0
$\begingroup$

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

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

$\endgroup$

closed as unclear what you're asking by Oray, JonMark Perry, D.A.G., Alexis, Deusovi Mar 11 '16 at 16:58

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ must we use all the numbers? $\endgroup$ – Marius Mar 11 '16 at 8:04
  • $\begingroup$ Please clarify if it is just one number per pair of parantheses? You say repeats are allowed, and the hint makes me think that you can use operators.. so can I put (9-7) + (..) and so on? $\endgroup$ – Arcane Mar 11 '16 at 8:08
  • $\begingroup$ i am not sure what you are asking, and i start to think u do not know either and i am voting for close... $\endgroup$ – Oray Mar 11 '16 at 15:39
  • $\begingroup$ @Oray If you're not sure, I'd suggest you look at existing answers. $\endgroup$ – dma1324 Mar 11 '16 at 15:43
  • $\begingroup$ What exactly are you unclear on? $\endgroup$ – dma1324 Mar 11 '16 at 19:23
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I edited the bonus part; check again. The first answer is correct though. $\endgroup$ – dma1324 Mar 11 '16 at 8:12
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

I remember solving similar question before:

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.