When a two-digit number is multiplied by the sum of its digits, the product is 913. When the number obtained by interchanging its digits is multiplied by the sum of the digits, the result is 418. The sum of the digits of the given number is :
1)5 2)4 3)7 4)6

• Let the two number be a b .
  According to the statement,
  (10a+b) * (a+b) = 913
  (10b+a) * (a+b) = 413
  
  Solving these two statements,
  10a+b/10b+a = 913/418
  10a+b/10b+a = 83/38
  Now, 83 is prime number cannot be divided further. 83 will be the number . sum will be 11 . The options are incorrect!! Please update.
  .
• 3 years agoHelpfull: Yes(2) No(2)
• Simply 83
  apply conditions
  (10a+b)*(a+b)=913 ---(1)
  (a+10b)*(a+b)=418 ---(2)
  divide (1)/(2) we get
  (10a+b)/(10b+a)=913/418=83/38 ---(3)
  ==> Just break equation into two
  10a+b=83
  a+10b=38
  solving the above equations we get a=8 , b=3..Hope the answer is 83. Prateek, your answer is correct. But you didn't understand the logic
• 3 years agoHelpfull: Yes(2) No(0)
• none of above
  83*11 = 913
  38*11 = 418
  so sum of digits od 38 = 3+8 = ans is 11
• 3 years agoHelpfull: Yes(1) No(2)
• Please ignore my typo : (10b+a) * (a+b) = 418 .
• 3 years agoHelpfull: Yes(0) No(1)
• (10a + b)*(a+b) = 913
  (10b + a)*(a+b) = 418
  solving
  a**2 = b**2 = 45
  solving
  9 + 36 = 45
  a = 3
  b = 6
• 2 years agoHelpfull: Yes(0) No(1)
• answer will be 11 = 3 + 8
  
  11*83 = 913
  11*38 = 418
  so here
  (10a + b) * (a+b)
  (10b + a) * (a + b)
• 1 year agoHelpfull: Yes(0) No(0)