@@The_Real_Sensei A well known value is 2^16 = 65536. Or more inaccurately, it is well known to those of us who played with 8 bit micros, where 65536 = 64 old kb which is the maximum amount of memory an 8 bit processor with a 16 bit address bus (eg 6502, 6800, Z80, 8080) could access (directly).
@@The_Real_Sensei Last night we were catching up on an Only Connect episode from 5 or 6 weeks ago. One sequence question was: What comes 4th in the sequence: 65536256, 25616, 164 The answer was 42 as each "number" is made up of a number and its square root, the square root of one starting the next. The first "number" 65536256 is 65536 (2^16) = 256^2 ((2^8)^2).
2 to the 18th is 2 to the 16th times 4 (or 2 squared). Anyone with a computer brain knows 2 to the 16th is 65536 (good old computer memory and bits). So times 4 is 262144, subtract 1 = 262143. Everyone should be able to do this in the modern world in their head.
Wrong! You wrote a stupid comment. Everyone should *not* able to do this in the modern world in their head! The large majority should not be able to do it. Get educated about people and don't make foolish posts such as this.
@@theyassinez1 Heresy! Blasphemer! Could not resist lol. I am old school, even know how to use a slide rule lol. It's amazing sometimes describing to some kids these days how the 8 bits are used in all the ASCII font tables. I love flipping a set of text into Ascii on them when they can't find an error and point out how that spot is not a 40 hence it's not a space, it's something else that merely displays as space like which is causing the data issue. They never think of that. They are so visual these days so dependent on all the code we wrote in the past to do everything for them. Sigh.
Children of the 8 bit era certainly know all the powers of 2 up to 16; So if you already know that 2^16 is 65536 you may as well go for the direct route. Either double it twice, or multiple by 4 (which ever you find easiest) and subtract 1.
@@vdamky Technically we actually move the digits through the place value columns left (multiply) or right (divide). However, on paper it is easier to move the _decimal point_ than to shift the digits as we implicitly know the place value columns and it is the decimal point that is visible[1] which tells us where the place value columns apply to the number. [1] If the decimal point is not visible (as the number is an integer) then it is taken to be after the last digit which is in the units place value column.
@@samueldeandrade8535 the thing is, in an exam, its completely impractical. You don't time to waste. yes i might use it as a last resort but doing this method is much faster.
Since there are many similar problems, most people probably know how to calculate them easily without making mistakes. You need not stick to the factorization formula. There are also concrete examples in this comment section. As a solution method, (1) decompose them into round numbers. That is, 2¹⁰ =1024=1000+24, 2⁹ =512=500+12, 2⁸ =256=250+6. (2) subtraction should be avoided, and addition should not carry over (no overlap of the same digits). It is unavoidable that there is a single overlap in this problem.
@@davidbornstein9197 You are right. But than the essence of this problem is brute calculation. There is nothing elegant about it. Besides, the power to which a person can calculate an exponent is subjective. Mathematic solutions are most powerful when the method minimize or completely remove the need for brute force calculations.
You're in the internet land. You should know all the powers of 2, up until 10. Because 1024 is an important round number in the world of computer science. :-)
In 2^n, for n = 1, unit place digit is 2 For n = 2, unit palce is 4 For n = 3, unit place is 8 For n = 4, unit place is 6 This cycle of 2, 4, 8, 6 will repeat for every power of form 4m+1, 4m+2, 4m+3, 4m+4. So if n = 18 it is of form 4m+2 so units place is 4, and subtracting 1 from it will give 3.
With the base numberof 2, it is easy to double the first at least10 - 12 times. And with pen and paper, you have no problem reaching 18. This is easily grinded out without fancy math shortcuts :)
It's Math in Norway - not in China or USA or Russia. Their approach is like - "choose the only topics in science you want to learn because we think about you mental health in childhood". Study in China or Russia: "Ok, it's 23:00 pm - you can go to sleep for 3 hours. We will continue tomorrow at 5pm. That's why these counties win Olympiads 😂
Some nice tricks, but what about, in this case, just calculating the thing? 2^18 = 2^10*2^8=1024*256=256000 + 24*256. The latter is 25*2^8-256=100*2^6-256=6144. So we get 262144; subtract one and you get 262143, with the "25 and we can make that 100, it's a power of 2" happening in one's head (if you don't get that, you can also easily calculate 1024*256 simply on paper). You would, of course, know the first 10 powers of 2 by heart (they teach that in school, just like the first 20 squares)... but then the solution represented also relies on the fact that you now 2^9=512 by heart. But then, squeezing in a binomian formular was rather beautiful. (I do mean that.)
Well ... first question should be, what kind of result is expected. Because 2^18-1 is pretty nice. 11111..11 binary is also good. It seems, that normal writing multiplying 513*511 is also trivial. And if I use pretty known fact, that 2^16=65536 ? Two addition and almost done.
Now, to chunk it in head, multiply (1000 x 256) + (2 x 10 x 256) + (4 x 256) - 1 If you practice chunking, you can do complex math REALLY fast in your head. You’ll also start to be able to memorize really long strings of numbers…like pi! As a math teacher, my students are often dazzled by this; it’s really VERY easy!
as a programmer, I have remembered many 2^x values without calculating them, for example, 2^24 is 16777216 and 2^16 is 65536. Oh yes, 2^18 is 262144, so if it is minused by 1😊
For those who are used to binary and computers, simply do 1024 (2^10) x 256 (2^10) = 262144, and subtract 1, or start at 65536 (2^16) and double it twice.
