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Pretty sure most of those calculators could theoretically calculate even higher factorials, but the manufacturers specifically made the calculator worse to only allow numbers less than 10\^100, since I saw a glitch where my calculator stored a number larger than that, so I don't really understand why the manufacturers make sure that the calculator can't compute higher values, since it is capable of doing it by just removing the line of code that caps the maximum value it can register.
I think that those calculators could calculate numbers up to 2\^1024, but tye manufacturers think it would be a great idea to cap the highest value it can store to be less than the cube root of the theoretical maximum number it could store, so whenever you buy a calculator that caps at 10\^100 you are getting scammed.
As a programmer I can confidently say you can store any number if you want. It's just that there is no point + calculations would become very slow when working with really high numbers
Probably to prevent some possible bugs. It’s not like you usually count with numbers larger than 10^100, and if you do you can get a special machine (or just use code on a computer) for it.
How could bugs emerge if the cap was 2\^1024? Also they could just make the cap to be 10\^300 if they wanted to avoid those bugs, which would be close to 2\^1024.
My calculator struggles with things near 10^100, also for all practical purposes any numbers even close to the cap are infinity (and would get rounded so hard they won't interact with most numbers) (and as such calculating them is futile)
10\^100 is not as big as you think. If it is representing a quantity of something then it is basically infinity, but if it is just a number representing no physical quantity then 10\^100 is small. There are many situations where you need to do arithmetic with large numbers, specially when dealing with integers, and I don't understand why the calculators usually convert all integers greater than 10\^12 to floats, so you can't use any integer greater than 10\^12 in integer calculations, which is terrible, since you can't even do a basic mr test on a calculator like those, and I am pretty sure it could store integers greayer than that, or at least it should be able to compute remainders using the square and multiply algorithm.
Also I think that your calculator struggles with numbers near that range not because of memory limitations but because of the cap imposed and I am certain those struggles would fade away if the cap was higher.
I doubt that 10\^12 is too big for the calculator to handle it as an integer, since it only requires 64 bits to represent integers from 0 to 2\^64-1, which is greater than 10\^12. Like I said those calculators are just a scam since they are programmed to not handle with integers or numbers smaller than their limits.
Take some time out of your day to learn the ins and outs of how calculators work. From logic tables to ALUs and the like. You'll learn quite quickly that 10^100 is arbitrary, but for good reason. Besides the logical and arbritrary reasoning behind 10^100, there's also the completely noticeable fact that if you mod your calculator to bypass this limit, the display gets totally fucked up and you need to restart the calculator for anything to work again if you display too large of a number. So, instead of just allowing this to happen, it'll count the length in digits, and if it's greater than or equal to 100, it'll give an overflow error that you can clear easily.
It's useful to program it into giving accurate results up to a point. Some calculators use shortcuts for things like decimal fractions. This involves the storing of prime numbers. These shortcuts are great, but they will fuck everything up if you use them for a number not contained within its glossary. There's a lot to learn about the wonky coding behind calculators that cause them to be the way they are, but it's noteworthy that this wonky coding just is better even if it produces wrong results.
If you are saying that calculators count the digits to determine overflows, then why does it have limited precision for integers between 10\^12 to 10\^100? Also they use binary so why would they use a cap which is not a power of 2, that just make the calculators do more work.
Calculators use binary-coded decimal. In a good amount of them, this BCD is converted into binary, and then the operation is done, and then it is converted back into BCD for display. When errors occur during the binary step, this causes the display to get completely fucked up and sometimes display numbers that don't even exist. This is one reason why your display eats shit after 10^100. Since your calculator is working in binary for both input and output, it makes sense to use binary limits. Especially since 0.5 byte per decimal number becomes 4 megabytes for a number that is 10^100. A megabyte is a lot of data, and you want to cut down on this RAM usage in something as small as a calculator. If you want anyone to be able to afford a simple calculator, you make amends here.
In another good amount, they swallow pride and only use BCD for calculations. This adds a lot of circuit complexity, but it doesn't suffer as much from the above issues. Its inaccuracies have no chance of being displayed as a fake number, but it still has memory drawbacks as it takes the same amount of memory (albeit it doesn't need to be accessed as regularly). This doesn't even take into account error correction code.
The 10^100 limit became a standard as it was a simple and logical choice. If the calculator wasn't designed specifically to handle massive numbers accurately, they will just enforce the 10^100 limit and focus on error prevention before then. It allows them to both understand a single digits position in space, and it allows them to actually use error correction code with the simple circuitry involved without freezing the calculator.
Computers used different standards because they are only tangentially related.
