4
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One has five.

Two has two.

Three and four only have one.

What does eight have?

hints? who needs hints? :)

Updated hints from comments:

@ibrahim_mahrir - A hint? I'd suggest to contemplate What, when, where but not why? But that's as likely to be taken the wrong way as the right one....... :) Anyway, it's yours and mine, and a bit superficial too.

@Adam - the decrease isn't an essential feature [in the sense that, one having more doesn't of itself mean the next ones must have less]. I think that's a safe enough comment :)

Comments on the puzzle and its hints + answer (ignore this bit!)

I found myself looking at an ordinary analog clock on the wall, and thinking about how one never really saw the "1" in "10". You only tend to see ten as a unit, not it terms of its digits "one" and "zero". A bit like how the eye skips over the the typos in some sentences. The puzzle came full fledged from that. I wasn't sure anyone would get it, I wanted to find out if someone could, on such abstract wording.

I suspect my comment on the 7-seg guess was too cluey, but it really was inspired for closeness. Its hard to hint without narrowing it down. The hints above, well....

what/where/when - what are they and where/when are they [found], would be more productive than looking for a formula. Asking why are they, wouldn't help though. Mine+yours = ours ("hours"), and anything superficial is talking about things on the face of something (a clock face).

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  • $\begingroup$ another related tag maybe .. $\endgroup$ – Amruth A Nov 13 '17 at 13:26
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    $\begingroup$ I want this to be fibbonaccian, but it doesn't fit (at all, not even convolutedly)! $\endgroup$ – Adam Nov 13 '17 at 13:34
  • $\begingroup$ @Adam - it's not :) so you're all safe there $\endgroup$ – Stilez Nov 13 '17 at 13:43
  • $\begingroup$ Either of mathematics or riddle tags are applicable for this? $\endgroup$ – Mea Culpa Nay Nov 13 '17 at 15:10
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    $\begingroup$ I hope the answer is not too broad. $\endgroup$ – prog_SAHIL Nov 13 '17 at 17:15
9
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I think that Eight has

1

because, each number has

the specified number of instances left of the colon on a standard 12 hour clock in one day

That is:

One has five (01:XX, 10:XX, 11:XX, 12:XX) Note: there are two in 11
Two has two (02:XX, 12:XX)
Three and Four only have 1 (03:XX, 04:XX)

Therefore Eight has:

1 (08:XX)


EDIT / ADDITION: While describing my solution in comments, I came to the obvious realization that the answer might be more succinctly written as:

The count of each number's digits in the months of the year.

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  • $\begingroup$ Bloody hell. Someone got it. Now that is impressive. For some bonus <something's>, I'd love to hear how you got there. $\endgroup$ – Stilez Nov 18 '17 at 18:05
  • $\begingroup$ My first one - woot! My not-so-exciting strategy was just to think about it on and off for a long time.. Binary / ASCII / letter / math approaches didn't seem to work out, and it seemed like single-digit and teen numbers had to be involved to get 1 to have more than 2, 3, or 4, and eventually it clicked. Out of curiosity, was my answer you intended solution - because I realize it might have been more succinctly answered using the edit at the bottom... $\endgroup$ – daroo Nov 20 '17 at 12:05
  • $\begingroup$ Oh, and the hints were not helpful, unfortunately for me.. I tried to use the "superficial" one for a while, but gave it up after getting nowhere with it.. $\endgroup$ – daroo Nov 20 '17 at 12:10
  • $\begingroup$ Your answer was indeed the intended one, see my update to the OP. I hadn't thought of that alternative which also uses the same values. But not many common things do, so it wouldn't have been too broad if you'd gone with that instead. Very impressed, and thank you also. $\endgroup$ – Stilez Nov 20 '17 at 22:08
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0

