### Ruslan Ciurca

Shared publicly -https://play.google.com/store/apps/details?id=com.actionsmicro.ezcast this app can do all that via chromecast ... a bit lag(ish) but looks cool ...

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Ruslan Ciurca

Lives in UK

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https://play.google.com/store/apps/details?id=com.actionsmicro.ezcast this app can do all that via chromecast ... a bit lag(ish) but looks cool ...

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Dan Guja

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Inca una: https://github.com/tacitknowledge/flip

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Oracle Java on Raspberry Pi: http://ow.ly/pkCkt We have now added the official hard-float Oracle Java 7 JDK to our repository, so if you're a fan, do check Eben's post out for more details.

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I guess there are many interesting open source projects already written in Java ;)

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This is the second post dedicated to the problems set posted on "Math, Math Education, Math Culture" LinkedIn group. Here is the original LinkedId discussion, again ... if you happen to have a LinkedIn account. Here is the pr...

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Very interesting. Almost certainly some government ss are behind that. Searching for the best crypto analysts...

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Download the world's first and only report focused purely on enterprise mobility cellular data and trends. You will learn:

Whether user behavior really changes when roaming

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Whether user behavior really changes when roaming

The average time employees spend on cellular per day

What sites/apps are hogging the most cellular data

The craziest bill shock stories we have heard in Q3

Download now: http://wandera.com/resources/mobilityreport3Q13/

#mobiledata #enterprisemobility

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NSA 'subverted random-number code' http://www.bbc.co.uk/news/technology-24048343

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Does anyone see the pattern?

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Ruslan Ciurca

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Another good one is ... if you look at the pattern in between 7 and 11, 13 and 17, 19 and 23 ... basically any consecutive primes p[k] and p[k+1] where p[k+1] - p[k] = 4. It happens that 2 | p[k+1] + 1, 3 | p[k+1] + 2 and 2 | p[k+1] + 3 and either 4 | p[k+1] + 1 or 4 | p[k+1] + 3 - easy to prove considering that any prime has one of the following form p=4*m + 1 or p=4*m + 3 and any 3 consecutive numbers contain a number divisible by 3

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Moldova

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