I did this in my head a little differently, but it could be done on paper the same way. Binary numbers are easy to work with and from computer experience, I know that 2^8 = 256, so 2^9 = 512. So, 512 is squared by (500 + 12) x(500 + 12), we know that (a + b)^2 = a^2 + 2ab + b^2. The rest is trivial, 500^2 = 250,000. 2 x 500 x 12 = 1,000 x 12 = 12,000, and 12 x 12 = 144. 250 + 12 = 262, so we have262,000 + 144 - 1 = 262,143. If you are not familiar with binary numbers, it is a simple matter to count up, 2, 4, 8,, 16, 32, 64, 128, 256, 512.
we might know 2^18 = 2^(6*3) as being the number of RGB colors you can encode when each Red, Green, Blue color component has 6bits resolution (64 levels) and that number is 262144 ... at least people should know 2^10 = 1024 and 2^16 = 65536 ;)
2^10 x 2^8 - 1 = 1024 x 256 - 1 = 262144 - 1 = 262143 OR 2^12 x 2^6 - 1 = 4096 x 64 - 1 = 262144 - 1 = 262143 I would think like that, because I'm used to calculate until 2^11, so 2^10 and 2^12 are easy. If I didn't feel confident multiplying 2 numbers with more than 3 algarisms, I can separate 18 in 12 and 6, because 2^6 is the last potence of 2 with 2 algarisms, so it's easier to calculate 4096 x 64. I wouldn't use factoration because it's not necessary, and my first thought was 1 = 2^0.
@@MyOneFiftiethOfADollar In your conception of God, he doesn't has a heart. And your belief shall be respected. In my conception (Gospel), Yeshua is the image of the invisible God and Yeshua has a heart, hence God has a heart. This is my belief and shall be respected. Also, this is a Math channel, better not talk about such things. (Blessing people is a common thing, to talk about god in a Math channel not.).
2^18=4^9=4x16^4= 4x256^2=4x(65,536) = 262,144. Subtract 1=262,143. Check video. Glad we agree, and glad it gives a numerical answer. not a logarithm to an obscure base.
I do not understand beauty of writing (512+1) and then explain that it is 513 verbally. Thanks God she did not come with a substitute formula for that.
If you remeber that 2^10=1024, 2^18 = 1024x1024/4. The division first will bring ...6 (divide 24 only by 4), then 6 multiplied by 1024 will bring 4 as the last digit. Can be done in the head like 5 seconds.
lol in India, we use an identity to do 511*513, it's basically, (x+a)(x+b)=x^2+(a+b)x+ab this eases out the calculation so you can do (500+11)(500+13) 500^2+ (11+13)500+ (11)(13) very easy calculations, you can do them in mind 250,000 + 24*500 + 143 = 250,000+12,000+143 =262,143 (we also have a trick to multiply numbers with 11 and rest of the calculations were pretty simple).
if you have to multiply 11 with 13 just write the first and last digit as it is, that is 1_3, now the middle digit is going to be the sum of the first and last digit, that is, 1 and 3, we know 1+3=4, so 143 is the ans to 11*13. It's a very easy trick, you can solve such problems instantly in seconds@@pacogutierrez2484
@@pacogutierrez2484 multiply the number by 10 and then add the number to it: example: 13 X 11 = 13*10 + 13 = 130 + 13 = 143; the idea is that multiplying with 10 is one of the easiest things to do and then addition is the next easy step - hope this helps
Probably the simplest and shortest way: 2^18-1=(2^9)^2-1=(512)^2-1 Applying Vedic Math: 512*512=524*5*100+144=262144 (which can mentally be calculated in about 15 sec) Therefore: 2^18-1=262144-1=262143
Not to me. Growing up in the 70s and 80s, I learnt 2^16=65536 (max amount of memory addressable by an 8-bit processor with a 16-bit address bus), and later 2^17=131072 (start location in memory of the first screen of the Sinclair QL). Thus 2^18-1 = 2^16×4 -1 = 65536×4 -1 = 262144 -1 = 262143 or = 2^17×2 -1 = 131072×2 -1 = 262144 -1 = 262143 (This latter being the easier of the two.)
@@cigmorfil4101 However, your solution work for this one particular exercice, his way works for every numbers simply. If i put you 7^6, having Sinclair QL memory benefit is useless However if i apply his Vedic Math : - 7^6 -1 = (7^3)^2 -1 = 343^2 -1 - 343*343 -1 = 386*3*100 +1 849 -1 = 1 158*100 +1849 -1 = 115 800 +1849 -1 = 117 649 -1 = 117 648 the addition is equal to the 43 of 343, squared, 43^2 = 1 849
Experts in computer hardware in 1990s to early 2000s should know 2^18=262,144 that's the amount of 256MB memory in KB that would show up during start up of a computer of Pentium II/III or Athlon era.
She has been trying to find the result of the 513*511 operation for exactly 2 minutes since 1:17 seconds of the video. If she writes these two numbers one under the other and multiplies them, she can get the result in 20 seconds.
It took me awhile, but I finally realized that this was quite an easy problem, and I thought that I might be able to do it in my head by simply doubling 18 times. Well, I didn't get that far on my first three attempts. but I now know that I can do it...eventually. How far did I get? Thirteen or fourteen times, but that was good enough for me. I now know that I can solve the problem in my head if I try long enough.
I just know - from years of doing computer science work - that 2^10 is 1024 and 2^8 is 256. 1024 * 256 is really easy, because 1, 2, and 4 are just doublings of 256 and adding them in different 10s places. then subtract 1.
