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 Post subject: Math Puzzles
PostPosted: September 23rd, 2009 02:30:55 am 
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Jamist
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does any1 have any math puzzles from like the AMC or somethin bcause this is the pl8ce to share them

I don't know if this is mathematically relevant (maybe Ben can tell me) but I came up with this problem a day ago and it was pretty fun to solve

basically it is this:

using only the numbers 1-9, using each digit only once, come up with an expression that has all the digits 1-9 in its answer.

the rules are: you have to just multiply, divide, add, subtract, or exponent straight numbers. like you could do (2*5*6)+8 to make 68 and use up your 2 5 6 and 8, but you couldn't just use up 6 and 8 and make 68

ALSO the answer can be a decimal, whatever just so long as it has 123456789 in it. so it could be like 95.8476213 if you can get it.


ARE YOU UP TO THE CHALLENGE???

also post your math problems here if you have some

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 Post subject: Re: Math Puzzles
PostPosted: September 23rd, 2009 03:45:25 am 
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Exploiting the many digits in sevenths:

3*(9*8 + 5*(4+6) + 1) + 2/7 = 369.285714285714285714285714285714285714285714...

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 Post subject: Re: Math Puzzles
PostPosted: September 24th, 2009 05:30:10 am 
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Jamist
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Ben Heaton wrote:
Exploiting the many digits in sevenths:

3*(9*8 + 5*(4+6) + 1) + 2/7 = 369.285714285714285714285714285714285714285714...



Awesome! I think I tried a similar method, on my TI-84 graphing calculator, and failed. But I did get some other cool solutions:

The first one I tried was (((((9^2)^3)*8)/7)/6)*5)=506134.2857, which, using my display, only yielded eight of the nine digits. But then I reworked it, and got (((((9^2)^3)/7)*8)/6)*5), which, now I realize, is the same exact number. Huh.

But I got two other answers. The ^5 part of this was found by guessing: 3((2^7+9*8)/6)^5-1=123456789.1

This was found by working backwards, it uses the least amount of digits (only six!): ((2*5)^9-7)/(3^4)=12345678.93

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 Post subject: Re: Math Puzzles
PostPosted: September 24th, 2009 21:36:28 pm 
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OK, I got one where the result is an integer, with no repeated digits:

(2*8)^7 - 5*6*9 + 3*4 - 1 = 268435197

My approach was to start by getting a nine-digit number with the first few digits all unique (and nonzero), and then use the remaining numbers to adjust the end of it a bit.


Here's a similar one using just 1 through 8, but still getting all nine digits in the result:

(5+6)^8 + 7*4*3*1 + 2 = 214358967


I'm curious what the smallest n would be such that the numbers from 1 to n are sufficient. Can anyone beat 8?

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 Post subject: Re: Math Puzzles
PostPosted: September 26th, 2009 01:17:17 am 
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I am going to beat 8, you wait!

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 Post subject: Re: Math Puzzles
PostPosted: September 26th, 2009 01:40:12 am 
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Got one:

(7^(1/2))+6^4=1298.6457513, although the graphing calculator doesn't show the 3 at the end in this form, it does show the 3 in just straight 7^(1/2).

Um, Ben, shouldn't most square roots just have a never-ending pattern of decimals, eventually hitting all 9 digits?

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 Post subject: Re: Math Puzzles
PostPosted: September 26th, 2009 01:53:43 am 
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Yeah, 3^(1/2) works that way. It'd take more digits than that for an integer solution, though.

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 Post subject: Re: Math Puzzles
PostPosted: September 26th, 2009 02:05:06 am 
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Using 1 to 5, integer result with some repeated digits:

2^(4^3+5-1) = 295147905179352825856

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 Post subject: Re: Math Puzzles
PostPosted: September 30th, 2009 01:16:57 am 
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awesome! I didn't say it had to be an integer, but oh well.

also does anyone have any other math puzzles or is this not a thing

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 Post subject: Re: Math Puzzles
PostPosted: September 30th, 2009 02:13:46 am 
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Yeah, I just like getting integers.

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 Post subject: Re: Math Puzzles
PostPosted: September 30th, 2009 21:42:18 pm 
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If you want some interesting maths puzzles, check this out:
http://www.m-a.org.uk/resources/sp/sp_1999.pdf

I'm going to be signing up for my MA membership pretty soon and I'm rather excited about the various publications I will be receiving!


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 Post subject: Re: Math Puzzles
PostPosted: October 19th, 2009 06:01:17 am 
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http://www.unl.edu/amc/a-activities/a7- ... mo2005.pdf

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