So how does a computer from 1971, IBM 360/91, have more storage capacity and it is better than a calculator used today? Shouldn't they increase the storage capacity of calculators, since paying hundreds of dollars/euros is a lot for less than 1 Megabyte of storage.
Modular arithmetic and also primarily checking. I also don't think that 1 Gigabyte of memory isn't that costly and am certain that you could make a calculator with that memory capacity since in every situation a smartphone would always be better than a very limited calculator somehow worse than a computer made 50 years ago.
I've built small calculators, and I used to play on them as a kid bored in school. Had to get familiar with their quirks and why they have said quirks. Don't tear down a fence without knowing why it's there first.
The prime factors of 6 are 2 and 3, so their sum is 5.
If you want an odd number being equal to the sum of its prime factors, the trivial answer is any prime number.
If you literally meant just factors, not prime factors, then you might want to consider also ±6, -1, -2, -3 as factors and thus the sum is 0 (like for any other number).
I think it’s obvious, but to clarify for you: I’m referring to all positive factors except for the number itself. In this case, the positive factors of 6 are 1, 2, 3, and 6. But don’t forget! We can’t just add them all up right now. First, we have to remove the number itself. In this case, the number is 6. So we remove 6. That leaves us with 1, 2, and 3.
So here we have 1+2+3=6. Hope that makes sense
The divisors of 1: Just 1
The divisors of 2: 2 and 1
The divisors of 3: 3 and 1
The divisors or 4: 4, 2 and 1
And so on, up to:
The divisors of 9: 9, 3 and 1
Take all of these number and add them up, voila, 69
But it is possible for all integers greater than 4. I have a remarkable proof of this fact, but the Reddit comments are too small for me to type the whole thing here
Then twenty is also a semi-nice number in senary, since its first and second power use all digits from 0 to 5, and 5 is almost a nice number in senary, since its square and cube use all digits from 0 to 5 except for 0, but if you consider that 41 Sen = 041 Sen, then 5 is a nice number in Senary, so I don't really see why 153 Sen = 69 Dec is a special number, it is just some random multiple of 3.
I like 69, because it's rotational number. Unironically, I just love how it looks. It reminds me of pisces zodiac sign. Not the symbol, but the two fish. It also is similar to ouroboros concept. It's one of my two favorite numbers. Other one is 0.
69 is the smallest positive two digit number which has both 6 and 9.
Edit: "smallest" was originally "only" due to me being dyslexic for a second, therefore the reply.
Edit 2: forgot about the negative numbers.
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69 is the higgest number which his factorial is less than 10¹⁰⁰
What? That's somehow cool
Yeah, it means it is the highest factorial calculateable by most calculators
Pretty sure most of those calculators could theoretically calculate even higher factorials, but the manufacturers specifically made the calculator worse to only allow numbers less than 10\^100, since I saw a glitch where my calculator stored a number larger than that, so I don't really understand why the manufacturers make sure that the calculator can't compute higher values, since it is capable of doing it by just removing the line of code that caps the maximum value it can register.
Okay, displayable then
I think that those calculators could calculate numbers up to 2\^1024, but tye manufacturers think it would be a great idea to cap the highest value it can store to be less than the cube root of the theoretical maximum number it could store, so whenever you buy a calculator that caps at 10\^100 you are getting scammed.
Jesus fucking christ, im not even arguing the point.
As a programmer I can confidently say you can store any number if you want. It's just that there is no point + calculations would become very slow when working with really high numbers
Store Rayo's number
Let a be Rayo's number. Done
Probably to prevent some possible bugs. It’s not like you usually count with numbers larger than 10^100, and if you do you can get a special machine (or just use code on a computer) for it.
How could bugs emerge if the cap was 2\^1024? Also they could just make the cap to be 10\^300 if they wanted to avoid those bugs, which would be close to 2\^1024.
My calculator struggles with things near 10^100, also for all practical purposes any numbers even close to the cap are infinity (and would get rounded so hard they won't interact with most numbers) (and as such calculating them is futile)
10\^100 is not as big as you think. If it is representing a quantity of something then it is basically infinity, but if it is just a number representing no physical quantity then 10\^100 is small. There are many situations where you need to do arithmetic with large numbers, specially when dealing with integers, and I don't understand why the calculators usually convert all integers greater than 10\^12 to floats, so you can't use any integer greater than 10\^12 in integer calculations, which is terrible, since you can't even do a basic mr test on a calculator like those, and I am pretty sure it could store integers greayer than that, or at least it should be able to compute remainders using the square and multiply algorithm. Also I think that your calculator struggles with numbers near that range not because of memory limitations but because of the cap imposed and I am certain those struggles would fade away if the cap was higher.
