A four-digit number gets reversed when it is multiplied by 4. find the sum of the digits of the number.

• let number is (1000x + 100 y + 10z + w)
  so, x must be 2. if it's greater then 2 then no. will be of 5 digit
  & x=1 is also not possible bcoz if no. is reversed then unit digit is 1 which can't be multiple of 4.
  
  from given condn
  4000x + 400y + 40z + 4w = 1000w + 100z + 10y + x
  => 3999x + 390y - 60z - 996w = 0
  => 1333x + 130y - 20z - 332w = 0-----(1)
  
  now w can be 3 or 8 coz last digit of reversed no. is 2 [3*4 & 8*4 gives 2]
  
  so, w=8, as 2yzw * 4 = wzy2
  
  put x=2,w=8 in (1)
  2666 + 130y - 20z - 2656 = 0
  => 10 + 130y - 20z = 0
  => 1 + 13y - 2z = 0
  => 2z = (13y +1)
  only possible value of y = 1 [ for y > 1 , z is not single digit]
  so, z=7
  
  so, no. is 2178 & 2178 * 4 = 8712
  
  sum of digits = x+y+z+w = 2+1+7+8 = 18
  
• 10 years agoHelpfull: Yes(8) No(1)
• Answer is 2+1+7+8= 18.
  number is 2178.
  2178*4= 8712
  
• 10 years agoHelpfull: Yes(4) No(2)
• 2178*4 = 8712
• 10 years agoHelpfull: Yes(3) No(4)
• Let the 4-digit number be x = 1000a+100b+10c+d
  
  4*(1000a+100b+10c+d) = 1000d+100c+10b+a ---(1)
  
  a = 1 or 2 (because 4*x is a 4 digit number, so a < 3)
  
  From equation (1)
  4*a = d ---(2)
  and
  4*d = a ---(3)
  
  To satisfy both (2) and (3)
  a = 2 and d = 8
  
  4*(2000+100b+10c+8) = 8000+100c+10b+2
  8032 + 400b + 40c = 8002 + 100c + 10b
  30 + 390b = 60c
  1 + 13b = 2c ---(4)
  
  From (4)
  b = 1 (max possible value of c is 9 for which RHS = 18 so, LHS
• 10 years agoHelpfull: Yes(2) No(2)
• 
  b = 1 (max possible value of c is 9 for which RHS = 18 so, LHS < 18 and hence b=1)
  
  2c = 13*1 + 1
  = 14
  c =14/2 = 7
  
  a+b+c+d = 2+1+7+8 = 18
  
  
• 10 years agoHelpfull: Yes(1) No(1)
• @dipin bro please tell the solution...
• 10 years agoHelpfull: Yes(0) No(2)