# Math number puzzle/riddle

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

$2+2=3$

$4+8=27$

$3+4=8$

$5+32=?$

• 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/… – edderiofer Jul 10 '17 at 14:44
• @edderiofer Well, this isn't quite a number-sequence puzzle. It's more of a pattern puzzle. – Rand al'Thor Jul 10 '17 at 14:46
• @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. – edderiofer Jul 10 '17 at 15:06
• i bet it has something to do with the powers of 2 (and/or 3) – Marius Jul 10 '17 at 15:07
• I kinda detest these types of questions which distort my sense of adding numbers :/ – ABcDexter Aug 9 '17 at 17:24

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.

• You can describe it as n + 2^(fib(n)) = x – geokavel Jul 22 '17 at 1:51

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 $5 * 31 + (-1) ^ 1 = 154$.

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

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

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