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August 23rd, 2013, 06:06 AM   #1
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Math Q&A challenge (Part 11)

You have an integer number where any digit appears at most twice.
The sum of all neighboring four digits in this number is a square.

Example: 205290 is such a number because no digit appears more
than twice, and 2+0+5+2, 0+5+2+9, and 5+2+9+0 are squares.

What is the maximum possible value for this number?
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August 24th, 2013, 02:34 PM   #2
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Re: Math Q&A challenge (Part 11)

This is interesting , I'll make the frst submission.

How about 9880013571357 ?

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August 24th, 2013, 07:03 PM   #3
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Re: Math Q&A challenge (Part 11)

Good try Agent...there is a higher number.
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August 24th, 2013, 07:37 PM   #4
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Re: Math Q&A challenge (Part 11)

Quote:
Originally Posted by Denis
Good try Agent...there is a higher number.
http://www.youtube.com/watch?v=n4ucO4xe28c&t=7s
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August 24th, 2013, 08:09 PM   #5
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Re: Math Q&A challenge (Part 11)



http://www.google.com/url?q=http://www. ... 7Am_YMK6_A

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August 24th, 2013, 08:13 PM   #6
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Re: Math Q&A challenge (Part 11)

Looked at both links. What do I not see?
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August 24th, 2013, 08:30 PM   #7
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Re: Math Q&A challenge (Part 11)

CRG gave me a 'ribbing' using Mario so I countered with Wario... you have to know about video game plumbers ... and bizzarro characters of a stienfool universe ...

Don't worry, you didn't miss anything important.

I tried a few more but can't get a bigger number.

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August 24th, 2013, 10:17 PM   #8
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81978100367936
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August 24th, 2013, 10:23 PM   #9
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Re: Math Q&A challenge (Part 11)

Quote:
Originally Posted by skipjack
81978100367936
That's what I get. I believe it's optimal, though I would wait for Denis's approval.

By the way, skipjack, if you don't have any question to ask then just pass it to someone for the game to continue, ok?
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August 24th, 2013, 10:44 PM   #10
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Re: Math Q&A challenge (Part 11)

And it looks like we forgot to post the rules AGAIN.

:P
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