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.

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

I literally multiplied 2 by itself 18 times and found the answer faster than her 💀💀

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.

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.

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.

The correct answer, in binary, is: 111111111111111111.

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.

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.

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.

This is how we make an easy mathematics question into a complicated one

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.

Why to make things so complicated- Just do 2^18= (2^9)^2 =((2^3)^3)^2= ((8)^3)^2 =(512)^2= (500+12)^2= 262144 =262144-1= 262143 ik it looks hard but believe me doing with this method in your notebook is the easiest solution....☺ And the fact is that no one straight away knows that 2^9=512

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.

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.)

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

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.

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.

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

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

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.

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).

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.

It can also be written as 2^18=2^0 is equal to 2^18-which is equal to 262144

I love Math, I love your channel and I love you. Many blessings for you, from the heart of God. ❤

This is where I learn my maths...

Much simpler to calculate 2*2*2*... iteratively in your head. Especially if you know that 1 Mb = 2^20 = 1048576 bytes

Wouldn't it be easier to multiply directly 512 * 512 = 256,000 + 5,120 + 1,024 = 262,144 and just to subtract 1?

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 ;)

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

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 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!

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.

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.

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.

Though there are many ways to arrive at a solution, this one really beat around the bush lol

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

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

2^20 is 1048576 , then divide by 4 and subtract one...

i just love how apparently 2 to the ninth power is common knowledge.

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

Outstanding presentation. Each step carefully shown. Precise terminology.

When I started in computing in 1965 we just new all the 2**N values up to n=32. and we knew that 2**N -1 was a stream of on bits.

Wouldn’t it be easier to multiply 512*512 using grade school math and subtract 1?

Prefiro multiplicar 512 por 512 e subtrair da unidade Em duas linhas o problema é resolvido

Who works in IT knows that 2^20 is 1048576. So divide by two twice and you have result :-). It would be 256*1024.

I just simplified in my head to 512^2 - 1. Pretty easy to multiply 512 x 512 on paper and subtract 1.

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

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

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

It's more complicated and longer. If i know that 2 to 9th is 512, just do the 512*512, minus 1.

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 clicked on one of these math problem videos and now I get more and more math problem videos from KZfaq that I have to solve omg

We could just do 513x511 and it would be done so much faster ;-;

A very roundabout way to approach this math problem.

This solution method is like using a chainsaw to cut bread...

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 actually thought you had some really nice smart approach to it, but this is just brute force tbh.

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.

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Appreciate your efforts.

262,143. I just doubled 2 seventeen times then subtracted 1. Did it in my head.

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.

It's a lot easier to solve if you just know your powers of 2 up to 18, as any decent computer programmer does.

*LKLogic* -- It is *not* solving for anything. It is simplifying a numerical expression down to one number.

for computer people, 2^16=65536, mult with 4, sub 1 makes 262143

2^8 = 256, 2^10 = 1024, 2^16 = 65536. They are all "round" numbers which every computer science student remembers

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?

maths which is the only thing can be done by various methods ❤

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.

De tête, la réponse est : 111 111 111 111 111 111 (en base 2). Beaucoup plus simple 😅

Already we knew 2 power of 9 is 512..then 512* 512 is 262144..finally subtract 1 then we will get 262143

Gostei muito!!!!

It can be quite easily calculated in mind

Why not use (2^10)*(2^8) and then subtract 1? or (2^9)^2 then subtract 1? These calculation methods are very simple and fast.

2 up too 18 is 524288, 2 minus 1 is half of 524288 is 262144

Very helpful

Let X,Y,Z and T be intregers with X

Calculators not allowed? Before I dropped my slide-rule it gave me 2^18-1, out to approximately 2.7 decimal places, but afterward only out to about 2.3 places. I'll just stick to the older, more reliable method of pencil and paper.

you can use congruence of numbers to solve this also a to the n and b to n identity

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^ 16 is 65536, then you can just multiply by 2 two more times and substract 1.

If I see (a-1)(a+1) I usually simplify TO a²-1, not the other way around. And if I see 512², my brain goes oh, "oh 2^9 squared, so 2^18, so 65536... 131072... 262144. The answer to 2^18-1 is obviousle 262144-1 = 262143. Was there any more context to this question, because apperently I would've failed horribly.

I did via the bit representation 0xFFFF is 16 1's and you need 2 more so 0xFFFF + 0x30000 = 0x3FFFF which is the answer in hex.

Every 2^10 is a k, so 2^20 is 1M (or 1024x1024). 2^18 is 1/4 M or 1024x256. Subtract 1, done. (You learn this working with binary or anything to do with computers - HDD space, addresses, etc, etc)

If this was really an Olympiad question, the level is decaying... 😅

The limes to infinite is a no brainer. 😊

Is this a math test for students will less than average iq? how can this compare to the chineese olympiad with the question of 9^x - 6^x=4^x

If you can compute 2^9 you can compute 2^18 and subtract 1

What is the wheather condition in Norway?

sir ya kon si class or school kon sa ha??

Based on the rules of exponents 1 = 2^0. So, my next step was 2^18 - 2^0 equaling 2^18, which is incorrect. Why does this method not wok?

Very instructive task 👍

"I'll calculate 1048576 divided by 4, and then subtract 1. This is because 2^20 equals 1048576 by my memory.

Too many Steps - You can just use binary math(base 2): Since 2^16 = 65536 and double it twice -> 131,072 ->262144 == 2^18 -1 =262,143

My calculator: you've underestimated my power

2^18 is 4^9. 4^9 is 64^3. Just solve for 64^3 and subtract 1

This seems like a very primitive task, what age is this for? Moreover, now many people know that 10**8 = 256 or that 10**10 = 1024. Therefore, you can just straight up 1024*256-1

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.