Congratulations, you just reinvented a shittier version of a black box. But I'm sure your idea is way more viable than military grade hardware gone through many iterations over decades to produce one of the most durable recording devices known to man. If we can't locate a black box with a freakin' transponder signal blasting out of it, what makes you think we're going to be able to locate your little LED light orange styrofoam ball?
I suppose the only fair thing to do would be to re-label the pro-choice groups as "pro-death" or "anti-life" just to keep the inflammatory rhetoric to a maximum.
Notable security professionals usually can line up a job in the country of their choice and being highly skilled in a necessary area will very often speed along the process. Most every country is looking for talented IT workers and the demand is only apt to increase.
It would be like the lottery if buying every ticket combination in that lottery guaranteed a minimum 10% net profit. With VC stuff you anticipate a 90% failure rate but with an approximate 2000% ROI on the one success out of ten.
You're not missing it. 2 and 5 themselves (in decimal, or any base where they are factors of the base) are the only prime numbers ending in 2 and 5. So while GP is correct in a logical sense (there are at least a billion numbers afterwards ending in neither 2 nor 5 that are also prime) there are, in fact, infinite numbers afterwards ending in neither 2 nor 5.
This is not true in base-16 for 5's however (0x25 = 37 is prime for example), but remains true for 2's. It's completely false for base-7 though (base-7(32) = 23, base-7(25) = 19). So...yeah. I'll stop there. Further playing around with bases is left as an exercise to the reader.
There are very few numerical properties that are base-dependent.
Some of the little tricks they teach you in school are strictly base-dependent, like if a decimal number ends in 5 or 0 it's divisible by 5 or 10 respectively. If a decimal number ends in a value divisible by 2 it's even else odd. Or if a decimal number's digits sum to a multiple of 3 or 9 then it's divisible by 3 or 9 respectively.
What they don't tell you is that is generalizable to other bases. Generically speaking if the final digit of a number in a given base is divisible by any factor of that base then the number itself is divisible by that factor (this should be fairly obvious) and if the digits of a number sum to a number divisible by a factor of (base-1) then that number itself is divisible by that factor (less obvious, but provable).
So for hex, for example, the factors of 16 are 2, 4, 8, 16. If a number in base-16 ends in 0 it's obviously divisible by 16, if it ends in 8 then it's divisible by 8 and so on. The factors of (16-1)=15 are 3, 5, and 15. So if the sum of digits of a hex number are divisible by 3, 5, or 15 then the number is also divisible by 3, 5, or 15 respectively as well.
I'm sure it's been considered but at least from a programming perspective I'd be more concerned about the latency on the port as regards the ability to process realtime high framerate graphics through there. When I was doing CUDA programming the most difficult (that is, time consuming) part was getting data from main memory to the graphics card. Would the Thunderbolt interface be as fast at shuttling data from main memory to an external graphics card? 40Gbps is great and all, but is the latency low enough?
Combinatorics is extraordinarily abstract. It's a senior level college math class at best, and I actually took a year of it in graduate school. I mean yeah, it's nice for explaining why the lottery is a scam, but it's very difficult for average folks to grasp. It doesn't feel intuitive at all.
It's not that simple. The drive isn't protected by a passcode, the *decryption key* is protected by a passcode. The drive is protected by encryption. Without the key it's basically just a bunch of random gibberish.
The iPhone's flash drive is encrypted. The key is securely stored. If you guess the lock code incorrectly 10 times then it's not the hard drive that's erased, it's the key that is irrevocably destroyed. At that point it doesn't matter if you have a bunch of copies of the disk, you have a bunch of garbage and the only key in the universe was just wiped out.
Even with MDM solutions you can't unlock the phone. You can *wipe* it, and if you control the email for the iCloud account you can even restore it to factory default and reuse the hardware. But you can't just open it and see what's inside.
