abcde is a 5-digit number such that abcde*4=edcba.....a,b,c,d,e are different digits...find abcde?

• 21978 will be be ans
• 11 years agoHelpfull: Yes(1) No(0)
• answer is 21978 * 4 = 87912
  Since EDCBA is a 5-digit number, we know ABCDE is < 1/4 × 100000.
  ABCDE < 25000, so A is 1 or 2.
  
  But EBCDA is a multiple of 4, so it is even, so A = 2.
  4 * ....E = ....2, so E is either 3 or 8.
  But, 4 × 2nnnn cannot be 3nnnn so E is 8.
  
  We have 2BCD8 * 4 = 8DCB2.
  and therefore 4 × BCD + 3 = DCB.
  Let's look at it case by case, for values of D:
  D = 1, B = 7
  D = 2, B = 1
  D = 3, B = 5
  D = 4, B = 9
  D = 5, B = 3
  D = 6, B = 7
  D = 7, B = 1
  D = 8, B = 5
  D = 9, B = 9
  D = 0, B = 3
  
  But B cannot be greater than 2, or there would be a carry, and B cannot
  be equal to 2, since A = 2 (and we assume each letter represents a
  different number).
  So, B = 1 and D = 2 or 7.
  But again, A is already 2, so D = 7.
  
  So we have 21C78 * 4 = 87C12
  We now have 4 * C + 3 = C (mod 10)
  3*C = 7 mod 10
  and therefore C = 9
  
  So the answer is 21978 * 4 = 87912
• 11 years agoHelpfull: Yes(1) No(1)
• @rishi..plz tell the logic..
• 11 years agoHelpfull: Yes(0) No(0)
• logic pls.
• 11 years agoHelpfull: Yes(0) No(0)
• pls tel d logic rishi
• 11 years agoHelpfull: Yes(0) No(0)
• piyush...ur answer is correct
• 11 years agoHelpfull: Yes(0) No(0)