19
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\begin{array}{l} &19\times 46=784\\ &56+43=102\\ &360\div 15=29\\ &196-55=114\\ \end{array}

Hint 1: For once, this is not an abuse of the equals symbol!
Hint 2: The clues are for the song only, the artist follows.
Hint 3: $26 + 89 = 115$
Hint 4: The third equation is potentially paradoxical, but I don't mind.
Hint 5: Released as a single (B side) in 1696!

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    $\begingroup$ If it's not an abuse of the equals symbol, why are the results not $874$, $99$, $25$, and $141$ respectively? $\endgroup$ – Ian MacDonald Dec 9 '15 at 22:01
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    $\begingroup$ @IanMacDonald I think you might mean $ 874, 99, 24,$ and $141\dots$ but you will have to guess the answer to see why! $\endgroup$ – martin Dec 9 '15 at 22:04
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    $\begingroup$ Well, I can make the maths work, but I'm not particularly familiar with the music of that era... but I guess I can't always get what I want. $\endgroup$ – Alconja Dec 10 '15 at 0:43
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    $\begingroup$ @Alconja you were nearly there, but Petter beat you to it! $\endgroup$ – martin Dec 10 '15 at 5:18
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    $\begingroup$ All good. :) I had a bit of a google and found nothing that seemed to fit (hence my hinted "you can't always get what you want", which was a b-side track from '69). If I had've stumbled across the song in the solution I'd have known for sure... but I guess my google-fu is not up to scratch. $\endgroup$ – Alconja Dec 10 '15 at 5:29
14
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The answer is

If 6 was 9 by Jimi Hendrix

Explanation

The numbers 6 and 9 have been switched in the equations, meaning that any 6 is actually a 9, and any 9 is a 6. Thus the equations turn into \begin{array}{l} &16\times 49=784\\&59+43=102\\&390\div 15=26\\&169-55=114\\\end{array} Which are valid equations!

Explanation of hints

Hint 3 becomes $29 + 86 = 115$ (though it was correct even before switching)
Hint 4 "I don't mind" are lyrics from the song If 6 was 9
Hint 5 The song was released in 1969
The other hints don't need explanation

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  • $\begingroup$ well done! I thought I was going to have to give another hint! :) $\endgroup$ – martin Dec 10 '15 at 5:11
  • $\begingroup$ note the answer to eq. three: $26\rightarrow 29$ $\endgroup$ – martin Dec 10 '15 at 5:14
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    $\begingroup$ Thanks! I thought the "paradoxical" thing should have some meaning, but I couldn't think of anything there :) Will need to mull it over a bit more! $\endgroup$ – Petter Dec 10 '15 at 5:17
  • $\begingroup$ Ah, got it now, will edit it in, thanks! $\endgroup$ – Petter Dec 10 '15 at 5:19

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