Archive for the ‘Computing’ Category

VPNs in China to access the external world internet

Monday, February 20th, 2017

BitCoin Core Software

Sunday, February 12th, 2017

Research Links

The first thing you will note is that the Bit Coin Core software will advise you it needs 100GB to download the blockchain. I bought a machine specifically for some new experiments including bitcoin and some other things.

VPN Software and Setup

Friday, February 10th, 2017

I am going to set up a VPN for my two locations.  This will be my page where I make notes about the process.

Research Links

HP ProDesk G1 Mini Desktop

Thursday, February 9th, 2017


-Processor: Intel® Core i5-4590T Quad Core 2.0 GHz com Turbo Boost to 3.0 GHz 
– Windows 10 Pro 64 Bits 
– Memory RAM: 8 GB DDR3 SDRAM 
– Hard Drive: 500 GB SATA 7200 RPM 
– Cache: 6 MB 
– Chipset: Intel® Q85 Express 
– Portas USB front: 2 (3.0) 
– Portas USB rear: 2 (2.0), 2 (3.0) 
– Placa de vídeo: Integrated: Intel® HD Graphics 4600 
– WIFI : No WiFi

Research Links

Although DisplayPort has much of the same functionality as HDMI, it is a complementary connection used in different scenarios. DisplayPort can emit an HDMI signal through the use of a passive adapter connected to a port that is designed for dual-mode.

DisplayPort Adaptation


  • DisplayPorts on the unit show the ++ version:  


Microsoft Hololens

Tuesday, February 7th, 2017

Research Links

RSA Code Made Easy

Tuesday, January 31st, 2017

A code is implemented with the following statement.  Decrypt the code.

 x^197 mod 3131  

The totient of 3131 is needed and is found below:

 3131 = 31 * 101    

 {psi} (3131) = psi(31) * psi(101) = (31-1)(101-1)=3000     because 31 and 101 are prime

Eulers Theorem a^{psi(n)} mod n =1     so:

 x^{3000m} mod (3131) = 1    Where m is an integer to give us more flexibility in generating an inverse in step XXX below.   Multiply both sides by x

 x^{3000m+1} mod (3131) = (x) mod (3131)

So if you can find a power d where:

 197d = 3000m + 1  or equivalently:

 (197d) mod(3000) = (1) mod(3000) then you can take the encrypted numbers to the power of d and out will pop the plain text original series

Now you use the Euclidean algorithm to find 1 in terms of 197 and 3000. This yields:

1 = 533(197) - 35(3000)     Taking the mod(3000) of both sides

(1)mod(3000) = (533)(197)  which shows d=533   is the decryption power we are looking for

This example used smaller numbers so as to make the example transparent.  However if the input character were to be X=31 or 101 I think the system breaks down.

Introduction for Cracking using OLLYDBG

Monday, January 30th, 2017

How to hack a Remote computer using Darkcomet(RAT) – Remote Access Tool

Monday, January 30th, 2017

Kali Linux Notes

Monday, January 30th, 2017

Research Links

Ulterior States – IamSatoshi Documentary

Sunday, January 29th, 2017