The puzzle essentially describes the function $f(x)=\lfloor \frac5x\rfloor$, or x => 5 // x in pseudo-code. The function is integer division with dividend 5. Thus:$$f(1)=5\\f(2)=2\\f(3)=1\\f(4)=1\\f(8)=0$$

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  • $\begingroup$ Clever, but I'm not sure that this really matches the non-arbitrary question wording :) ... that said, I'm curious what solutions will be found :) $\endgroup$ – Stilez Nov 13 '17 at 15:32
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    $\begingroup$ @Stilez Remember that if it has many possible solutions, it's very likely too broad. $\endgroup$ – EKons Nov 13 '17 at 15:40
  • $\begingroup$ I'm kind of hoping it doesn't. After all there are an infinity of formulae for a mathematical riddle, if a formulaic answer is looked for.... but rarely are there many answers having that click of certainty and elegance :) If it turns out I'm wrong, and there are multiple elegant answers, I'll have to add few more hints to narrow it down. But that makes it less elegant, so if possible I'll leave as-is. $\endgroup$ – Stilez Nov 13 '17 at 15:53
  • $\begingroup$ I don't think the verb have applies to this example. We don't say one has five if 5 // one == five. $\endgroup$ – ibrahim mahrir Nov 13 '17 at 20:48
1
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If we view the numbers as displays on something like a digital clock, with 7 segments, the answers refer to the number of unlit segments.
The number 1 uses 2/7 segments, leaving 5. 2 uses 5 segments. I'm not sure about 3 and 4 though, if the two displays were overlapping it would leave one unlit. 8 would have 0 in this pattern, as it uses every segment.

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  • $\begingroup$ This is clever. Yes, it is a lateral thinking puzzle, I was wondering when lateral thinkers would click the answer button. No, it's not this, but right now you're closer than anyone else to the right kind of mindset / angle. In at least two ways! For one thing, it would indeed be a property of a number (on at least most standard 7 seg displays). For another, it's, well.. lateral :) Howeverrrrr..... that still leaves you quite the wide range of thinking to find it in! :) And it pains me to say, but 7 seg's won't help much more ;-) $\endgroup$ – Stilez Nov 15 '17 at 20:13
0
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$-9$

Polynomial fit: $y = - \frac{1}{6}x^3 + 2x^2 - \frac{47}{6} x + 11$

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  • $\begingroup$ Some imagination puh-lease :) its a riddle not homework,....... :) $\endgroup$ – Stilez Nov 14 '17 at 7:51
0
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Perhaps, these are corresponding possible

anagrams of given words

and thereby

Eight has only one.

As,

ONE has - EON, ENO, NEO, ONE, (and one more, I could not get it right now)
TWO has - TWO, OWT
THREE and FOUR - THREE and FOUR

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    $\begingroup$ THREE - ETHER / THERE $\endgroup$ – prog_SAHIL Nov 15 '17 at 12:24
  • $\begingroup$ Clever again. I like lateral! (i.e., No. ;-) ) $\endgroup$ – Stilez Nov 15 '17 at 20:14
  • $\begingroup$ And TWO has TOW $\endgroup$ – Martin Smith Nov 19 '17 at 14:39
0
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Eight has

$0$

Because what a number has is the

Number of its non-zero multiples you can show with the fingers of one hand 1 has five multiples: 1,2,3,4 and 5 2 has two: 2 and 4 3 has only one: 3 4 has only one: 4 8 doesn't have any

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  • $\begingroup$ This is essentially the same answer that was already given. $\endgroup$ – Jason V Nov 15 '17 at 14:10
  • $\begingroup$ @Jason V where is the earlier given answer? $\endgroup$ – Mea Culpa Nay Nov 16 '17 at 3:03
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    $\begingroup$ The first answer (by the Greek name I can't type on a phone) the floor function of 5/x. This is the same answer presented a slightly different way $\endgroup$ – Jason V Nov 16 '17 at 3:07
0
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What about -

Multiples under 6

Answer would be

Eight has

Zero

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