Ideally Apple will change this and allow an option to lock down recovery mode, maybe with the option of "if you forgot you pin, you can erase everything first," and refuse to load even signed software until the pin is entered or the device is erased.
My understanding is that they did change this in iPhone 5S and above models. The vulnerability only exists in this form on the 5C and regular 5 models. The terrorist's phone was specifically a 5C. The downside is if you comply with this order then it's possible to crack every single 5C out there with this method and that's a horrendous precedent to set.
Apple owns the endpoint. It is very difficult (though not theoretically impossible) to guarantee security when what you considered a trusted endpoint is compromised. It's like if you're Alice sending a message to Bob but now Bob has turned state's evidence on you. You're boned.
Not the phone. The iCloud account. They had access to the terrorists' email address and initiated a password reset in hopes of gaining access to the phone. That didn't work.
Well now there's an interesting thought. I wonder if Apple could just reconfigure his iCloud account to accept whatever password the phone provides as valid and accept the backup from there...? It would be like a man-in-the-middle attack except Apple is itself a trusted endpoint so maybe they could futz with the iCloud security to allow this phone to backup to it. Could Apple spoof an endpoint that way?
It's totally cool. Just like the FBI wants a backdoor for iPhones they can use this backdoor for surveillance systems! I'm sure nothing bad will ever happen from having this backdoor in place!
Well, unless you're shooting that off into deep space the sun will forcibly exceed the temperature tolerances for all such discs held on Earth after about another 4-5 billion years or so...
You can't download the 6u113 update unless you have a support contract with Oracle. Without one the latest version you can get in Java 6 is 6u45, from 2013, when it officially went end of life.
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Congratulations, you just reinvented a shittier version of a black box. But I'm sure your idea is way more viable than military grade hardware gone through many iterations over decades to produce one of the most durable recording devices known to man. If we can't locate a black box with a freakin' transponder signal blasting out of it, what makes you think we're going to be able to locate your little LED light orange styrofoam ball?
I suppose the only fair thing to do would be to re-label the pro-choice groups as "pro-death" or "anti-life" just to keep the inflammatory rhetoric to a maximum.
Notable security professionals usually can line up a job in the country of their choice and being highly skilled in a necessary area will very often speed along the process. Most every country is looking for talented IT workers and the demand is only apt to increase.
I'm thinking Spike more likely.
... I hear the passengers thought he was with the Al-Gebra network - and he was holding potential weapons of math instruction.
*slow clap*
Autorun has changed as of Windows 7. Non-optical media can no longer auto-start a program on the media.
That was some epically beautiful nerdiness right there.
That attack vector can be blocked by squirting epoxy into the USB ports.
It would be like the lottery if buying every ticket combination in that lottery guaranteed a minimum 10% net profit. With VC stuff you anticipate a 90% failure rate but with an approximate 2000% ROI on the one success out of ten.
01 is not a prime.
Just like in base-10 the number 1 is not considered a prime.
You're not missing it. 2 and 5 themselves (in decimal, or any base where they are factors of the base) are the only prime numbers ending in 2 and 5. So while GP is correct in a logical sense (there are at least a billion numbers afterwards ending in neither 2 nor 5 that are also prime) there are, in fact, infinite numbers afterwards ending in neither 2 nor 5.
This is not true in base-16 for 5's however (0x25 = 37 is prime for example), but remains true for 2's. It's completely false for base-7 though (base-7(32) = 23, base-7(25) = 19). So...yeah. I'll stop there. Further playing around with bases is left as an exercise to the reader.
There are very few numerical properties that are base-dependent.
Some of the little tricks they teach you in school are strictly base-dependent, like if a decimal number ends in 5 or 0 it's divisible by 5 or 10 respectively. If a decimal number ends in a value divisible by 2 it's even else odd. Or if a decimal number's digits sum to a multiple of 3 or 9 then it's divisible by 3 or 9 respectively.
