15
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Found this in my brother's math book, we tried to solve it but didn't manage to find any solution. Apology if it it too easy, but we really tried to find it.

Find the odd number out:

$\frac{3}{8}$; $\frac{4}{9}$; $0.37$; $0.4\overline{8}$; $\frac{7}{15}$; $\frac{9}{11}$

The only thing we found is that $0.4\overline{8}$ is the only one with a written $\overline{x}$, but we don't like this solution.

Is there a nicer one?

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    $\begingroup$ Source: Fokus Mathematik 6, Cornelsen (German Math book) $\endgroup$ – palsch Jun 2 '16 at 17:19
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    $\begingroup$ Well, the "oddest" one, in terms of their $2$-adic valuations is $\frac{3}8$, which has a valuation of $-3$, where the others have valuations of $0$ (and are thus more "even"). That's probably not what the question is asking though. $\endgroup$ – Milo Brandt Jun 2 '16 at 19:33
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    $\begingroup$ Considering this came from a textbook, what is the curriculum material from the rest of the chapter? What specific details about fractions is this particular chapter trying to teach them? $\endgroup$ – Bulldogg6404 Jun 2 '16 at 20:20
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    $\begingroup$ If this was a lateral-thinking question, you might answer "5" - it's the only digit occurring only once in the question. $\endgroup$ – Lawrence Jun 3 '16 at 2:47
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    $\begingroup$ @Lawrence or $2$ as it's the only digit not used (there is an upside down $6$, honest) $\endgroup$ – Jonathan Allan Jun 4 '16 at 5:20

11 Answers 11

20
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Perhaps:

$\frac{9}{11}$, because it's the only one more than $\frac{1}{2}$

Another possibility:

$0.37$ ($\frac{37}{100}$), because it's the only one whose rational $\frac{p}{q}$ representation does not have 3 as a factor in either term, and also the only one whose denominator has 3 digits.

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    $\begingroup$ I'd say the probability is more than $.5\overline0$ that this answer is correct, taking into account that the puzzle is from a 6th grade math book that could be teaching fractions $\endgroup$ – humn Jun 2 '16 at 17:28
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    $\begingroup$ Good job on keeping it simple. $\endgroup$ – Wildcard Jun 3 '16 at 0:20
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    $\begingroup$ There is always going to be a highest number... what makes that 'odd'? $\endgroup$ – Lamar Latrell Jun 3 '16 at 9:13
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    $\begingroup$ @LamarLatrell: Good point. $\endgroup$ – KeyboardWielder Jun 3 '16 at 15:52
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    $\begingroup$ @LamarLatrell, there's always going to be a highest-ranked user on puzzling.stackexchange.com, but that doesn't mean that it is not remarkable to be that user! $\endgroup$ – LSpice Jun 3 '16 at 18:19
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Due to the occurrence of $0.4\overline{8}$ rather than $\frac{22}{45}$ I think it could be

$\frac9{11}$

Because

If we write them all explicitly with their recurring digits:
$\frac{3}{8}=0.375\overline{0}$, $\frac{4}{9}=0.\overline{4}$, $0.37=0.37\overline{0}$, $0.4\overline{8}=0.4\overline{8}$, $\frac{7}{15}=0.4\overline{6}$, $\frac{9}{11}=0.\overline{81}$
it is the only one with a period not equal to one.


An alternative, quirky one that I kind of like, but definitely not the intended one:

$0.4\overline{8}$

Because

writing them in irreducible form and factorising we see it is the only one that has neither a square or a prime as at least one of it's numerator or denominator: \begin{align}\frac{3}{8}&=\frac{3}{2^3}&\frac{\text{prime}}{\text{cube}} \\\frac{4}{9}&=\frac{2^2}{3^2}&\frac{\text{square}}{\text{square}} \\0.37=\frac{37}{100}&=\frac{37}{(2\cdot5)^2}&\frac{\text{prime}}{\text{square}} \\0.4\overline{8}=\frac{22}{45}&=\frac{2\cdot11}{3^2\cdot5}&\text{odd one out} \\\frac{7}{15}&=\frac{7}{3\cdot5}&\frac{\text{prime}}{\text{prime}\cdot\text{prime}} \\\frac{9}{11}&=\frac{3^2}{11}&\frac{\text{square}}{\text{prime}}\\\end{align}


