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This not-so-simple puzzle asked by my teacher today, given:

2 + 2 = 3

4 + 8 = 27

3 + 4 = 8

So, the answer for:

5 + 32 = ?

I'm new here so pardon my mistakes. Please let me know what you guys thinks about this riddle.

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    $\begingroup$ This sort of puzzle can have an infinite number of answers and as such doesn't make for a good puzzle. See here: puzzling.meta.stackexchange.com/questions/5712/… $\endgroup$
    – edderiofer
    Jul 10 '17 at 14:44
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    $\begingroup$ @edderiofer Well, this isn't quite a number-sequence puzzle. It's more of a pattern puzzle. $\endgroup$
    – Rand al'Thor
    Jul 10 '17 at 14:46
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    $\begingroup$ @Randal'Thor Pretty sure the same objection still applies though. There are a potentially infinite number of functions that could be represented by "+", or it might not even be a function but three somewhat-related sequences. $\endgroup$
    – edderiofer
    Jul 10 '17 at 15:06
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    $\begingroup$ I kinda detest these types of questions which distort my sense of adding numbers :/ $\endgroup$
    – ABcDexter
    Aug 9 '17 at 17:24
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    $\begingroup$ @ABcDexter Study some abstract algebra; might change your mind ;) $\endgroup$
    – 2.7182818284590452353602874713
    Apr 25 '20 at 5:26
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Let each equation be represented as $a+b=c$

I observed that in the first equation $a * (b - 1)$ gives us $2 * 1 = 2$,
and in the second it gives us $4 * 7 = 28$,
and finally in the third it gives us $3 * 3 = 9$.

These numbers all differ from their respective answers by 1, and I thought that could be represented by adding a term to the above multiple, $(-1) ^ x$ where $x$ represents the boolean value of $a < b$.

Using this logic, $5 + 32$ would equal $5 * 31 + (-1) ^ 1 = 154$.

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Rearranging the statements:

$2+2=3$
$3+4=8$
$4+8=27$

We see a pattern develop:

$(n+1) + 2^n = x$

If the first equation resulted in $1$, then $x = n^3$ is a solution. If the second equation resulted in $9$, then $x = 3^n$ is a solution. Alas, neither of these is the case.

Additionally, the final request pairs $5$ with $32$, which would have ordinarily been paired with $16$ if following the suggested pattern.

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    $\begingroup$ You can describe it as n + 2^(fib(n)) = x $\endgroup$
    – geokavel
    Jul 22 '17 at 1:51
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One convoluted potential answer:

$2+2=3$
$4+8=27$
$3+4=8$

Step one: Express teach term as its most reduced exponential

$2^1 + 2^1 = 3^1$
$2^2 + 2^3 = 3^3$
$3^1 + 2^2 = 2^3$

Step two: Multiply instead of exponentiate

$2*1 + 2*1 = 3*1$
$2*2 + 2*3 = 3*3$
$3*1 + 2*2 = 2*3$

or

$2+2=3$
$4+6=9$
$3+4=6$

Step 3: Note the left side is always 1 greater than the right

SO:

5 + 32= 49

because

$5^1 + 2^5 =(7^2) + 1$
$5*1 + 2*5 =(7*2) + 1$
$5+10=(14) + 1$

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  • $\begingroup$ $2^7$ could also be a solution since exponent might be bigger than base like in the third equation, right? $\endgroup$
    – SomeUser
    Oct 29 '20 at 16:25
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My answer to $5+32$ is $135$

Reasoning:

The given numbers can be written as follows:

$2*2^1-1^2=3$

$3*2^2-2^2=8$

$4*2^3-3^2=27$

$5*2^5-5^2=135$

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  • $\begingroup$ Similar: $5*2^4-5^2=55$ $\endgroup$
    – z100
    Oct 27 '20 at 18:32

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