I tired everything like G.P series Limits (calculus) Logarithms Binomial At the end all the results were in the power of 2 some even exceeding 2^18 😅 so its better to learn some power table as 2^10 always helps
Whenever you have x², you will get the same result if you multiply (x-t) by (x+t) and add t². Therefore, a way that I consider to be the fastest way to solve problems like this mentally is to look for a convenient (x-t) (or (x+t)). In the case, for example, to solve 512², the convenient (x-t) is 500. Therefore, the (x+t) is 524, and the t is 12. Therefore, 512²=500*524+12². Knowing that 524*1000 is 524000, just divide this by 2 to get 262000. Then, knowing that 12² is 144, we have that 512² is 262144. Finally, just subtract 1 and get to the final answer to the problem.
Programmers who developed on the PDP-10 which had a 36 bit processor would know that 2^18 is 262144, 1000000 octal and 40000 hex. Especially useful when doing some register arithmetic on the left and right halves of the register in your head. I used to have to do that a lot at CompuServe in the 90's.
This one is hilarious. When I saw it I thought, surely you just solve it using arithmetic? But hey if there's a trick to working with exponents of 2 I've love to know it. No - you go through a convoluted process which gets you nowhere, then just solve using arithmetic anyway.
Eu fiz 17 contas, mas multiplicar por 2 é muito fácil e rápido, saiu bem mais rápido do que no vídeo, e de certa forma meus neurônios ainda estão intactos
For us that knows our powers of 2 it can be simplified as: ( 2^18) - 1 = ((2^16) * (2^2) ) - 1 = (65,536 * 4) - 1 = 262,144 - 1 = 262,143 Of course one could also solve it using logs or even by using a slide rule!
Am I the only person that just did the math in my head and calculated the same result? I understand the method behind that, but you can easily just calculate the result in less than 1 minute.
Зная степени двойки(а в Российских школах дети это знают, так же как и таблицу умножения) вычислить этот можно за минуту. 2^10*2^8 - 1= 1024*256 - 1= 262143. Привет из России!
Explain something to me. How people with such abilities, educated, can support invasion on a neighbouring country? Or wait, maybe it doesn't matter how well one's educated in math... Regards from Poland.
Instead of calculating 2^9, you could just use rules of exponents and rewrite it as 8^3. In my opinion, easier to calculate. This is only if you don’t know the powers of 2 up until 9 or above
It is surprising that back in the 80-90s of the last century, almost all schoolchildren (at least in my country) in grades 6-7 were able to multiply 513 by 511 without a calculator and expansion. I think that in 20 years, schoolchildren will not be able to even multiply 500 by 500 without gadgets.
30 seconds, 6 lines and 6 columns of digits, it's all this calculus takes yet the number of children/students who can't make it as simply as that is exponentiating, so I'm afraid your conclusion is right.
2^18 = 2^20 / 4 = 1048576 / 4 = 524288 / 2 = 262144. Actually, multiplying by 2 is very simple, so if you have a good trained memory, you can multiply by 2 in memory. Also if there is no rule that requires to provide answer in base 10, you can just write that (2^18 - 1) base 10 = (1000000000000000000 - 1) base 2 = 111111111111111111 base 2 😊
Its very simple you can solve it in seconds by a simple and no need of using formulas etc First method Power 18 Like 6*3 2 power 3 = 8 ( Now multiply 8 six times to get an answer) 8*8*8*8*8*8= 262,144-1= 262,143 Second method 2 power 18 6*3= 18 2 power 6 = 64 ( Now multiply 64 by three times to get an answer) 64*64*64 = 262,144-1=262,143
2power 12 calculate krna aasan hai Usk bad 2ki power 6 calculate kro. Aapas main simple multiply krk -1 krdo jo tarika aapne bstaya hai usse assan hoga (2^12×2^6)-1 (4056×64)-1 I have done this orally
Any old school computer scientist can answer it without by without even drawing on paper. 2^18 is close to 2^16, which is the limit of array size in an 16 bit OS. So its just 4x65536 - 1.
I do not see how this is more simple than 512x512-1. We all know 2^16=65536, so double it twice in a row and substract 1. More over 513x511 is quite obvious, 1x513 is obvious and everybody knows 13x5=65
I know a much easier solution for this. When I was a child I found out how can I square any number only using my head. So 512x512-1 can be calculated with the following method 512-12 is 500 and 512+12 is 524 so just to be able to multiply with a round number. 500*524+12*12-1 equals 250000 + 12000 + 144 - 1 which is the same result but easier and faster to solve your example.
For people from the C64-era this question is a very easy task, because they all know that 2^16 = 65536. So all they have to calculate is 65536 * 2 * 2 and then subtract 1. A matter of seconds.
Even without knowing powers of 2. It is very easy to produce them: 2^1: 2 2^2: 4 2^4: 16 2^8: 256 2^16: 65535 Then we use a combination of the above. 2^(16+2)=65535*4=262144 Subtract 1. This procedure scales quite well. Because its complexity is logarithmic. And we can easily combine any desired exponent. Proceeding with exponent 9 is also good. But it doesn't really bring advantages.
Why all the complicated brackets and splitting, adding, subtracting, changing powers? I don't understand any of it. Why doesn't she just go 2*2 18 times and the take one away? I got the answer that way by about 90 seconds and just needed my fingers to keep track of the multiples and my head to multiply by 2. Why make it so hard?
