What solution would you come up with for this problem?

Question

So the question is: put numbers $1, 3, 5, 7, 9, 11, 13$ and $15$ into gaps in the following expression:

$$\_\_ + \_\_ + \_\_ = 30$$

The most naive approach to use summation in the group of integers will fail, since given numbers are odd, and when adding three odd integers, you get an odd integer, which $30$ is not.

I came up with the following ideas:

Turn $9$ upside down and get a $6$ and for instance $6 + 11 + 13 = 30$ works.
In the field of characteristic $5$ holds $5+5+5=30$.
If we take base $11$ (i.e. add a symbol $A$ for $10$) then $30_{11} = 33_{10}$ and then holds $13+7+9=30$.

All these approaches are pretty straightforward, so I was wondering, if maybe someone could come up with another more sophisticated and yet reasonable solution for this puzzle?

Thanks in advance.

This is a comment because it is far from being sophisticated: $5.7 + 11.3 + 13$? — Silvia Ghinassi, Aug 31 '15 at 20:20
I am not quite sure if usage of decimal point is permitted, but thanks for your contribution. — nakajuice, Aug 31 '15 at 20:21
Decimal points not permitted, but turning 9 upside down to get 6 is!!! And base 11 is? Who is making the rules here? — BruceET, Aug 31 '15 at 21:20

score 0 · Answer 1 · answered Aug 31 '15 at 23:01

0

You could horizontally flip the $5$, making a $2$.

$$2+13+15=30$$

You could also vertically flip $15$, making $12$.

$$7+11+12=30$$

answered Aug 31 '15 at 23:01

Leonidas Lanier

4,149

I don't think this can be considered a 2. – Zach Gates Aug 31 '15 at 23:14
Depends on the font. Using a digital clock font, a 5 can be flipped to make a 2. – Leonidas Lanier Sep 01 '15 at 01:09

What solution would you come up with for this problem?

1 Answers1