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Topic: 5th root to any 2-digit number
Message: Posted by: roi_tau (Jun 22, 2006 01:56PM)
Hello my friends.

This is my first post in this subject so Im a little nervous...

There is a way to calculate the 5th root of a number.

Effect:
you tell a member of the audience to think of a 2 digit number and multiplie it by itself 5 times.
for example ,if the number was 23 so the result would be 6436343.

Now he tells you the result and you figure out the original number(23)

the first digit is the first digit in the result.

What is the formula to the second digit?

Have fun
ROi
Message: Posted by: Munseys_Magic (Jun 22, 2006 05:37PM)
I'm confused. In your example, the first digit (2) is NOT the first digit in the result (6).
Message: Posted by: McCritical (Jun 22, 2006 05:48PM)
[quote]
On 2006-06-22 18:37, CuriousMath.com writes:
The trick works like this:

The last digit is same as the last digit of the 5th power.

The tens place of the root can be determined by neglecting the last 5 digit of the fifth power.

Now if the remaining number lies in the following range, then it is the following number (ie, tens place of the root):

1-30-------------->1
20-230----------->2
230-1000------->3
1000-3000------>4
3000-7500------>5
7500-16000----->6
16000-32000---->7
32000-57000---->8
57000-99999----->9

Example: 130691232

Units place: 2

For tens place, neglecting the last five digits, we get 1306 which lies in the range 1000-3000. Hence tens place is 4.

Therefore the 5th root of 130691232 is 42
[/quote]

http://www.curiousmath.com/modules.php?op=modload&name=News&file=article&sid=74
Message: Posted by: roi_tau (Jun 23, 2006 12:28AM)
Thanks you very much.

Are you a relative of Martin Gardner?...

The next step in rapid math is to calculate a sine(x) in a show.


Have fun
ROi
Message: Posted by: Nir Dahan (Jun 23, 2006 02:41AM)
I've actually did some real heavy root calculations while in high school.
it pays of to remember log2 and log3 - a lot can be done with it once you are familiar with logarithm rules.

as for sin(x) for very small x it is very easy...

nir
Message: Posted by: roi_tau (Jun 23, 2006 06:32AM)
Hi Nir.

We once met at Roei Zaltzman hoem with Nimrod Harel.

You did some great things there.(I remember the supermarket list with the change)

I tried to solve it with matlab but there is too many options.

Im also a friend of Shlomi from the forums.I guess you know him.

How are you in Germany ? How is your kid?

Richard Feynman has a chapter on fast calculation in his book regarding calculate e^x

very funny.

Have fun and Shabat Shalom

ROi
Message: Posted by: Nir Dahan (Jun 23, 2006 07:19AM)
Hi Roi,

glad this effect left an impression on you. it also fooled some big names (some of which won't admit it :) )

try to get a hold of a book by Wallace Lee on fast math... it is THE best (but don't tell anybody I told you)

Nir.

P.S. I now have 2 kids, and in October the 3rd is coming...
Message: Posted by: Scott Cram (Jun 23, 2006 12:09PM)
I cover [url=http://members.cox.net/beagenius/root2.html]how to do 5th roots on my website[/url] (as well as [url=http://members.cox.net/beagenius/root.html]cube roots[/url] and [url=http://members.cox.net/beagenius/root3.html]square roots[/url]), and you can even [url=http://members.cox.net/beagenius/rootquiz.html]quiz yourself on fifth roots[/url], too!

To figure out the tens digit (which I usually do first, since the ones digit is so easy), first, you need to memorize the 5th powers of each of the 1-digit numbers:

1^5 = 1
2^5 = 32
3^5 = 243
4^5 = 1,024
5^5 = 3,125
6^5 = 7,776
7^5 = 16,807
8^5 = 32,768
9^5 = 59,049

To figure out the 10s digit, mentally drop or ignore the 5 rightmost digits, and focus on the remaining digits. If the number you're given is 11,881,376, for example, you drop the rightmost five digits, and look only at the remaining ones: 118. 118 is greater than 32 (2^5), but less than 243 (3^5), so the tens digit must be 2. The ones digit is a 6, so the fifth root is 26!

Let's try another one: 601,692,057

Drop the rightmost 5 digits for now: 6,016

6,016 is greater than 3,125 (5^5), but less than 7,776 (6^5). Therefore, the tens digit is 5.

The ones digit in the number is 7, so the ones digit in the answer is 7.

Therefore, the fifth root of 601,692,057 is 57.


Nir, are you talking about "Math Miracles" by W. Wallace Lee? Great book!

Speaking of books, [url=http://gmvlog.blogspot.com/2006/05/arthur-benjamin.html]Arthur Benjamin[/url] has a [url=http://headinside.blogspot.com/2006/03/notes-from-math-class_17.html]new mathemagic book coming out in August[/url]!
Message: Posted by: Nir Dahan (Jun 23, 2006 02:09PM)
Scott,

thanks for the update on arthur benjamin - I got his first book, can't wait for the second to come out already...

Nir
Message: Posted by: roi_tau (Jun 24, 2006 02:01PM)
Hi Scot and Nir.

I don't know where you found those books but I can't find to buy(don't mention download) niether Math Miracles nor Benjamin's book (even so ,Scot,you have a great movie of him.)

Help me.(after all Im younger than both of you-it's your job to educate me...)

Thanks in advance

ROi
Message: Posted by: TomasB (Jun 24, 2006 02:34PM)
If find it easier to memorize this table for the ten-value digit:

1-(2) 30
2-(3) 230
3-(4) 1000
4-(5) 3000
5-(6) 7600
6-(7) 16000
7-(8) 32000
8-(9) 56000

It's probably easier for me since I can see so many connections between the values on the right and on the left.

/Tomas
Message: Posted by: idris (Jun 29, 2006 11:38AM)
Math Miracles was originally published in 1976 by Mickey Hades, a Canadian publisher and dealer. If I recall correctly it was about 50 pages (8.5" by 11") and comb bound. I can double check when I get home tonight.

Mickey Hades has a web site (http://www.mickyhades.com/) but I don't find any reference to his back list there. His phone number and email are listed so you could contact him about the book.

Your best bet for will probably be a used book dealer. You might try H&R Magic Books at http://www.magicbookshop.com. I don't see it currently listed there, but you never know when he might get a copy.

Good luck,
Message: Posted by: idris (Jul 1, 2006 10:02PM)
As an update to my previous message, the 1976 edition is actually the second edition of Math Miracles. That is the copy I have. According to the title page, the original copyright was from 1960. I don't know, but suspect the first edition was self published by W Wallace Lee.

My second edition has 85 numbered pages, and a few blank pages on the end.
Message: Posted by: Scott Cram (Jul 2, 2006 04:51AM)
Arthur Benjamin's first book, [url=http://www.amazon.com/gp/product/1565651189/sr=8-1/qid=1151833784/ref=astonishmenlinks/102-2878197-5034544]Mathemagics[/url] is still in print, and not difficult to find or order from bookstores.

His second book, however, won't be available until next month.
Message: Posted by: stanalger (Jul 27, 2006 01:46PM)
[quote]
On 2006-06-23 13:09, Scott Cram wrote:
Speaking of books, [url=http://gmvlog.blogspot.com/2006/05/arthur-benjamin.html]Arthur Benjamin[/url] has a [url=http://headinside.blogspot.com/2006/03/notes-from-math-class_17.html]new mathemagic book coming out in August[/url]!
[/quote]

Not really a new book. It's a reprint of Art's
"Mathemagics: How to Look Like a Genius Without Really Trying."
The new publisher insisted on a new title.