19
$\begingroup$

\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!

$\endgroup$
7
  • 1
    $\begingroup$ If it's not an abuse of the equals symbol, why are the results not $874$, $99$, $25$, and $141$ respectively? $\endgroup$ Dec 9, 2015 at 22:01
  • 1
    $\begingroup$ @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, 2015 at 22:04
  • 1
    $\begingroup$ 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, 2015 at 0:43
  • 1
    $\begingroup$ @Alconja you were nearly there, but Petter beat you to it! $\endgroup$
    – martin
    Dec 10, 2015 at 5:18
  • 1
    $\begingroup$ 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, 2015 at 5:29

1 Answer 1

14
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ well done! I thought I was going to have to give another hint! :) $\endgroup$
    – martin
    Dec 10, 2015 at 5:11
  • $\begingroup$ note the answer to eq. three: $26\rightarrow 29$ $\endgroup$
    – martin
    Dec 10, 2015 at 5:14
  • 1
    $\begingroup$ 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, 2015 at 5:17
  • $\begingroup$ Ah, got it now, will edit it in, thanks! $\endgroup$
    – Petter
    Dec 10, 2015 at 5:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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