Fill in the parentheses [closed]

Question

$()+()+()+()+()=30$

How can you fill in the parentheses with the numbers $1, 3, 5, 7, 9, 11, 13$ to make the above equation true? Repeats are allowed, but you must fill in all of the parentheses!

Bonus if you already solved this:

Can you do it without extra symbols? (no +, -, ÷, ｘ, <<, !, etc.)

(EDIT: You don't have to use all of the numbers.)

EDIT2: Okay, the answer is not that there is no answer. I'm not looking for a mathematical proof here.

EDIT3: The above information is all you need to solve the problem. Be creative!

One possible solution:

$(5+1)+(5+1)+(5+1)+(5+1)+(5+1)=30$

My favorite solution:

${{5}\choose{3}}$+(5)+(5)+(5)+(5)=30

Answer 1

Answer is:

(3!) + (9) + (7) + (3) + (5) = 30

For Bonus:

sum of 5 odd numbers is ALWAYS an odd number so it is not possible unless something is done. What I am doing is using the number 9 as BASE.

so;

(13) + (11) + (5) + (3) + (1) = 33 , base9 (33) = base10 (30)

or a more direct way with that logic;

(7+5+5+5+5)base9 = (30)base9

Answer 2

One answer:

Using base 11: 1 + 3 + 3 + 11 + 13 = 30
Equivalent in base 10
1 + 3 + 3 + 12 + 14 = 33

Second one:

Using base 13: 1 + 3 + 5 + 11 + 13 = 30
Equivalent in base 10
1 + 3 + 5 + 14 + 16 = 39

And a third one:

Because you cannot add 5 odd numbers to get an even number in base 10 you have to use tricks.
3 + 5 + 3! + 7 + 9 = 30
equivalent to
3 + 5 + 6 + 7 + 9 = 30

[EDIT]
Bonus one or two:

Using base 9: 1 + 1 + 3 + 11 + 13 = 30
Equivalent in base 10:
1 + 1 + 3 + 10 + 12 = 27

Using base 7: 1 + 1 + 1 + 11 + 13 = 30
Equivalent in base 10:
1 + 1 + 1 + 8 + 10 = 21

Answer 3

I remember solving similar question before:

By putting number 9 upside down, i get (13)+(7)+(6)+(3)+(1)=30