1)in how many ways can the letters of english alphabet be arranged so that there are 7 letters between A & B n no repeated.
2)Find no.of zeroes in 15*32*25*22*40*75*98*112*125
3)what must be added to 5678 to get remainder 35 when divided by 460?
4) f(x)=1+x+x^2+......x^6.
Find remainder when f(x^7) is divided by f(x).
5)How many integers satisfy (x^2-x-1)^(x+2)=1?
6)there are 3 single digit distinct numbers A,B,C in geometric progression.
how many possible different values are for abs(A+B-C).

• Sol 1 : A and B are fixed and can be arranged by 2!
  remaining letters=24
  Arrange 7 letters out of 24= 24P7
  Now consider A +7 letters+B as 1 letter, now remaining letters=26-9=17
  Then 1letter and 17 letters can be arranged by 18!
  i.e ans : 24P7*2!*18!
  
  Sol 2 : (3*5)*(2^5)*(5^2)*(11*2)*(5*2^3)*(3*5^2)*(49*2)*(7*2^4)*(5^3)
  so no of 5 = 9
  no of 2=14
  5*2=10
  hence no of zero=9
  
  sol :3 firstly take 5678 and divide it with 460
  reminder = 158
  now, we have to add a no. in 158 that can make vale 35 more then 460 i.e. 495
  hence simply we have to add 337 in 158 to get reminder 35
  hence
  ans : 337
  
  sol : 4 put x=0
  so f(x)=1
  f(x^7)=1
  1/1=0
  so 0 is answer
  
  sol: 5 at x = 0 => (-1)^2 = 1
  x = -1 => (1 + 1 -1)^1 = 1
  x = -2 => (4 + 2 - 1)^0 = 1
  x = 2 => (4 - 2 - 1)^4 = 1
  hence only 4 can satisfy
  ans : 4
  
  sol : 6 A,B,C may be
  (1,2,4) & (4,2,1)
  (1,3,9) & (9,3,1)
  (2,4,8) & (8,4,2)
  (4,6,9) & (9,6,4)
  
  find abs(A+B-c) for these 8 GPs
  1, 5, 5, 11 , 2, 10, 1 , 11
  
  we get 5 different values (1,2,5,10,11)
  
  ans : 5
  
• 9 years agoHelpfull: Yes(5) No(1)
• solution to question 4 will be 7.
  
  f(x)= 1+ x+ x^2+ x^3+......+x^n for x>1
  
  now when f(x^(n+1)) is divided by f(x) remainder is simply (n+1)
  
  so, in this case remainder is (6+1)= 7
• 9 years agoHelpfull: Yes(5) No(3)
• @ vibk
  approach is wright but logic is slightly mistaken.....
  u said that u have taken 7 letters , a & b ans shuffled them in 3! ways.......
  but see the condition given in the question... i.e. a 7 letters arranged in between a and b
  hence 7 digits shuffled in 24P7 ways....
  now.... A _ _ _ _ _ _ _ B hence A & B only shuffled in two ways i.e.
  A _ _ _ _ _ _ _ B itself
  ans B _ _ _ _ _ _ _ A
  i.e. 2! ways
  and once this operation completed consider all of the digits as a single digit to be shuffled in rest 17 letters....
  ........got my point
• 9 years agoHelpfull: Yes(2) No(0)
• A -B or B - A --------------------> 2 WAYS
  from rest 24 letters, 7 can be arranged in 24P7 = 24! ways
  also position of A-------B can be 1-------9, 2--------10, ..., 18-------26 ( 18 ways )
  total no. of ways = 24P7 * 2 * 18! ways.
  
• 9 years agoHelpfull: Yes(2) No(1)
• davish thanku bt cn u tell in sol 1 y are treating A+7letters+B as one letter
• 9 years agoHelpfull: Yes(1) No(3)
• simply in solution 1 you have taken 24P7*2! for the three of these i.e. 24P7 for all 7 letters and 2! for a & b hence these three are treated as a single letter for our convenience... got
• 9 years agoHelpfull: Yes(1) No(1)
• @davish i have dont regarding first solution...i think it shud be 24P7*3!*7!
  consider 7 letters as 1 nd we get three arrangements to make i.e. A+1letter+B i.e. 3! nd rest 7! for 7 letters..plz correct me if ma approach iz wrong
• 9 years agoHelpfull: Yes(1) No(0)
• how you r selecting digits of the number for qs 6
• 9 years agoHelpfull: Yes(0) No(1)