# Math puzzle

**URL:** <https://boards.straightdope.com/t/math-puzzle/559931>\
**Category:** The Game Room\
**Created:** [November 8, 2010, 5:38pm UTC](https://boards.straightdope.com/t/math-puzzle/559931 "2010-11-08T17:38:46Z")\
**Posts on this page:** 15\
**Page:** 1

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**Author:** ![Grey](https://avatars.discourse-cdn.com/v4/letter/g/b782af/32.png) [@Grey](https://boards.straightdope.com/u/Grey)\
**Post date:** [November 8, 2010, 5:38pm UTC](https://boards.straightdope.com/t/math-puzzle/559931/1 "2010-11-08T17:38:46Z")

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This is one of those math puzzles I can never figure out and always slap my head once they’ve been explained.

Take 2 numbers a and b, but a =/= b  
Take the mean to be x

x= (a+b)/2  
2x=(a+b)  
2x(a-b)=(a+b)(a-b)  
2xa-2xb=a[sup]2[/sup]-b[sup]2[/sup]  
a[sup]2[/sup]-2xa=b[sup]2[/sup]-2xb  
a[sup]2[/sup]-2xa+x[sup]2[/sup]=b[sup]2[/sup]-2xb+x[sup]2[/sup]  
(a-x)[sup]2[/sup]=(b-x)[sup]2[/sup]  
a-x=b-x  
a=b

But we defined a=/=b

What am I missing?

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**Author:** ![hdc\_bst](https://avatars.discourse-cdn.com/v4/letter/h/9f8e36/32.png) [@hdc\_bst](https://boards.straightdope.com/u/hdc_bst)\
**Post date:** [November 8, 2010, 5:44pm UTC](https://boards.straightdope.com/t/math-puzzle/559931/2 "2010-11-08T17:44:25Z")

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> [@Grey](#):
>
> (a-x)[sup]2[/sup]=(b-x)[sup]2[/sup]  
> a-x=b-x

Your error occurs in this line. There are two roots to consider, the positive and negative, so…

either a-x = b - x … or … a - x = x - b

Because a /= b, we know that

a - x = x - b … or in other words x is the arithmetic mean of a and b (as defined).

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**Author:** ![Joey\_P](https://avatars.discourse-cdn.com/v4/letter/j/919ad9/32.png) [@Joey\_P](https://boards.straightdope.com/u/Joey_P)\
**Post date:** [November 8, 2010, 5:45pm UTC](https://boards.straightdope.com/t/math-puzzle/559931/3 "2010-11-08T17:45:40Z")

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nm

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**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\
**Post date:** [November 8, 2010, 6:07pm UTC](https://boards.straightdope.com/t/math-puzzle/559931/4 "2010-11-08T18:07:49Z")

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Yup. In these puzzles, typically one of two things happens: division by zero, or taking square roots and ignoring the existence of two possible roots. As **hdc\_bst** found, in this case it’s the latter here.

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**Author:** ![Grey](https://avatars.discourse-cdn.com/v4/letter/g/b782af/32.png) [@Grey](https://boards.straightdope.com/u/Grey)\
**Post date:** [November 8, 2010, 6:32pm UTC](https://boards.straightdope.com/t/math-puzzle/559931/5 "2010-11-08T18:32:02Z")

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Figured as much. Thanks folks.

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**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\
**Post date:** [November 8, 2010, 9:45pm UTC](https://boards.straightdope.com/t/math-puzzle/559931/6 "2010-11-08T21:45:13Z")

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> [@](#):
>
> Yup. In these puzzles, typically one of two things happens: division by zero, or taking square roots and ignoring the existence of two possible roots.

I’ve also seen them based on omitting a constant of integration. IIRC, it made use of logs, since for most integrals, leaving off the constant will give you the integral from 0 to your point, but for the integral of 1/x, you find the integral from 1 to your point (since the integral from 0 diverges).

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**Author:** ![Enderw24](https://avatars.discourse-cdn.com/v4/letter/e/ba9def/32.png) [@Enderw24](https://boards.straightdope.com/u/Enderw24)\
**Post date:** [November 8, 2010, 10:16pm UTC](https://boards.straightdope.com/t/math-puzzle/559931/7 "2010-11-08T22:16:11Z")

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> [@Grey](#):
>
> 2xa-2xb=a[sup]2[/sup]-b[sup]2[/sup]  
> a[sup]2[/sup]-2xa=b[sup]2[/sup]-2xb

Also, this line is incorrect. So to move the 2xa over to the other side, you subtract it and that works out.  
But to move the 2xb to the other side you ADD it, which means it’s b[sup]2[/sup]+2xb

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**Author:** ![Arnold\_Winkelried](https://avatars.discourse-cdn.com/v4/letter/a/3d9bf3/32.png) [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)\
**Post date:** [November 8, 2010, 10:19pm UTC](https://boards.straightdope.com/t/math-puzzle/559931/8 "2010-11-08T22:19:36Z")

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> [@Enderw24](#):
>
> Also, this line is incorrect.

