I saw this on Instagram with no solution and was wondering what the answer is. I got $33$.
$$1+4=5$$
$$2+5=12$$
$$3+6=21$$
$$8+11=?$$
I saw this on Instagram with no solution and was wondering what the answer is. I got $33$.
$$1+4=5$$
$$2+5=12$$
$$3+6=21$$
$$8+11=?$$
On
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*}
On
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 :)
On
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.
On
I got 40...
But I guess there could be any number, because you can use the Lagrange Interpolation Polynomial
On
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
$a_n + b_n = c_n$
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 :)