This one's different to all the previous ones...
\begin{align}7+53&=28\\16+6&=21\\5+19&=97\\15+8&=\,?\end{align}
Can you find the value of the question mark?
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4$\begingroup$ I found something interesting. Let's represent your equations as $A+B=C$. Let $A\| B$ mean "A concatenate B" and $d(x)$ be the "digits of some assigned $x$". Also, let $n$ be the number that corresponds to the position of each equation ($n=1$ for first equation, $n=2$ for second equation, etc) then for some value $D$, if $$D=\frac{A\| B-d(A)\cdot d(B)}{\frac{1}{2}|A - B|+2-n}$$ then $D$ is an integer, and close to $C$ :) $\endgroup$– Mr PieJul 6, 2019 at 6:46
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1 Answer
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answer
84
step1
convert all numbers to chemical element symbol
step2:
concatenate them to get R.H.S.
step3:
convert result to atomic number
e.g.
7+53=N+I=Ni=28
16+6=S+C=Sc=21
5+19=B+K=Bk=97
answered Jul 3, 2019 at 9:27
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