There is a 7 digit number X in which the units digit and the first digit is divisible by 4. The middle digit is divisible by 3. A number X' is formed by reversing the digits of the number X (eg. If X=1234567 the X' =7654321). Y=X-X'. What is the remainder when Y is divided by 18.
a)R < 3 b)3

• ans: Rem = 0
  
  unit & last digit of X is multiple of 4, can be 0,4,8
  so,Y=X-X' is always an even no.
  => Y is divisible by 2
  
  also , difference of a no. & its reverse is always a multiple of 9
  [for 2 digit no., 10x+y-(10y+x)= 9(x-y)
  for 3 digit no., 100x+10y+z-(100z+10y+x)=99(x-z) etc]
  => Y=X-X' is divisible by 9
  
  Y is multiple of both 2 & 9 , so Y is multiple of 18
  so, when Y is divided by 18 remainder is always zero.
  
• 10 years agoHelpfull: Yes(130) No(1)
• If we take a no. let
  X=8009004(since the first and last digits are divisible by 4)
  X'=4009008(reverse)
  Y=X-X'=3999996
  The no is divisible by 18 and leaves zero remainder
• 10 years agoHelpfull: Yes(10) No(3)
• 1st and last digits are divisible by 4 means 8*****4
  middle number divisible by 3 then consider as x=8123574 x'=4753218
  y is divisible by 18 and remainder is 0.
• 10 years agoHelpfull: Yes(2) No(0)
• between 9 and 17
• 10 years agoHelpfull: Yes(0) No(13)
• ans:- 0
  Can u explain aditya how it will be between 9 and 17?
• 10 years agoHelpfull: Yes(0) No(3)
• R
• 10 years agoHelpfull: Yes(0) No(3)
• ans 0
  4123454 x
  4543214 y
  y-x=41976/18=2332
  remainder=0
  
• 10 years agoHelpfull: Yes(0) No(2)
• 6845906704956709/76904760947/do u understand
• 10 years agoHelpfull: Yes(0) No(2)