_n = 384     ar^n-1 = 384

 3 x 2^{n - 1} = 384

 2^n-1 = 128 = 2⁷

 n - 1 = 7

 n = 8

 Number of terms = 8.

The smallest 5-digit number = 10000.

 41) 10000 (243
     82
     ---
     180
     164
     ----
      160
      123
      ---
       37
      --- 

 Required number = 10000 + (41 - 37)
                 = 10004.    

Given Exp. =(a³ + b³)= (a + b) = (768 + 232) = 1000(a² - ab + b²)

Largest 5-digit number = 99999

 91) 99999 (1098
     91
     ---
     899
     819
     ----
      809
      728
      ---
       81
      ---
 
 Required number = (99999 - 81)
                 = 99918.

The smallest 6-digit number 100000.

 111) 100000 (900
      999
      -----
        100
        ---

 Required number = 100000 + (111 - 100)
                 = 100011.       

Unit digit in 7⁹⁵ = Unit digit in [(7⁴)²³ x 7³]
= Unit digit in [(Unit digit in(2401))²³ x (343)]
= Unit digit in (1²³ x 343)
= Unit digit in (343)
= 3

Unit digit in 3⁵⁸ = Unit digit in [(3⁴)¹⁴ x 3²]
= Unit digit in [Unit digit in (81)¹⁴ x 3²]
= Unit digit in [(1)¹⁴ x 3²]
= Unit digit in (1 x 9)
= Unit digit in (9)
= 9

Unit digit in (7⁹⁵ - 3⁵⁸) = Unit digit in (343 - 9) = Unit digit in (334) = 4.

So, Option B is the answer.

This is an A.P. in which a = 51, l = 100 and n = 50.

Sum =	n	(a + l)	=	50	x (51 + 100)   = (25 x 151)   = 3775.
	2			2