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I've recently come accross this puzzle.

2,5,8 = 122348

3,5,4 = 181224

7,4,6 = 352230

5,3,10 = 203340

3,9,4 = ?

I was able to figure out 30**40 but couldn't proceed further. How are the middle numbers in the right related to the left ones?

Source:

https://brainly.in/question/16666300

https://brainly.in/question/16515827

I'm not sure if the tags/title is apt and if this comes under puzzles. Edits are welcome.

WhatsApp forward

Edit 1: Arriving at 30**40 is me trying. 30 and 40 may not be in the answer. Shown my effort because they downvote your post into oblivion in stackoverflow if you don't.

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    $\begingroup$ Unfortunately, around here we're sticklers for proper attribution, so I'm going to put this on hold until it contains something more definite than "I've recently come across...". Ideally, a link to where it came from; if at all possible, enough information to enable readers to find the original source for themselves. Thanks! $\endgroup$ – Gareth McCaughan Apr 21 at 18:38
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    $\begingroup$ @gareth-mccaughan I don't have any source for this. This is a WhatsApp forward. I love solving puzzles and wanted to know the reasoning behind this. I've seen the same question in brainly as well. brainly.in/question/16666300 and brainly.in/question/16515827. I think they don't count as attributes. But I've seen a lot of questions in this community that fall under my category. Thanks. $\endgroup$ – duplex143 Apr 22 at 5:59
  • $\begingroup$ @duplex143 does all numbers are right ? because i tried many solution one solution is closest . but,for 3,9,4 =33 its giving 34 . $\endgroup$ – Swati Apr 25 at 13:24
  • $\begingroup$ @Swati I've edited the post to clear any confusion. You're not supposed to deviate from the info already given. But if think the question can be changed, I'm interested to know. $\endgroup$ – duplex143 Apr 25 at 16:36
  • $\begingroup$ @duplex143 Sorry im saying about 5,3,10=33 not 3,9,4 . Also, I don't know if the question has some mistake maybe i'm wrong . $\endgroup$ – Swati Apr 25 at 18:40
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Possibly solved? Be the judge yourselves.

Everyone has possibly found the rule that generates the first couple of numbers and the last couple of numbers. The rule is :

$(a,b,c)\mapsto (d,e,f) \Rightarrow d=a(b+1),\ f=c(b+1)$. Thus $(3,9,4)\mapsto (30,e,40)$. But then, what is the rule that generates $e$ ?

Be warned though, the argument that I am going to provide will not work for all triples, but hinges on a very crucial observation provided in the $4$ values given to us, and will only hold for triples following that pattern.

Always works now...

We don't need this observation now.

The observation is that:

$c$ is always even! Once we take the assumption that $c$ will always have to be even, we can proceed.

So far, we have established one assumption

Now we look for patterns.

We observe that since $c$ is always even, $c/2$ is an integer. Since $e$ has to be a two digit number, call it $XY$ where $X$ and $Y$ represent the digits. There are two observations now. One is that $X+Y=\lfloor(c/2)\rfloor+1$ in all the values that is given. The second is that either $X$ and $Y$ are consecutive with $Y>X$ or $X=Y$. $\lfloor \cdot\rfloor$ is the floor function, which sends any number to the integer just less than it.

How does that help us? Here's how.

Take $c$, divide it by $2$ and add $1$ to it. Now we have $\lfloor(c/2)\rfloor+1$. Call it $x$. If $x$ is even, divide $x$ by $2$ and set $X=Y=x/2$. Else, set $X=(x-1)/2,\ Y=(x+1)/2$. We check that in the former case $X=Y$ and in the latter case $X<Y$ with $X$ and $Y$ consecutive. Also $X+Y=x$ in both cases, so both our observations hold. And there's a nice symmetry to it!

What does this tell us?

$(a,b,c)\mapsto (d,X,Y,f)$, a $4$-tuple rather than a $3$-tuple.

Is this true?

