Drexel dragonThe Math ForumDonate to the Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Math Forum » Discussions » Inactive » math-history-list

Topic: An Odd Numbers' Puzzle from an Ancient Egyptian Binary Puzzle.
Replies: 11   Last Post: Jan 15, 2011 5:14 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Hossam Aboulfotouh

Posts: 161
From: Egypt
Registered: 7/20/07
An Odd Numbers' Puzzle from an Ancient Egyptian Binary Puzzle.
Posted: Jan 7, 2011 2:58 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

Greetings and Happy New Year to the Moderators, the List Members and the Readers.

Today is January 7; and many people likes 7 and any thing that includes or starts with 7. I therefore would like to introduce the following note that seems it created a puzzle to me.

160 is the total sum of the following sequential 10 odd numbers;

7, 9, 11, 13, 15, 17, 19, 21, 23, and 25.

They form a progression series, where 16 is its hidden average; both 9 and 25 are terms/parts of this series; and the relation between these three numbers were said to be uncomprehended before the days of Pythagoras. The difference "2" between any two terms equals 1/8 of their average 16.

If one multiplied each of its 10 terms by 1/16, the results would be the 10 terms of the series in the solution of the ancient Egyptian puzzle that were numbers Problem#64 in the so-called "Rhind Mathematical Papyrus", i.e., in the papyrus, the total sum of the ten terms is 10. Its smallest term was (7 * 1/16) = 1/4+1/8+1/16 and its largest term was (25 * 1/16) = 1 + 1/2+ 1/16. See the 10 terms of RMPp#64 via this link:

My note here is that, in the spectrum of odd numbers, the total sum of the series of numbers from 7 to 25 equal 160, i.e., it equals 16 * 10 = 2^4 * 10^1.

One could rewrite it in the form [2^n * 10^(n-3)], were n = 4 for the above series.

The next series of odd numbers that its total sum comply with the condition [2^n * 10^(n-3)] is the series of odd numbers from 7 to 505, where its sum is 64000 = (2^6 * 10^3). It?s the total sum of 250 odd numbers.

That is, 505 is the number that comes after 25. Using the normal ways of calculation in Excel spread sheet, I was not able to find the number(s) that come(s) after 505, because the numbers become bigger and bigger.

My question here, is there any way (an equation or so) to identify the intervals for the coming but working odd numbers? Taking into consideration that the sum of the numbers missed the interval of 3200 = 2^5*10^2, the case might be false at other intervals too.

Or if you can get those numbers using your own computer program, is it possible to write them here?

Best regards.

Hossam Aboulfotouh

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum 1994-2015. All Rights Reserved.