Question 1008650
Let's take the digits in the number to be A B and C.
  From the question,
  A + B + C = 11

A=> Hundreds,
B=> Tens
C= Units.

 B = 3A ( tens digit is three times the hundreds digit.. As in the question)
 B = 2C (tens digit is twice the unit digit.. As in the question)

   This therefore means that 3A = 2C = B.
To relate all elements together, we have.
C = 3A/2
A= 2C/3
B= 3A or B = 2C.

  From the first equation, (A+B+C=11)
If we substitute the values of B and C in terms of A, we have the following;

{{{A + 3A + 3A/2 = 11}}}

   
{{{4A+3A/2 = 11}}}

 
{{{11A/2 =11}}}

{{{11A = 22}}}

{{{A = 2}}}

If A = 2, Then B = 6, Making C = 3.

  The original number is 263


(Engr. Terry, Nigeria)