If abcde is a five digit number and abcde*4=edcba, then what is the value of b+c+d?

• hy .. can you tell me that these questions are of infosys recently ?
  
• 8 years agoHelpfull: Yes(15) No(5)
• how can we find that number 21978???
• 8 years agoHelpfull: Yes(9) No(2)
• A must be 1 or 2 since any other number would means that the answer would have 6 digit. A also have to be even since 4 time any number is even. thus A is 2. E would have to be 3 or 8. but E cannot be 3 since the last digit have to be 8 or 9. thus E is 8. now 4xB must be less than 10 thus B is one or 2. but A is already 2 thus B is one. 4 x D +3 must end with 1 two possible digit 2 and 7 but 2 already used thus D is 7. 4xC+2 must end with C. two possible numbers 0 and 9. now verify the options
  
• 8 years agoHelpfull: Yes(9) No(1)
• 21978*4=87912 then b+c+d=1+9+7=17
• 8 years agoHelpfull: Yes(6) No(3)
• 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
• 6 years agoHelpfull: Yes(2) No(0)
• A-2,B -1,C-9,D-7,E-8
  Therefore B+C+D= 17
• 8 years agoHelpfull: Yes(1) No(2)
• 21978
  *4
  ________
  87912
  
  so b+c+d=17
• 8 years agoHelpfull: Yes(1) No(4)