Thanks, but It isnt olimpic way to do It. The better way I Saw the resolution is: 2^18 = 2*9 * 2*9 = 512*512 = (500+12)(500+12) = 500² + 2*12*500 + 12² = 250.000 + 12.000 + 144. Subtracting -1, we have 262.143 without a lot of multiplications
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Much simpler to just calculate 2^18 first, then minus 1. I don't see why 511*513 is any easier to calculate than 512*512.
It's far much easier to use the fact that 2^17 is 131072 (address of start of screen memory in a Sinclair QL), double it and subtract 1...
Exactly what i was thinking 😊
How are you going to calculate 2^18 though?
@@The_Real_Sensei
A well known value is 2^16 = 65536.
Or more inaccurately, it is well known to those of us who played with 8 bit micros, where 65536 = 64 old kb which is the maximum amount of memory an 8 bit processor with a 16 bit address bus (eg 6502, 6800, Z80, 8080) could access (directly).
@@The_Real_Sensei
Last night we were catching up on an Only Connect episode from 5 or 6 weeks ago. One sequence question was:
What comes 4th in the sequence: 65536256, 25616, 164
The answer was 42 as each "number" is made up of a number and its square root, the square root of one starting the next.
The first "number" 65536256 is 65536 (2^16) = 256^2 ((2^8)^2).
Just wanted to share a different approach:
2^18 - 1 = 2^10 x 2^2 x 2^6 - 1 = 1024 x 4 x 64 - 1 = 4096 x 64 = 262 144 - 1 = 262 143
@bangel7513no!!
or just 2^9*2^9-1=512x512-1 :P
@@makalaga56Yes, that’s more straightforward and even easier to calculate than 513x511.
Fuck dis shit
@bangel7513 only typo
2 to the 18th is 2 to the 16th times 4 (or 2 squared). Anyone with a computer brain knows 2 to the 16th is 65536 (good old computer memory and bits). So times 4 is 262144, subtract 1 = 262143. Everyone should be able to do this in the modern world in their head.
Wrong! You wrote a stupid comment. Everyone should *not* able to do this in the modern world in their head! The large majority should not be able to do it. Get educated about people and don't make foolish posts such as this.
I'm an software engineer and I still don't remember all the powers of two from the top of my head nor should I have to...
@@theyassinez1 Heresy! Blasphemer! Could not resist lol. I am old school, even know how to use a slide rule lol. It's amazing sometimes describing to some kids these days how the 8 bits are used in all the ASCII font tables. I love flipping a set of text into Ascii on them when they can't find an error and point out how that spot is not a 40 hence it's not a space, it's something else that merely displays as space like which is causing the data issue. They never think of that. They are so visual these days so dependent on all the code we wrote in the past to do everything for them. Sigh.
Agreed, however the idea is to be able to follow a process more than just express a result, I think.
@@theyassinez1As a software engineer you should know 2 to the 16 because it’s the size of a short.
Children of the 8 bit era certainly know all the powers of 2 up to 16; So if you already know that 2^16 is 65536 you may as well go for the direct route. Either double it twice, or multiple by 4 (which ever you find easiest) and subtract 1.
@@vdamkyна 2 просто умножить...
@@vdamky
Technically we actually move the digits through the place value columns left (multiply) or right (divide). However, on paper it is easier to move the _decimal point_ than to shift the digits as we implicitly know the place value columns and it is the decimal point that is visible[1] which tells us where the place value columns apply to the number.
[1] If the decimal point is not visible (as the number is an integer) then it is taken to be after the last digit which is in the units place value column.
even starting from only 8bit 256 it takes shorter time, that was spent in the video
I'm one of them C64's 6502 assembler's magic limit was 65535 or 0xFFFF (next number 2^16).
65536*4-1
That's how I did it Paul!
The correct answer, in binary, is: 111111111111111111.
Thats big brain move
This is wrong answer
Mukundsir
No specific easy answer.
Antora islam riya
Yeah I thought this is what the answer was gonna be when I saw the thumbnail - it’s a super boring problem if it’s just base 10
As a computer enthusiast I remember powers of 2 up to 2^16 at least which is 65536. Then I multiply it by 4 and subtract 1.
So I am not the only one. 🙂
Excellent! My math brain told me 2^10 = 1024, then multiply by 256 and subtract one. Easy.
Same initial reasoning... But then 2^9 is like the max byte size times two which is 512, square that and subtract 1. Rather easy.
@@chrisw4562 that's what I did as well
Yes, if you remember 2^16 = 65536 then multiply by 4 and subtract 1 is quicker than the method in the video.
I literally multiplied 2 by itself 18 times and found the answer faster than her 💀💀
They probably wanted to see how you got to the solution. Just by multiplying 2 by 18 times is not enough for an actual solution.
@@boldibrown8533it is exactly the solution. Unless they stated that you needed to use something specific. Computing it is a way to solve
@@boldibrown8533 hahahaha. What? Sucessive multiplication is a solution just fine. Stop talking silly things.
@@samueldeandrade8535 the thing is, in an exam, its completely impractical. You don't time to waste. yes i might use it as a last resort but doing this method is much faster.
@@huntbat man, some people are very fast with multiplications. Also, calculating 2¹⁸-1 shouldn't even be in a test.
Since there are many similar problems, most people probably know how to calculate them easily without making mistakes. You need not stick to the factorization formula. There are also concrete examples in this comment section. As a solution method, (1) decompose them into round numbers. That is, 2¹⁰ =1024=1000+24, 2⁹ =512=500+12, 2⁸ =256=250+6. (2) subtraction should be avoided, and addition should not carry over (no overlap of the same digits). It is unavoidable that there is a single overlap in this problem.
Is there a reason why we would know what 2^9 is and not know what 2^18 is? That is really the essence of this problem.
