SOLUTION: What is a five digit number in which the first digit is twice the second, the third is half the fourth, the fifth is double the second, and the sum of all of them is 21?

The first and fifth digit are the same because "twice the second" 
and "double the second" mean the same.  

So we have one of these four possibilities:  

A    8 4 _ _ 8    which has sum of digits 8+4+8=20
B    6 3 _ _ 6    which has sum of digits 6+3+6=15
C    4 2 _ _ 4    which has sum of digits 4+2+4=10
D    2 1 _ _ 2    which has sum of digits 2+1+2=5

the third is half the fourth, so we must fill in the two 
remaining blanks with one of these pairs of digits

0&0, 1&2, 2&4, 3&6 or 4&8, 

which have sums 

0+0=0, 1+2=3, 2+4=6, 3+6=9, 4+8=12.

The sum of all five digits must be 21.  Therefore:

A is ruled out because we cannot add 0,3,6,9 or 12 to 20 and get 21.
B is possible because we can add 6 to 15 and get 21.
C is ruled out because we cannot add 0,3,6,9 or 12 to 10 and get 21.
D is ruled out because we cannot add 0,3,6,9 or 12 to 5 and get 21.

So B is the only possibility and the 3rd and 4th digits are 2&4.

Answer: 

      6 3 2 4 6
 
Edwin