What they don't tell you is that is generalizable to other bases. Generically speaking if the final digit of a number in a given base is divisible by any factor of that base then the number itself is divisible by that factor (this should be fairly obvious) and if the digits of a number sum to a number divisible by a factor of (base-1) then that number itself is divisible by that factor (less obvious, but provable).
So for hex, for example, the factors of 16 are 2, 4, 8, 16. If a number in base-16 ends in 0 it's obviously divisible by 16, if it ends in 8 then it's divisible by 8 and so on. The factors of (16-1)=15 are 3, 5, and 15. So if the sum of digits of a hex number are divisible by 3, 5, or 15 then the number is also divisible by 3, 5, or 15 respectively as well.
Fun little math quirks on bases.
The twin prime conjecture is independent of the base, so the base doesn't matter for it to be true or false.
I would find this surprising, since in a base 2 system every prime number ending in 1 is followed by a prime number ending in 1.
In base 2, every prime ending in 0 is also followed by a prime ending in 1.
I'm sure it's been considered but at least from a programming perspective I'd be more concerned about the latency on the port as regards the ability to process realtime high framerate graphics through there. When I was doing CUDA programming the most difficult (that is, time consuming) part was getting data from main memory to the graphics card. Would the Thunderbolt interface be as fast at shuttling data from main memory to an external graphics card? 40Gbps is great and all, but is the latency low enough?
Combinatorics is extraordinarily abstract. It's a senior level college math class at best, and I actually took a year of it in graduate school. I mean yeah, it's nice for explaining why the lottery is a scam, but it's very difficult for average folks to grasp. It doesn't feel intuitive at all.
It's not that simple. The drive isn't protected by a passcode, the *decryption key* is protected by a passcode. The drive is protected by encryption. Without the key it's basically just a bunch of random gibberish.
The iPhone's flash drive is encrypted. The key is securely stored. If you guess the lock code incorrectly 10 times then it's not the hard drive that's erased, it's the key that is irrevocably destroyed. At that point it doesn't matter if you have a bunch of copies of the disk, you have a bunch of garbage and the only key in the universe was just wiped out.
Even with MDM solutions you can't unlock the phone. You can *wipe* it, and if you control the email for the iCloud account you can even restore it to factory default and reuse the hardware. But you can't just open it and see what's inside.
Ideally Apple will change this and allow an option to lock down recovery mode, maybe with the option of "if you forgot you pin, you can erase everything first," and refuse to load even signed software until the pin is entered or the device is erased.
My understanding is that they did change this in iPhone 5S and above models. The vulnerability only exists in this form on the 5C and regular 5 models. The terrorist's phone was specifically a 5C. The downside is if you comply with this order then it's possible to crack every single 5C out there with this method and that's a horrendous precedent to set.
Apple owns the endpoint. It is very difficult (though not theoretically impossible) to guarantee security when what you considered a trusted endpoint is compromised. It's like if you're Alice sending a message to Bob but now Bob has turned state's evidence on you. You're boned.
Not the phone. The iCloud account. They had access to the terrorists' email address and initiated a password reset in hopes of gaining access to the phone. That didn't work.
Well now there's an interesting thought. I wonder if Apple could just reconfigure his iCloud account to accept whatever password the phone provides as valid and accept the backup from there...? It would be like a man-in-the-middle attack except Apple is itself a trusted endpoint so maybe they could futz with the iCloud security to allow this phone to backup to it. Could Apple spoof an endpoint that way?
It's totally cool. Just like the FBI wants a backdoor for iPhones they can use this backdoor for surveillance systems! I'm sure nothing bad will ever happen from having this backdoor in place!
Well, unless you're shooting that off into deep space the sun will forcibly exceed the temperature tolerances for all such discs held on Earth after about another 4-5 billion years or so...
You can't download the 6u113 update unless you have a support contract with Oracle. Without one the latest version you can get in Java 6 is 6u45, from 2013, when it officially went end of life.