Yes 2 to the 9 you can do in your head
@@davidbornstein9197 You are right. But than the essence of this problem is brute calculation. There is nothing elegant about it. Besides, the power to which a person can calculate an exponent is subjective. Mathematic solutions are most powerful when the method minimize or completely remove the need for brute force calculations.
U r right. It is very simple. 512 x 512 - 1 = 2,62,143
Simply do: 512 x 512 - 1 = 2,62,143
What is the point in beating around the bush?
You're in the internet land. You should know all the powers of 2, up until 10. Because 1024 is an important round number in the world of computer science. :-)
In 2^n, for n = 1, unit place digit is 2
For n = 2, unit palce is 4
For n = 3, unit place is 8
For n = 4, unit place is 6
This cycle of 2, 4, 8, 6 will repeat for every power of form 4m+1, 4m+2, 4m+3, 4m+4.
So if n = 18 it is of form 4m+2 so units place is 4, and subtracting 1 from it will give 3.
With the base numberof 2, it is easy to double the first at least10 - 12 times. And with pen and paper, you have no problem reaching 18. This is easily grinded out without fancy math shortcuts :)
This cannot possibly be a math Olympiad question. This is way too easy.
It's Math in Norway - not in China or USA or Russia. Their approach is like - "choose the only topics in science you want to learn because we think about you mental health in childhood". Study in China or Russia: "Ok, it's 23:00 pm - you can go to sleep for 3 hours. We will continue tomorrow at 5pm. That's why these counties win Olympiads 😂
@@cppdeveloper yeah true, in my country, they teach us unnecessary subjects
This is how we make an easy mathematics question into a complicated one
Some nice tricks, but what about, in this case, just calculating the thing? 2^18 = 2^10*2^8=1024*256=256000 + 24*256. The latter is 25*2^8-256=100*2^6-256=6144. So we get 262144; subtract one and you get 262143, with the "25 and we can make that 100, it's a power of 2" happening in one's head (if you don't get that, you can also easily calculate 1024*256 simply on paper).
You would, of course, know the first 10 powers of 2 by heart (they teach that in school, just like the first 20 squares)... but then the solution represented also relies on the fact that you now 2^9=512 by heart.
But then, squeezing in a binomian formular was rather beautiful. (I do mean that.)
Well ... first question should be, what kind of result is expected. Because 2^18-1 is pretty nice. 11111..11 binary is also good.
It seems, that normal writing multiplying 513*511 is also trivial.
And if I use pretty known fact, that 2^16=65536 ?
Two addition and almost done.
Now, to chunk it in head, multiply (1000 x 256) + (2 x 10 x 256) + (4 x 256) - 1 If you practice chunking, you can do complex math REALLY fast in your head. You’ll also start to be able to memorize really long strings of numbers…like pi! As a math teacher, my students are often dazzled by this; it’s really VERY easy!
as a programmer, I have remembered many 2^x values without calculating them, for example, 2^24 is 16777216 and 2^16 is 65536. Oh yes, 2^18 is 262144, so if it is minused by 1😊
For those who are used to binary and computers, simply do 1024 (2^10) x 256 (2^10) = 262144, and subtract 1, or start at 65536 (2^16) and double it twice.
Wouldn't it be easier to multiply directly 512 * 512 = 256,000 + 5,120 + 1,024 = 262,144 and just to subtract 1?
I did this in my head a little differently, but it could be done on paper the same way. Binary numbers are easy to work with and from computer experience, I know that 2^8 = 256, so 2^9 = 512. So, 512 is squared by (500 + 12) x(500 + 12), we know that (a + b)^2 = a^2 + 2ab + b^2. The rest is trivial, 500^2 = 250,000. 2 x 500 x 12 = 1,000 x 12 = 12,000, and 12 x 12 = 144. 250 + 12 = 262, so we have262,000 + 144 - 1 = 262,143. If you are not familiar with binary numbers, it is a simple matter to count up, 2, 4, 8,, 16, 32, 64, 128, 256, 512.
we might know 2^18 = 2^(6*3) as being the number of RGB colors you can encode when each Red, Green, Blue color component has 6bits resolution (64 levels) and that number is 262144 ... at least people should know 2^10 = 1024 and 2^16 = 65536 ;)
Pragmatic reasoning.
65537 is 2^16+1 is well know prime number.
2^10 x 2^8 - 1 = 1024 x 256 - 1 = 262144 - 1 = 262143
OR
2^12 x 2^6 - 1 = 4096 x 64 - 1 = 262144 - 1 = 262143
I would think like that, because I'm used to calculate until 2^11, so 2^10 and 2^12 are easy. If I didn't feel confident multiplying 2 numbers with more than 3 algarisms, I can separate 18 in 12 and 6, because 2^6 is the last potence of 2 with 2 algarisms, so it's easier to calculate 4096 x 64. I wouldn't use factoration because it's not necessary, and my first thought was 1 = 2^0.
I love Math, I love your channel and I love you. Many blessings for you, from the heart of God. ❤
❤️🥰
Did not know God had a physical beating heart.
God is widely advertised as being beyond time and space.
@@MyOneFiftiethOfADollar In your conception of God, he doesn't has a heart. And your belief shall be respected.
In my conception (Gospel), Yeshua is the image of the invisible God and Yeshua has a heart, hence God has a heart. This is my belief and shall be respected.
Also, this is a Math channel, better not talk about such things. (Blessing people is a common thing, to talk about god in a Math channel not.).
@@ulisses_nicolau_barros
You were the one who first mentioned "god".