Are you sure about that? :dubious:

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**Author:** ![Enderw24](https://avatars.discourse-cdn.com/v4/letter/e/ba9def/32.png) [@Enderw24](https://boards.straightdope.com/u/Enderw24)\
**Post date:** [November 8, 2010, 11:12pm UTC](https://boards.straightdope.com/t/math-puzzle/559931/9 "2010-11-08T23:12:37Z")

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> [@Arnold\_Winkelried](#):
>
> Are you sure about that? :dubious:

What would be incorrect?  
To move -2xb to the other side, you have to add it on both ends.

Actually, what would happen is you’d get -b[sup]2[/sup]+2xb, which you could convert to b[sup]2[/sup]-2xb but only if you put a -1 on the other side as well. Either way it doesn’t work.

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**Author:** ![Malacandra](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@Malacandra](https://boards.straightdope.com/u/Malacandra)\
**Post date:** [November 8, 2010, 11:19pm UTC](https://boards.straightdope.com/t/math-puzzle/559931/10 "2010-11-08T23:19:17Z")

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> [@Enderw24](#):
>
> What would be incorrect?  
> To move -2xb to the other side, you have to add it on both ends.
> 
> Actually, what would happen is you’d get -b[sup]2[/sup]+2xb, which you could convert to b[sup]2[/sup]-2xb but only if you put a -1 on the other side as well. Either way it doesn’t work.

You’re wrong. He’s added b[sup]2[/sup] to both sides, and left the -2xb where it was…

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**Author:** ![Arnold\_Winkelried](https://avatars.discourse-cdn.com/v4/letter/a/3d9bf3/32.png) [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)\
**Post date:** [November 8, 2010, 11:20pm UTC](https://boards.straightdope.com/t/math-puzzle/559931/11 "2010-11-08T23:20:00Z")

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2xa-2xb=a[sup]2[/sup]-b[sup]2[/sup] \<— line from the OP  
add -2xa to both sides, gives

-2xb=a[sup]2[/sup]-b[sup]2[/sup]-2xa  
re-arrange terms on right side, gives

-2xb=a[sup]2[/sup]-2xa-b[sup]2[/sup]  
add b[sup]2[/sup] to both sides, gives

b[sup]2[/sup]-2xb=a[sup]2[/sup]-2xa  
swap sides, gives

a[sup]2[/sup]-2xa=b[sup]2[/sup]-2xb \<— line from the OP

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**Author:** ![Bryan\_Ekers](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bryan_ekers/32/183_2.png) [@Bryan\_Ekers](https://boards.straightdope.com/u/Bryan_Ekers)\
**Post date:** [November 9, 2010, 1:26am UTC](https://boards.straightdope.com/t/math-puzzle/559931/12 "2010-11-09T01:26:29Z")

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A trivial observation on hdc’s solution: If X is the mean of A and B and A!=B, X must be halfway between A and B, i.e. smaller than one of them but bigger than the other, and by the same amount each time.

Therefore, either: A - X = X - B (and B - X = X - A).

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**Author:** ![Colibri](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/colibri/32/1841_2.png) [@Colibri](https://boards.straightdope.com/u/Colibri)\
**Post date:** [November 9, 2010, 2:12am UTC](https://boards.straightdope.com/t/math-puzzle/559931/13 "2010-11-09T02:12:24Z")

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Off to The Game Room.

Colibri  
General Questions Moderator

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<div class="post-metadata">

**Author:** ![Enderw24](https://avatars.discourse-cdn.com/v4/letter/e/ba9def/32.png) [@Enderw24](https://boards.straightdope.com/u/Enderw24)\
**Post date:** [November 9, 2010, 3:01am UTC](https://boards.straightdope.com/t/math-puzzle/559931/14 "2010-11-09T03:01:40Z")

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> [@Malacandra](#):
>
> You’re wrong. He’s added b[sup]2[/sup] to both sides, and left the -2xb where it was…

Oh  
Well how about that?

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**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [November 10, 2010, 1:22am UTC](https://boards.straightdope.com/t/math-puzzle/559931/15 "2010-11-10T01:22:58Z")

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In general, an easy way to find a step where things go wrong: Plug in actual numbers for a and b and walk through the lines till you find the first equality that isn’t actually true with those specific values. If you did this, it would readily have been apparent that the step moving from (a - x)^2 = (b- x)^2 to a - x = b - x was the source of the problem.