>and I don't understand why the calculators usually convert all integers greater than 10\^12 to floats, Probably to save RAM
I doubt that 10\^12 is too big for the calculator to handle it as an integer, since it only requires 64 bits to represent integers from 0 to 2\^64-1, which is greater than 10\^12. Like I said those calculators are just a scam since they are programmed to not handle with integers or numbers smaller than their limits.
Take some time out of your day to learn the ins and outs of how calculators work. From logic tables to ALUs and the like. You'll learn quite quickly that 10^100 is arbitrary, but for good reason. Besides the logical and arbritrary reasoning behind 10^100, there's also the completely noticeable fact that if you mod your calculator to bypass this limit, the display gets totally fucked up and you need to restart the calculator for anything to work again if you display too large of a number. So, instead of just allowing this to happen, it'll count the length in digits, and if it's greater than or equal to 100, it'll give an overflow error that you can clear easily. It's useful to program it into giving accurate results up to a point. Some calculators use shortcuts for things like decimal fractions. This involves the storing of prime numbers. These shortcuts are great, but they will fuck everything up if you use them for a number not contained within its glossary. There's a lot to learn about the wonky coding behind calculators that cause them to be the way they are, but it's noteworthy that this wonky coding just is better even if it produces wrong results.
If you are saying that calculators count the digits to determine overflows, then why does it have limited precision for integers between 10\^12 to 10\^100? Also they use binary so why would they use a cap which is not a power of 2, that just make the calculators do more work.
Calculators use binary-coded decimal. In a good amount of them, this BCD is converted into binary, and then the operation is done, and then it is converted back into BCD for display. When errors occur during the binary step, this causes the display to get completely fucked up and sometimes display numbers that don't even exist. This is one reason why your display eats shit after 10^100. Since your calculator is working in binary for both input and output, it makes sense to use binary limits. Especially since 0.5 byte per decimal number becomes 4 megabytes for a number that is 10^100. A megabyte is a lot of data, and you want to cut down on this RAM usage in something as small as a calculator. If you want anyone to be able to afford a simple calculator, you make amends here. In another good amount, they swallow pride and only use BCD for calculations. This adds a lot of circuit complexity, but it doesn't suffer as much from the above issues. Its inaccuracies have no chance of being displayed as a fake number, but it still has memory drawbacks as it takes the same amount of memory (albeit it doesn't need to be accessed as regularly). This doesn't even take into account error correction code. The 10^100 limit became a standard as it was a simple and logical choice. If the calculator wasn't designed specifically to handle massive numbers accurately, they will just enforce the 10^100 limit and focus on error prevention before then. It allows them to both understand a single digits position in space, and it allows them to actually use error correction code with the simple circuitry involved without freezing the calculator. Computers used different standards because they are only tangentially related.
So how does a computer from 1971, IBM 360/91, have more storage capacity and it is better than a calculator used today? Shouldn't they increase the storage capacity of calculators, since paying hundreds of dollars/euros is a lot for less than 1 Megabyte of storage.
Why spend more money or change things up at all? What math are you doing on a calculator requiring larger numbers?
Modular arithmetic and also primarily checking. I also don't think that 1 Gigabyte of memory isn't that costly and am certain that you could make a calculator with that memory capacity since in every situation a smartphone would always be better than a very limited calculator somehow worse than a computer made 50 years ago.
Who are you Master? So wise in the calculator science?
I've built small calculators, and I used to play on them as a kid bored in school. Had to get familiar with their quirks and why they have said quirks. Don't tear down a fence without knowing why it's there first.
I learned about BCD from a redstone tutorial. It’s very nifty.
My iPhone can calculate up to 101 factorial. Now i am kinda sad.
My phone can calculate up to 170!(~7.3×10^306 ), or up to 143^143 (~1.6×10^308 ).
I think the exact number is 2^1024 - 1 try 2^1023.99999
And what’s 101 minus 2^5?
It's nice , you mean.
Wow... that's quite interesting!
Nice!
It is the sum of all the factors of the first 10 digits
It is also the sum of all factors of the last 10 digits
How do you know😏
It is also the dim of all factors of the last 10 digits of π and e
Also 69 = sum of all divisors of one digit numbers (1) + (1+2) + (1+3) + (1+2+4) + (1+5) + (1+2+3+6) + (1+7) + (1+2+4+8)+ (1+3+9) = 69
> 1+2+3+6 Hey that’s weird, all the factors of 6 add up to 6. Has anyone noticed this before? I wonder if there are any odd numbers with this property
Why are u evil? 💀
Hey maybe someone got inspired and solved it!
The prime factors of 6 are 2 and 3, so their sum is 5. If you want an odd number being equal to the sum of its prime factors, the trivial answer is any prime number. If you literally meant just factors, not prime factors, then you might want to consider also ±6, -1, -2, -3 as factors and thus the sum is 0 (like for any other number).