2^18=4^9=4x16^4= 4x256^2=4x(65,536) = 262,144. Subtract 1=262,143. Check video. Glad we agree, and glad it gives a numerical answer. not a logarithm to an obscure base.
How many people did not understand the beauty of the presented solution! 😢 Thank you for what you are doing! That was pure mind pleasure! ❤
I did not understand. 2^18=262144 is not less known fact, that 2^9=512
I do not understand beauty of writing (512+1) and then explain that it is 513 verbally. Thanks God she did not come with a substitute formula for that.
If you remeber that 2^10=1024, 2^18 = 1024x1024/4. The division first will bring ...6 (divide 24 only by 4), then 6 multiplied by 1024 will bring 4 as the last digit. Can be done in the head like 5 seconds.
lol in India, we use an identity to do 511*513, it's basically, (x+a)(x+b)=x^2+(a+b)x+ab
this eases out the calculation
so you can do (500+11)(500+13)
500^2+ (11+13)500+ (11)(13)
very easy calculations, you can do them in mind
250,000 + 24*500 + 143
= 250,000+12,000+143
=262,143
(we also have a trick to multiply numbers with 11 and rest of the calculations were pretty simple).
how is the trick of multiplying with 11?
if you have to multiply 11 with 13 just write the first and last digit as it is, that is 1_3, now the middle digit is going to be the sum of the first and last digit, that is, 1 and 3, we know 1+3=4, so 143 is the ans to 11*13. It's a very easy trick, you can solve such problems instantly in seconds@@pacogutierrez2484
@@pacogutierrez2484 multiply the number by 10 and then add the number to it: example: 13 X 11 = 13*10 + 13 = 130 + 13 = 143; the idea is that multiplying with 10 is one of the easiest things to do and then addition is the next easy step - hope this helps
@@soumyasbh thank you very much
Probably the simplest and shortest way:
2^18-1=(2^9)^2-1=(512)^2-1
Applying Vedic Math:
512*512=524*5*100+144=262144 (which can mentally be calculated in about 15 sec)
Therefore: 2^18-1=262144-1=262143
Not to me.
Growing up in the 70s and 80s, I learnt 2^16=65536 (max amount of memory addressable by an 8-bit processor with a 16-bit address bus), and later 2^17=131072 (start location in memory of the first screen of the Sinclair QL).
Thus 2^18-1
= 2^16×4 -1
= 65536×4 -1
= 262144 -1
= 262143
or
= 2^17×2 -1
= 131072×2 -1
= 262144 -1
= 262143
(This latter being the easier of the two.)
@@cigmorfil4101 However, your solution work for this one particular exercice, his way works for every numbers simply.
If i put you 7^6, having Sinclair QL memory benefit is useless
However if i apply his Vedic Math :
- 7^6 -1 = (7^3)^2 -1 = 343^2 -1
- 343*343 -1 = 386*3*100 +1 849 -1 = 1 158*100 +1849 -1 = 115 800 +1849 -1 = 117 649 -1 = 117 648
the addition is equal to the 43 of 343, squared, 43^2 = 1 849
Experts in computer hardware in 1990s to early 2000s should know 2^18=262,144 that's the amount of 256MB memory in KB that would show up during start up of a computer of Pentium II/III or Athlon era.
it's 256 kiB or 262.144 kB.
She has been trying to find the result of the 513*511 operation for exactly 2 minutes since 1:17 seconds of the video. If she writes these two numbers one under the other and multiplies them, she can get the result in 20 seconds.
This is where I learn my maths...
Hey.
Just take the tenth power of 2 i.e. 1024, multiply it by itself, and divide the result by 4.... subtract 1 from the resulting value to get 262143
❤
It took me awhile, but I finally realized that this was quite an easy problem, and I thought that I might be able to do it in my head by simply doubling 18 times. Well, I didn't get that far on my first three attempts. but I now know that I can do it...eventually. How far did I get? Thirteen or fourteen times, but that was good enough for me. I now know that I can solve the problem in my head if I try long enough.
I just know - from years of doing computer science work - that 2^10 is 1024 and 2^8 is 256. 1024 * 256 is really easy, because 1, 2, and 4 are just doublings of 256 and adding them in different 10s places. then subtract 1.
i just love how apparently 2 to the ninth power is common knowledge.
I don't know about common knowledge, but you can probably count it on your fingers in less than 5 seconds.
2^10 is pretty common knowledge (it is how many bytes are in a kilobyte), so dividing that by 2 is pretty easy
I tired everything like
G.P series
Limits (calculus)
Logarithms
Binomial
At the end all the results were in the power of 2 some even exceeding 2^18 😅 so its better to learn some power table as 2^10 always helps
I wish you were my maths teacher when I was little! I would have definitely developed interest on Maths. You make the problems look so easy.
Whenever you have x², you will get the same result if you multiply (x-t) by (x+t) and add t². Therefore, a way that I consider to be the fastest way to solve problems like this mentally is to look for a convenient (x-t) (or (x+t)). In the case, for example, to solve 512², the convenient (x-t) is 500. Therefore, the (x+t) is 524, and the t is 12. Therefore, 512²=500*524+12². Knowing that 524*1000 is 524000, just divide this by 2 to get 262000. Then, knowing that 12² is 144, we have that 512² is 262144. Finally, just subtract 1 and get to the final answer to the problem.
You can simply do a difference of squares : 2^18 - 1 = (2^9 + 1) (2^9 - 1) = 513 X 511 = 262 143.
I just simplified in my head to 512^2 - 1. Pretty easy to multiply 512 x 512 on paper and subtract 1.
