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My niece is having problems with the attached and I can't figure it out. I think I'm overcomplicating it. Does anyone have thoughts on how to solve it?

Math puzzle

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    $\begingroup$ I voted for no source but then retracted because of "Golden Key Russian School LLC © 2021. Sources available upon request." Not sure whether this counts. $\endgroup$
    – WhatsUp
    Dec 2 '21 at 15:17
  • $\begingroup$ Could it be related to how those numbers are written fully spelled out in your language? Like the number of some letter or something. $\endgroup$
    – Nathaniel
    Dec 2 '21 at 18:29
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I think I got it. The answer is 1.
EDIT: or possibly 2 to 4; see the insightful comments of @Nathaniel and @WhatsUp below.
Still pretty sure about the "got it" part though :-)

It's simply just

Count the twos and threes. (Or alternatively, count the threes and nines.)

But there's a twist:

The number 73,133 has a formatting error in it, accidentally splitting it at the thousands separator, so it's now displayed as two separate numbers.

"What on earth are you blabbering about!", I hear you exclaim.

Look!

enter image description here

Correcting for that error, we are now faced with another problem: 3 is a given already. It makes no sense to pose that number as the question. (Well spotted, @Nathaniel!)

But we won't need to add many new assumptions to account for that: If the number in the question isn't a 3, then what is it? Well, anything between 3,000 and 3,999 would work. If such a number were printed with the same software that mangled the 73,133, it would display the leading 3 as a separate digit on its own line, with the remaining three digits on the next line, which is unfortunately completely off-page.

This error is so common it has a name (several, actually): wikipedia calls it delimiter collision.

Accounting for @WhatsUp's astute observation (we don't know if we're counting twos or nines in addition to the threes), we can further assume that the 3 off-page digits don't include any 2s or 9s, or if they do, there's exactly one of each. The former possibility seems like the more likely one, since that would match both the clarity and difficulty level of the puzzle with the other puzzles seen on the same page.

Sadly, there's no way to be sure what the three off-page digits are, so the final answer must remain a mystery.


Post Scriptum

In the chat (link in the comments below) @DavidA.Craven raises a very interesting possibility: the samples given are also consistent with another error type: if the layout program, instead of misinterpreting delimiters, actually split each number after the first 3 digits, then the intended solution rule to the original puzzle could have been "count the threes", which is way more satisfying as a puzzle solution.

Deciding which explanation is more likely is largely a matter of aesthetics and speculation, so we decided to leave it at "it's a problem having to do with extraneous line breaks either way." (Let the record state that the more I think about it, the more I like his solution)

Might be interesting to contact the publisher and find out for certain :-)

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    $\begingroup$ I like your idea, but would'nt it be weird if the unknown association is the same as an already existing one? $\endgroup$
    – Nathaniel
    Dec 2 '21 at 20:26
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    $\begingroup$ Oh, you are absolutely right! So the answer is 1 + whatever the 3-digit number on the next line (off-page) adds to it. This also explains the error: it's not actually a typo, but the numbers have been formatted with a thousands separator that has been accidentally translated into a record separator. So the original input was probably "73 133". (Or 73,133, if the input was in comma-separated values form.) $\endgroup$
    – Bass
    Dec 2 '21 at 20:33
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    $\begingroup$ It's also quite weird that $2$ only appears once. If we assume that some of the lines are broken, then it might be more natural to think that the rule is just counting the number of $3$'s (with more broken lines). $\endgroup$
    – WhatsUp
    Dec 2 '21 at 20:46
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    $\begingroup$ Wikipedia calls this bug delimiter collision, and it's definitely something that could to happen if you input numeric values onto a template with a fixed layout. (Notice how all the visible puzzles have the exact same number of arrows in them.) The wikipedia article even gives this exact case as an example: for example, field collision can occur whenever an author attempts to include a comma as part of a field value (e.g., salary = "$30,000") $\endgroup$
    – Bass
    Dec 2 '21 at 21:24
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    $\begingroup$ I agree with @WhatsUp. I think the left hand column should be 334, 100, 65, 2393, 73, 1333, and the actual question isn't printed at all. Then the answer is completely simple, and was what I thought it was when I started reading down the list until I got partly confused by 239 and really confused by 73. $\endgroup$ Dec 2 '21 at 22:56
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Find the rule puzzles can tend towards "Read the puzzle-maker's mind."

I found that

Number of 2's plus number of 3's plus twice number of 7's
Number of 9's plus number of 3's plus twice number of 7's

both work, giving

2

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    $\begingroup$ The other puzzles visible on the page are much less arbitrary though. $\endgroup$
    – Bass
    Dec 2 '21 at 20:00
  • $\begingroup$ Assuming each digit has a non-negative integer value to sum, it can be solved by deduction. 3 must be 1, then 0, 1, 4, 6, 5 must be 0, 2+9 must be 1, and 7 must be 2. It all fits. So 133 must give 2. $\endgroup$
    – Florian F
    Dec 3 '21 at 23:06
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Although the "3" below the page is best explained by Bass's answer, let me just say that such puzzles are not very good because too little data is given to make the official answer convincingly simpler than alternatives. For example, here is a formula that works for the numbers requested as well as the hypothesized intended pattern!!

n ↦ n%5−n%4+n%2   (where % is modulo; x%y is the remainder when x is divided by y)

For inputs 334,100,65,239,3,73,73133 it produces the outputs 2,0,0,2,1,3,3 as desired.

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