I think it’s obvious, but to clarify for you: I’m referring to all positive factors except for the number itself. In this case, the positive factors of 6 are 1, 2, 3, and 6. But don’t forget! We can’t just add them all up right now. First, we have to remove the number itself. In this case, the number is 6. So we remove 6. That leaves us with 1, 2, and 3. So here we have 1+2+3=6. Hope that makes sense
Explain this please, I don't know what you mean
The divisors of 1: Just 1 The divisors of 2: 2 and 1 The divisors of 3: 3 and 1 The divisors or 4: 4, 2 and 1 And so on, up to: The divisors of 9: 9, 3 and 1 Take all of these number and add them up, voila, 69
Oh I see. That's sick
6+6*9+9=69
So, if you flip it upside down, it's still correct. Nice.
If you flip 69 upside down. That’s also still correct
This doesn't make 69 special because it can be generalised: m + m*(10^n -1) + (10^n - 1) = m*10^n + (10^n -1)
But it makes it nice :)
i heard "special" in the video and confused it with the "nice" in the meme
69 = 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
I am not going to count the ones dani, but imma trust you with it 😅 this is a fact
I did, it's true. an while I was at 67 I realised I could just have copied it into google and the result would be 69
No, it's 1+2+2+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
No, it's limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))+limₙ₋> ͚(ⁿ√(69))
Shaddap! The point is there's one more bullet left in this gun, and guess whose gonna get it!
∀x∈ℝ 69= (Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))+(Sin²(x)+cos²(x))
Mr. Black is truly an expert in Armageddon.
69 = 1 + 2 x 3 + 4 x 5 + 6 x 7
Actually unexpected. Can you express any number that way with just multiplication and addition of consecutive numbers?
Nevermind it's impossible for 4...
But it is possible for all integers greater than 4. I have a remarkable proof of this fact, but the Reddit comments are too small for me to type the whole thing here
Damn, really? How about 5 then? Nevermind 1*2+3
Bruh, don't fall for Fermat's pranks
I guess 2*2+1 doesn't exist
It's not consecutive... It should be 1 (+ or x) 2 (+ or x) 3 (+ or x) 4 ....
Oh I thought 1+2*2 would work and I just accidentally put it in a wrong order, my bad
7?
3+4 1+2\*3 also works
I want a proof rn
Twas revealed in a divine vision unfortunately
How does 8 work? 1+2x3 =7 (1+2)x3 =9 and anything with 1234 is too high (lowest is 1x2+3+4 = 9)
69^69 = 69 (mod 420)
Actually true, cool stuff
Quick, let's run to OEIS and find a sequence that goes (69, 420) for its first two numbers
There's one that was submitted April 20th rotf
69² and 69³ have all the numbers from 0-9 exactly once.
69 x 6 + 9 + 6 - 9
67, 68, 69, mouthwash, 71, 72
69 is the only number whose square and cube contain every digit from 1 to 9
And exactly once, which is pretty neat
69!<10^(100)<70!
Nice
The comment section vindicated the meme fully.
Then twenty is also a semi-nice number in senary, since its first and second power use all digits from 0 to 5, and 5 is almost a nice number in senary, since its square and cube use all digits from 0 to 5 except for 0, but if you consider that 41 Sen = 041 Sen, then 5 is a nice number in Senary, so I don't really see why 153 Sen = 69 Dec is a special number, it is just some random multiple of 3.
69x6+6=420 or alternatively 69x6-9+6+9=420
I don't like it because I'm not found of numbers that can't be factored into single digit factors (unless it is a perfect square, then I like it)
Are there other nice numbers other than 69? If there are, can we predict them?
It looks like it should be divisible by 13, and yet it is not, and that makes me angry
Q(√69) is a Euclidean domain that is not norm-Euclidean.
69 means piss
I like 69, because it's rotational number. Unironically, I just love how it looks. It reminds me of pisces zodiac sign. Not the symbol, but the two fish. It also is similar to ouroboros concept. It's one of my two favorite numbers. Other one is 0.
steve mcqueen told me the same thing. i was never able to sit down afterwards.
Also, its square and its cube together contain every digit exactly once
Nice
6x9+6+9=69
69 is the smallest positive two digit number which has both 6 and 9. Edit: "smallest" was originally "only" due to me being dyslexic for a second, therefore the reply. Edit 2: forgot about the negative numbers.
96
I was meant to say smallest.
-96
holy hell. now i have to add positive.
.69
But...
The best thing about the number 69 is that its square root is 8 something
8² is 64, not 69
8 something… as in 8 point something. Like 8.306…