It's a lot easier to solve if you just know your powers of 2 up to 18, as any decent computer programmer does.
My point exactly
Programmers who developed on the PDP-10 which had a 36 bit processor would know that 2^18 is 262144, 1000000 octal and 40000 hex. Especially useful when doing some register arithmetic on the left and right halves of the register in your head. I used to have to do that a lot at CompuServe in the 90's.
@@beragis3
I guess you were au fait with 6-bit (sixbit) encoding which allowed 6 characters in a 36 bit word?
@@cigmorfil4101Yep 6 bit, 7 bit, 8 bit and 9 bit characters
Even if you don’t know them by heart… working out a power table for two is easy and I doubt they mark off for showing work.
For a smaller power like 18, it's better to count until 2^18, if you have practiced that.
For larger power its useful.
This one is hilarious. When I saw it I thought, surely you just solve it using arithmetic? But hey if there's a trick to working with exponents of 2 I've love to know it. No - you go through a convoluted process which gets you nowhere, then just solve using arithmetic anyway.
I'd write 2^18 = (2^3)^2^3 =(8^2)^3 = 64^3 = 64 x 64 x 64 = 262,144 ----> then subtract one unit, resulting to 262,143 :)
maths which is the only thing can be done by various methods ❤
Time to get a girlfriend
@@delanym already have !! 🫠
See if you play 2048 you’ll already have all powers of 2 memorized up until 2^11, so just take 2^9 as 512, square it, and subtract 1
elegant but Id rather just calculate 2^18 and then -1
at most, do 512x512 -1
Outstanding presentation.
Each step carefully shown.
Precise terminology.
Prefiro multiplicar 512 por 512 e subtrair da unidade
Em duas linhas o problema é resolvido
Ué... não dá mais de 2 linhas amigo? 4 se não me engano
@@editorx2023 Sim, dá mais de 2 linhas, mas é muito mais rápido que a solução do vídeo. Dá pra fazer em menos de 1 minuto
Eu fiz 17 contas, mas multiplicar por 2 é muito fácil e rápido, saiu bem mais rápido do que no vídeo, e de certa forma meus neurônios ainda estão intactos
@@evandroa4845 inacreditível
We could just do 513x511 and it would be done so much faster ;-;
I'd better do (2⁹)² - 1 = 512² - 1 = 262144 - 1 = 262144
The rest is not necessary.
@eblan7689: Dude, you're off by 1, which is a common (and often fatal) programming error!
Easier decomposing 2^18-1 as a difference of squares, then as a difference of cubes times a sum of cubes . At the end: 63 times 73 times 57= 262,143
A very nice solution! Thanks.
For us that knows our powers of 2 it can be simplified as: ( 2^18) - 1 = ((2^16) * (2^2) ) - 1 = (65,536 * 4) - 1 = 262,144 - 1 = 262,143
Of course one could also solve it using logs or even by using a slide rule!
1:23 you are done. Every fifth grader can do a multiplication on paper, come on.
This is cool; using the mathematical principles that we already know in unorthodox ways is a good way to strengthen one's skills
Am I the only person that just did the math in my head and calculated the same result? I understand the method behind that, but you can easily just calculate the result in less than 1 minute.
Did you just 512*512?
I did ut in my head too; I'm a programmer 😊
Well she just made that 10x more difficult for zero reason. People really love to make themselves seem relevant. She must be a teacher.
Зная степени двойки(а в Российских школах дети это знают, так же как и таблицу умножения) вычислить этот можно за минуту. 2^10*2^8 - 1= 1024*256 - 1= 262143. Привет из России!
Даже не зная степени двойки, проще на 2 умножить 18 раз
Идиот😂
Explain something to me. How people with such abilities, educated, can support invasion on a neighbouring country? Or wait, maybe it doesn't matter how well one's educated in math... Regards from Poland.
@@rafakrasicki914 are you asking me?
можно еще сложить 2 + 2^2 + 2^3 + ... + 2^17, это гораздо математичней
Instead of calculating 2^9, you could just use rules of exponents and rewrite it as 8^3. In my opinion, easier to calculate. This is only if you don’t know the powers of 2 up until 9 or above
It is surprising that back in the 80-90s of the last century, almost all schoolchildren (at least in my country) in grades 6-7 were able to multiply 513 by 511 without a calculator and expansion. I think that in 20 years, schoolchildren will not be able to even multiply 500 by 500 without gadgets.
i don't think that you are good at analyzing.
I think you must have written your comment back in 2000...
In my country, most kids in high school can't solve this.
30 seconds, 6 lines and 6 columns of digits, it's all this calculus takes yet the number of children/students who can't make it as simply as that is exponentiating, so I'm afraid your conclusion is right.
yes that's true ....easily multiplied
It can also be written as 2^18=2^0 is equal to 2^18-which is equal to 262144
Very helpful
2^18 = 2^20 / 4 = 1048576 / 4 = 524288 / 2 = 262144.
Actually, multiplying by 2 is very simple, so if you have a good trained memory, you can multiply by 2 in memory.
Also if there is no rule that requires to provide answer in base 10, you can just write that
(2^18 - 1) base 10 = (1000000000000000000 - 1) base 2 = 111111111111111111 base 2 😊
U make it more harder than normal solution 😂😂
2^18-2^0
First : 2^18 = 2^0.2^18
Second :2^0(2^18-1)
(2^9)^2 = 512^2
Finally:
(262.144-1).1
262,143.1
262,143
Mam... It is difficult to understand.please explain easily.