Doing a quick check that our rule works: $$(2,5,8)\to \lfloor8/2\rfloor=4\to 4+1=5\to X=(5-1)/2=2,\ Y=(5+1)/2=3\to XY=23$$ $$(3,5,4)\to \lfloor4/2\rfloor=2\to 2+1=3\to X=(3-1)/2=1,\ Y=(3+1)/2=2\to XY=12$$ $$(7,4,6)\to \lfloor 6/2\rfloor=3\to 3+1=4\to X=Y=4/2=2\to XY=22$$ $$(5,3,10)\to \lfloor10/2\rfloor=5\to 5+1=6\to X=Y=6/2=3\to XY=33$$

Finally, the moment has come :

$$(3,9,4)\to \lfloor4/2\rfloor=2\to 2+1=3\to X=(3-1)/2=1,\ Y=(3+1)/2=2\to XY=12$$

Putting everything together:

$$(3,9,4)\mapsto (30,12,40) \Rightarrow 3,9,4=301240$$

And that concludes the problem.

If I am right, I do owe an apology to Galen. He has done such a wonderful job and I would hate to prove him wrong. I wholeheartedly believed he was right, but in the end, I came up with something that does not agree with him. Please accept my humble apologies.

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  • $\begingroup$ I'm actually quite happy that you found a human-understandable solution. That's the point of puzzles, which is why I insisted that no one take my answer seriously. $\endgroup$ – Galen Apr 30 at 0:20
  • $\begingroup$ But as far as I'm concerned, you rose to the occasion against the artificial neural network! Congratulations! $\endgroup$ – Galen Apr 30 at 0:22
  • $\begingroup$ The rule you inducted is in agreement with my brute-force inference that using only $*, /, +, -$ is not enough within 7 operations to satisfy all example cases, since you used floor rounding. $\endgroup$ – Galen Apr 30 at 0:26
  • $\begingroup$ No need to apologize. I've upvoted your answer. $\endgroup$ – Galen Apr 30 at 0:28
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    $\begingroup$ @Galen Thanks! I did enjoy the puzzle, and I liked your idea of using a neural network to solve this, and I have actually started researching on how neural networks and multiple agent networks can solve such riddles. $\endgroup$ – Yuzuriha Inori Apr 30 at 3:00
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$3,9,4=$

$301240$

because

if we label the inputs $a,b,c$, then the output is $a(b+1):\lfloor\frac{c}{5}\rfloor+1:\lfloor\frac{c}{4}\rfloor+1:(b+1)c$, where $\lfloor x\rfloor$ is the 'floor' function, and returns the integer part of its operand.

The examples:
$2,5,8\to 2(5+1:\lfloor\frac{8}{5}\rfloor+1:\lfloor\frac{8}{4}\rfloor+1:(5+1)8=122340$.
$3,5,4\to 3(5+1):\lfloor\frac{4}{5}\rfloor+1:\lfloor\frac{4}{4}\rfloor+1:(5+1)4=181224$.
$7,4,6\to 7(4+1):\lfloor\frac{6}{5}\rfloor+1:\lfloor\frac{6}{4}\rfloor+1:(4+1)6=352230$.
$5,3,10\to 5(3+1):\lfloor\frac{10}{5}\rfloor+1:\lfloor\frac{10}{4}\rfloor+1:(3+1)10=203340$.

So,
$3,9,4\to 3(9+1):\lfloor\frac{4}{5}\rfloor+1:\lfloor\frac{4}{4}\rfloor+1:(9+1)4=301240$.

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Don't upvote this answer (at least while there's a bounty on the question), and don't accept it as an answer. It is a partial answer, and ends with what I hope is a fun or amusing prediction.

Here's what I figured out by hand:

For $a,b,c = d,e,f$, I was able to induct that $d = a + ab$ and $f = bc+c$, which has a nice symmetry to it. This suggests, as the OP says, that $3,9,4=30XY40$.

