I don't think there is a way to get $30$ using the numbers from that list because all those numbers are odd. We know that odd + odd = even (that's the sum of any two numbers from the list) and even + odd = odd (we add one more number from the list to the sum which is even). So, the sum is always going to be odd, but $30$ is even!
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It's actually impossible since all numbers are odd and $30$ is even.
Notice that every odd number can be represented as $2n+1$. Thus $$(2n+1)+(2k+1)+(2t+1)=2·(n+k+t+1)+1$$ which is odd and not even
I don't think there is a way to get $30$ using the numbers from that list because all those numbers are odd. We know that
odd + odd = even(that's the sum of any two numbers from the list) andeven + odd = odd(we add one more number from the list to the sum which is even). So, the sum is always going to be odd, but $30$ is even!