Or how about

$\text{All of them!}$

Because

\begin{align}\text{This}& &\text{is the only one written using:}\\\frac38&; &\text{an even denominator} \\\frac49&; &\text{an even numerator} \\0.37&; &\text{no power of 2}&\text{ }^* \\0.4\overline{8}&; &\text{only even digits} \\\frac7{15}&; &\text{a triangle denominator} \\\frac9{11}&; &\text{a prime denominator} \\\end{align}
$\text{* }2^0=1\text{, which appears in the last two}$


I'll squeeze another in here...

$0.37$

Because

it is the only one which, once removed, make the set of irreducible forms, $\frac{n}{d}$, have $$\frac{\prod{n}}{\sum{d}}\in \Bbb{Z}$$
\begin{align}\frac38:\frac{4\times 37\times 22\times 7\times 9}{9+100+45+15+11}&=\frac{5698}{5} \\[2ex]\frac49:\frac{3\times 37\times 22\times 7\times 9}{8+100+45+15+11}&=\frac{153846}{179} \\[2ex]0.37:\frac{3\times 4\times 22\times 7\times 9}{8+9+45+15+11}&=\frac{189}{1}=189 \\[2ex]0.4\overline{8}:\frac{3\times 4\times 37\times 7\times 9}{8+9+100+15+11}&=\frac{27972}{143} \\[2ex]\frac7{15}:\frac{3\times 4\times 37\times 22\times 9}{8+9+100+45+11}&=\frac{87912}{173} \\[2ex]\frac9{11}:\frac{3\times 4\times 37\times 22\times 7}{8+9+100+45+15}&=\frac{22792}{59}\end{align}

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    $\begingroup$ Possibly better than the intended answer! $\endgroup$ – humn Jun 2 '16 at 17:39
  • $\begingroup$ Sounds good. My brother will ask his teacher on Monday and then I'll accept the answer she suggests. $\endgroup$ – palsch Jun 2 '16 at 17:48
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    $\begingroup$ I'd love to be around to see a kid come up with this answer :) $\endgroup$ – Lamar Latrell Jun 3 '16 at 9:14
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    $\begingroup$ I'd like to give another +1 for "All of them". That is exactly the list that I wanted to give my brother to give it his teacher. ;-) $\endgroup$ – palsch Jun 3 '16 at 15:48
  • $\begingroup$ OMG, there will be so much confusion in my brother's class. $\endgroup$ – palsch Jun 5 '16 at 14:11
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The answer is

0.37

Because

It's the only one not containing a horizontal bar.

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    $\begingroup$ I really like this outside the box solution :) $\endgroup$ – Jonathan Allan Jun 3 '16 at 19:16
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    $\begingroup$ I really like this outside the "bar" solution ;D $\endgroup$ – Simply Beautiful Art Jun 3 '16 at 20:23
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9/11

because

this is the only number in the list whose denominator is a prime when the number is expressed in the simplest fractional form possible.

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  • $\begingroup$ Why the downvotes? $\endgroup$ – Lawrence Jun 3 '16 at 2:42
  • $\begingroup$ @Lawrence, I didn't downvote (and I think that the answer is as plausible as any of the others), but perhaps it is because the other answers use spoilers, and this one doesn't? $\endgroup$ – LSpice Jun 3 '16 at 18:20
  • $\begingroup$ @LSpice Thanks for the suggestion. That'd be harsh to pin on someone new to Puzzling. Anyway, it's +6 vs -2 now. I agree with you about plausibility. Given Jonathan Allan's "all of them" answer, I'm curious about the textbook's (or teacher's) standard answer. $\endgroup$ – Lawrence Jun 4 '16 at 3:35
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I believe the answer is:

$$3/8$$

Because:

It is the only number that requires more than 2 significant digits. $$3/8 = 0.375\text{ requires 3 significant digits}$$ $$4/9 = 0.\overline{4}\text{ can be written with one significant digit}$$ $$0.37\text{ has 2 significant digits}$$ $$22/45 = 0.4\overline{8}\text{ can be written with 2 significant digits}$$ $$7/15 = 0.4\overline{6}\text{ can be written with 2 significant digits}$$ $$9/11 = 0.\overline{81}\text { can be written with 2 significant digits}$$

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5
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Maybe it is

$\frac{4}{9}$

because

it's the only square of a rational number.