Then I tried:

A breadth first search of all arrangments () of $*,/,+,-$ with $a,b,c$ with up to and including 7 operations... amounting to 773514 explored nodes. This didn't result in finding $XY$, suggesting something less obvious than simple arithmetic.

Finally, I tried this (it gets silly here):

I trained a dense and fully-connected (3,2,3,3,1 for layer widths) artificial neural network using rectified linear units for all layers, stochastic gradient descent (lr=0.0001), mean squared error for loss, 1000 epochs, pseudoreplicated data (n=1000), with a batch size of 4. The training features were the input triplets, and the training output were the middle column numbers. The final training errors by row were 0.02843285, -0.03020573, 0.02318954, -0.01794052, so the predictions were pretty close (); within an integer. If I round round the predictions on the training set to the nearest integer, here are the predictions of the for each case $$ (2, 5, 8) \rightarrow 23 $$ $$ (3, 5, 4) \rightarrow 12 $$ $$ (7, 4, 6) \rightarrow 22 $$ $$ (5, 3, 10) \rightarrow 33 $$ So the model learned some pattern that approximates the well to the examples because after rounding it didn't get even one of them wrong. General life advice, just because you can train a model like this doesn't mean you should trust what it has to say... Even so, I'll share what it predicted in the next box.

THE CHALLENGE

Cue the boss music, the machine learning predicted that $XY=09$ (well, more specifically $9.006071$). That is to say, the model predicts $ (3, 9, 4) \rightarrow 09$. Defeat the machine by actually solving this puzzle like you're supposed to! Go team humans!

Bonus Predictions

The following tuples were also predicted to map to 09 for the middle digits by the machine learning model. $$(2,3,3)$$ $$(2,4,3)$$ $$(2,8,4)$$ $$(3,5,3)$$ $$(3,10,4)$$ $$(4,2,2)$$ $$(4,6,3)$$ $$(4,10,4)$$ $$(5,3,2)$$ $$(5,7,3)$$ $$(5,8,3)$$ $$(6,4,2)$$ $$(6,5,2)$$ $$(6,9,3)$$ $$(7,5,2)$$ $$(7,6,2)$$ $$(7,10,3)$$ $$(8,7,2)$$ $$(9,8,2)$$ $$(10,9,2)$$ $$(10,10,2)$$ It would be nice to show all the predictions, but there are just too many. If there's a specific prediction you'd like to see, put that 3-tuple in the comments of this answer.

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    $\begingroup$ Here's one more added which probably is useful. The middle column numbers are not a linear combination of any 4 of $a,b,c,ab,bc,ca,abc,a^2,b^2,c^2$. $\endgroup$ – Yuzuriha Inori Apr 26 at 11:06
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2,5,8    122348    2*(8-2)=12      (8+5)+(2*5)=23     (5*8)+8=48

3,5,4    181224    (3*5)+3=18       (3*4)=12       (5*4)+4=24

7,4,6    352230   (6*7)-7=35       (4*7)-6=22       (4*6)+6=30

5,3,10   203340  (3*10)-10=20   (3*10)+3=33    (5*10)-10=40

3,9,4   392133   (4+9)*3=39       (3*4)+9=21      (4*9)-3=33

Your number is 392133.
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  • $\begingroup$ @Galen.I can not use that symbol. $\endgroup$ – Vassilis Parassidis Apr 24 at 22:25
  • $\begingroup$ galen I never use it before you can tell me how. $\endgroup$ – Vassilis Parassidis Apr 24 at 22:29
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    $\begingroup$ I don't understand your answer. Can you edit it to further explain what you think the underlying relationships/rules are? $\endgroup$ – Galen Apr 25 at 0:07
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    $\begingroup$ Please explain your answer within your answer by editing it. Take us through how you fragmented the number (and define what that means), and how you came up with each equation. $\endgroup$ – Galen Apr 25 at 1:22
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    $\begingroup$ Your answer to a puzzle should not itself be a puzzle; I don't understand all of your equations. $\endgroup$ – Galen Apr 25 at 1:55

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