She only needs to put = under = and not bring a and b into it
Which part you did not understand?
Its very simple you can solve it in seconds by a simple and no need of using formulas etc
First method
Power 18
Like 6*3
2 power 3 = 8
( Now multiply 8 six times to get an answer)
8*8*8*8*8*8= 262,144-1= 262,143
Second method
2 power 18
6*3= 18
2 power 6 = 64
( Now multiply 64 by three times to get an answer)
64*64*64 = 262,144-1=262,143
took 2 steps.. 2 to 18 and subtract 1 from total. simple and quick
Who works in IT knows that 2^20 is 1048576. So divide by two twice and you have result :-). It would be 256*1024.
My answer.
I would do ((2^16)*4)-1 2^16 = 65536. mutilply by four and subtract 1.
In my assember years, we used to have 2 to the power of 16 by heart.
Much simpler to calculate 2*2*2*... iteratively in your head. Especially if you know that 1 Mb = 2^20 = 1048576 bytes
2power 12 calculate krna aasan hai Usk bad 2ki power 6 calculate kro.
Aapas main simple multiply krk -1 krdo jo tarika aapne bstaya hai usse assan hoga
(2^12×2^6)-1
(4056×64)-1
I have done this orally
Wouldn’t it be easier to multiply 512*512 using grade school math and subtract 1?
There is a way to find out what the last number is when dealing with equations that use large numbers. I vaguely remember it from hs
Any old school computer scientist can answer it without by without even drawing on paper. 2^18 is close to 2^16, which is the limit of array size in an 16 bit OS. So its just 4x65536 - 1.
I doubt "old school computer scientists" participate in math olympiad :)
I do not see how this is more simple than 512x512-1. We all know 2^16=65536, so double it twice in a row and substract 1. More over 513x511 is quite obvious, 1x513 is obvious and everybody knows 13x5=65
you can use congruence of numbers to solve this also a to the n and b to n identity
I have math channel,watch please
I know a much easier solution for this. When I was a child I found out how can I square any number only using my head. So 512x512-1 can be calculated with the following method 512-12 is 500 and 512+12 is 524 so just to be able to multiply with a round number. 500*524+12*12-1 equals 250000 + 12000 + 144 - 1 which is the same result but easier and faster to solve your example.
For people from the C64-era this question is a very easy task, because they all know that 2^16 = 65536. So all they have to calculate is 65536 * 2 * 2 and then subtract 1. A matter of seconds.
I actually thought you had some really nice smart approach to it, but this is just brute force tbh.
Even without knowing powers of 2.
It is very easy to produce them:
2^1: 2
2^2: 4
2^4: 16
2^8: 256
2^16: 65535
Then we use a combination of the above.
2^(16+2)=65535*4=262144
Subtract 1.
This procedure scales quite well. Because its complexity is logarithmic. And we can easily combine any desired exponent.
Proceeding with exponent 9 is also good. But it doesn't really bring advantages.
It's more complicated and longer. If i know that 2 to 9th is 512, just do the 512*512, minus 1.
So we "know" that 2^9=512, but we cannot find out 2^18=512^2 so we need to write a dissertation to calculate 2^18-1
2^18=4^9=64^3=64*64*64=262,144
That minus 1=262,143
Multiplication is cool, it doesn't bite you..
Try it for small numbers.
😉
2^20 is 1048576 , then divide by 4 and subtract one...
262,143. I just doubled 2 seventeen times then subtracted 1. Did it in my head.
Instead of going for FOIl 1:43 method.. just multiply 513*511 = 262143
I prefer (2^10)^2 / 4 = (1000^2+2*24*1000+24^2)/4 = 1,000,000/4+12000+12^2 -> 262143 (minus 1)
*LKLogic* -- It is *not* solving for anything. It is simplifying a numerical expression down to one number.
Why all the complicated brackets and splitting, adding, subtracting, changing powers? I don't understand any of it. Why doesn't she just go 2*2 18 times and the take one away? I got the answer that way by about 90 seconds and just needed my fingers to keep track of the multiples and my head to multiply by 2. Why make it so hard?
From Venezuela:!!!! Very, very wonderful your class!!!!!!!!
Cool! Pretty elegant solution! :)
Is it that much more work to just multiply it out? You should a good chunk of the binary digits to give yourself a running start
On my own I got exactly where you got, using the identity: a^2-b^2 = (a-b) (a+b) = 511x513....then I just calculated this easy multiplication on paper
2^10 = 1024.
2^8 = 256
2^18-1= 1024x256 -1= 262144-1= 262143
I see this using binary numbers.
2^4 -1 is 111 I’m binary.
Likewise 2^18 -1 is 1……1(18 1s) in binary.
Convert from binary to decimal!
for computer people, 2^16=65536, mult with 4, sub 1 makes 262143
Seems easy to me.
18 = 3*3*2, so 2^18 = (((2^3)^3)^2) = ((8^3)^2) = ((64*8)^2) = 512^2 = 500^2 + 2*500*12 + 12^2 = 250,000 + 12,000 + 144 = 262,144
262,144 - 1 = 262,143
Solved by mental calculation in less than 30 seconds.
Thanks, but It isnt olimpic way to do It. The better way I Saw the resolution is:
2^18 = 2*9 * 2*9 = 512*512 = (500+12)(500+12) = 500² + 2*12*500 + 12² = 250.000 + 12.000 + 144. Subtracting -1, we have 262.143 without a lot of multiplications
2^18 is easy. 500 times 512 is 256,000. 10 times 512 is 5120. 2 times 512 is 1024. Add them all in mind minus 1