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4
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I'm gonna go with

4/9

because

it's the only one whose reciprocal does not have a repeating portion.

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  • $\begingroup$ 9/4??? There is no 9/4 in the sequence. $\endgroup$ – palsch Jun 4 '16 at 21:43
  • $\begingroup$ @palsch There is a $\frac49$ the reciprocal of which is $\frac94$ $\endgroup$ – Jonathan Allan Jun 4 '16 at 22:34
  • $\begingroup$ Oh, now I understand. +1 $\endgroup$ – palsch Jun 4 '16 at 22:36
3
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Perhaps it is

$\frac{4}{9}$

because

Its fractional part (represented in base 10) contains only 1 unique digit, whereas all the others contain multiple unique digits.

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2
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First, write all numbers with the smallest common denominator (19800). Now compare the numerators. A close look will reveal that the answer is

3/8

because

its numerator is the only one with an odd number of ones in binary representation! (And it also the only odd numerator, but that's no big surprise.)

This is the most literal interpretation of odd as a criterion yet I believe. Of course from this point of view the second oddest number is

22/45

because

after taking 3/8 from the list, there are two numerators with four ones in their binary representation and two with eight ones. But this one is the only one with six ones.

After that, we can separate these two clusters, but there is not any single "odd" number left.

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    $\begingroup$ 6th grade math book. No puzzling book. Of course this is the anwer. ;-) :D $\endgroup$ – palsch Jun 4 '16 at 19:23
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    $\begingroup$ @palsch: Children can be amazingly competent lateral thinkers, so who knows... ;-) $\endgroup$ – MarLinn Jun 4 '16 at 19:30
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    $\begingroup$ And sorry, the interpretation doesn't work (at least as an answer to the math book because it's German and doesn't say anything about "odd". P.S.: How does one come up with such a genius answer? :) $\endgroup$ – palsch Jun 4 '16 at 19:32
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    $\begingroup$ I'm flattered, but it's actually really simple. XD First I remembered that it was quite common in my school years that such puzzle questions had something to do with the (repeated) sum of digits. I have no idea why, but it was a crazy common trope. That alone didn't bring up anything interesting enough, but as an IT-guy it's one of my first instincts to try "something in binary" - because what is so special about base 10 anyway. And the sum of digits in binary is just the number of ones, so there you go. Oh and of course the initial normalization is also the one we used most often back when. $\endgroup$ – MarLinn Jun 4 '16 at 20:17
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I think you all are thinking waaay too hard on this.

9/11 is the only one that's greater than a half.

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  • $\begingroup$ not only is it greater than one half it's greater than 0.8 where all the others are between 0.3 and 0.5. An extreme outlier $\endgroup$ – Jasen Jun 6 '16 at 4:25
  • $\begingroup$ You are right, although @KeyboardWielder was faster. I accepted his answer and upvoted yours. $\endgroup$ – palsch Jun 7 '16 at 4:52
1
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It is:

0.37

because

It is the only number that has an exact decimal form

This answer is wrong because I am bad at math

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    $\begingroup$ And 3/8 hasn't? $\endgroup$ – M Oehm Jun 2 '16 at 17:29
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    $\begingroup$ Oh shoot I was thinking 8/3 $\endgroup$ – gannolloy Jun 2 '16 at 17:31
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    $\begingroup$ I had the same thought, but I'm bad at maths, too: I've checked with a calculator. :( $\endgroup$ – M Oehm Jun 2 '16 at 17:33
  • $\begingroup$ @StewieGriffin I've never deleted an answer because it was wrong before, is that what I should do? $\endgroup$ – gannolloy Jun 3 '16 at 16:12

protected by Aza Jun 6 '16 at 3:15

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