I saw this on Instagram with no solution and was wondering what the answer is. I got $33$.
$$1+4=5$$
$$2+5=12$$
$$3+6=21$$
$$8+11=?$$
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5This would essentially be a misuse/abuse of the + notation, but it appears as though $$ x + y := x(1 + y)$$ – The Chaz 2.0 Apr 26 '16 at 20:00
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4(Hence $ 8 + 11$ would equal $8(1 + 11) = 8(12) = 96$ – The Chaz 2.0 Apr 26 '16 at 20:01
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@TheChaz2.0 Mark it as an answer...+1 – imranfat Apr 26 '16 at 20:19
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5It's a mathematical Rorschach test; I got $19$ (my favorite flavor is vanilla too). – pjs36 Apr 26 '16 at 20:29
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1I can get $40$: We start with $\color{brown}{0}$ then $1+4 + \color{brown}{0}=\color{red}{5}$, $2+5+\color{red}{5}=\color{blue}{12}$, $3+6+\color{blue}{12} = \color{pink}{21}$, $8+11+\color{pink}{21} = 40$ – Winther Apr 26 '16 at 21:32
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2It should be clear from what people have found that there is no "obvious" unique solution to this puzzle - there are plenty (well atleast 3) of plausible solutions. Thus this is not a math question and more suitable for Puzzling SE imo. – Winther Apr 26 '16 at 21:37
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@Winther Puzzling SE will close it as too broad. – manshu Apr 26 '16 at 21:58
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3@manshu It's definately too broad (and off-topic) for this site too imo. – Winther Apr 26 '16 at 21:59
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I understand if you guys decide to close it. I thought it would be more straightforward. – nullByteMe Apr 26 '16 at 22:00
6 Answers
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Just for kicks, define \begin{equation} "a+b=c"\quad\Longleftrightarrow\quad a=\left\lfloor \frac{c}{b} \right\rfloor \end{equation} and I propose $88$ as a result :)
marco trevi
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A possible law is $$a+b=ab+a$$ (this is valid for the three first sums) so an answer is $$8+11=8\cdot11+8=96$$
It could be possible maybe another law satisfying the three first sums.
Piquito
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Another law giving the sum for the $n-th$ term is $\frac{n(n+1)}{2}+\frac{(n+3)(n+4)}{2}−\frac{3\cdot 4}{2}=n^2+4n$ which gives for the $8−th$ term 96 too. – Piquito Apr 26 '16 at 22:00
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Piquito, I understand your answer, but 8 + 11 $\neq$ 8.11 + 8. Because 8 + 11 = 19 while 8.11+8 = 96. I agree that this is the answer for this puzzle but $a + b $ is not equal to ab + a in general assuming a and b is greater than 2. – user777 Apr 26 '16 at 22:57
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$a_n + b_n = c_n$
should in fact be interpreted as
$a_n \cdot b_n + n = c_n$
with
$b_n = a_n + 3$
and the following recursive relation for $a_n$
$a_n = a_{n-1} \cdot a_{n-2} + (n-2),\:\:$ $n \geq 3$
$a_0 = 0, \:\:\:\:a_1 = 1, \:\:\:\:a_2 =2$
$a_1 \cdot b_1 + 1 = c_1$
$a_{2} \cdot b_{2} + 2 = c_{2}$
$a_{3} \cdot b_{3} + 3 = c_{3}$
$a_{4} \cdot b_{4} + 4 = c_{4}$
give us
$1+4=5$
$2+5=12$
$3+6=21$
$8+11=c_4$
and
$c_{4}$
$= a_{4} \cdot b_{4} + 4 = 92$
or to demonstrate the recursion
$c_{4}$
$= (a_{3} \cdot a_{2} + 2) \cdot b_{4} + 4 $
$= (3 \cdot 2 + 2) \cdot 11 + 4$
$= 8 \cdot 11 + 4 = 92$
This results in the following sequence
$1+4=5$
$2+5=12$
$3+6=21$
$8+11=92$
$200 + 203 = 40605$
$4824 + 4827 = 23285454$
$964808 + 964811 = 930857371296$
$\vdots$ $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\vdots$
Im pretty sure this is what they had in mind :)
JKnecht
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For future reference, note the possibility of self-deleting, editing, then self-undeleting answers. – hardmath Apr 26 '16 at 21:44
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If you number your ‘equations’ as $a_n+b_n =c_n$, you have the relation $c_n=a_n+b_n+c_{n-1}$. Hence \begin{align*} 4+7&=32,\\5+8&=45,\\6+9&=60,\\7+10&=77,\\8+11&=96.\end{align*}
Bernard
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I got 40...
But I guess there could be any number, because you can use the Lagrange Interpolation Polynomial
Luísa Borsato
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96 For each line add the two numbers and store it in D. Add D to the C from the previous step.
A+B=C D
1+4=5 5 (4+1 gives 5)
2+5=12 7 (5+2 is 7. Add 7 to 5 = 12 from previous step)
3+6=21 9 (6+3 is 9. Add 9 to 12 = 21 from previous step)
4+7=32 11 (4+7 is 11. Add 11 to 21 = 32 from previous step)
5+8=45 13
6+9=60 15
7+10=77 17
8+11=96 19
sunny
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