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2¹⁰⁰ - 2⁹⁹ = (2 - 1)*2⁹⁹ = 2⁹⁹ . But I'm not expanding that. I only know up to 2³² . I suppose I can figure out 2³³ , and cube it with a calculator: (2³³)³ = 8,589,934,592³ = 633,825,300,114,114,700,748,351,602,688 .
I am trying to get another brain cell working. That way they can both get some company. And your videos are helping with just the right amount of explanation. Thank you your majesty. 🇬🇧
Multiplying something by 2, doubles the number and when you subtract a number from its double, you get back the same number. In this case 2^100 is simply double of 2^99, so if you subtract 2^99 from it you will end up with 2^99.
My solution was exactly like yours, though the outcome initially surprised me. Note that the problem can be generalized to a^(n+1) - a^n = aa^n -a^n = a^n(a - 1). It's convenient that when a = 2, the (a - 1) term is just one, so it disappears.
Another way is to use x = 2^100 - 2^99. Divide both sides of the equation by 2^99 to get x / 2^99 = 2^100 / 2^99 - 2^99 / 2^99 which becomes x / 2^99 = 2 -1 or x = 2^99 = 1 then x = 2^99.
I watch these videos mostly to get to the end where she smiles and wishes us a wonderful day. 😁Okay, I'm exaggerating a little, but it is a very welcome touch.
Fifty years ago I would have got this in an instant but time hasn’t been on my side…..applying the same principle 2cubed minus 2squared = 2squared….i can sleep easy now.
2^100 = 2^(99+1)=2*2^99. So 2^99 is 1/2 of 2^100. Something minus one half of something will be one half of something. So answer is 2^99 or 0b1(0….0) with 99 zeros
I didn't utilize the a^m * a^n rule and just recognized that because the bases were the same I could just factor it to 2^99(2 - 1). (2-1) is (1) then x * 1 is just x. (x is the portion factored out. 2^99 in this case)
I solved that much your way and then, knowing that log(2)=0.301 (I'm an audio engineer, this is one of those things we just know, 'cos decibels), easily worked out that 2exp99=10exp29.8 which is roughly 6.5x10exp29. Anything much more accurate than that I'm clearly going to need a calculator or log tables, or some tricky longhand.
I checked 2^4 - 2^3 = 16 - 8 = 8 = 2^3 and did it again for 2^5 - 2^4 = 2^4 and realized that the patten is that the answer would always be the 2^smaller power, meaning 2^99. Thanks for reminding me of the x^(a+b) = x^a * x^b.
It's easy if you visualize the numbers in binary. 2^99 is 1 followed by 99 zeroes, 2^100 is 1 followed by 100 zeroes. Adding 2^99 + 2^99 carries the leading 1 bit to the next position, equaling 2^100...
I looked at 2^3 - 2^2 and noted that this was 8 - 4, giving 4, ie 2^2, the second term. Did the same with 2^4 - 2^3, making 16 - 8, ie 8, which is 2^3, again the second term. So, in that basis, the answer had to be 2^99.
You are the one that does not scribble The one that rightly puts = under = But no need for the second line Just take out the 2 to power 99 from the first line to get 2 ^99[2 -1 = 2^99
Hello, new sub here. Love the channel and the material you present. All the best to you! 3:20 I solved it using this alternative method, but the factoring method works very well too.
in my head I came up with 2^99 = (2^10)^9*(2^9) = 1024^9*512. 1024 is approx 1000+2%, 512 is approx 500+2%, so 2^99 = 500 followed by 27 zeros +a little over 20% so a little over 600,000,000,000,000,000,000,000,000,000 , which is not far off.
I think that it would be really helpful for maths students if, instead of just presenting it as a rule, you’d explain WHY x^(a+b) = x^a . x^b. Teaching maths should be about ensuring that students understand why things are as they are, otherwise it’ll go in one ear and out the other.
But then she should explain that 2^99 = 2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2 and then explain the commutative property of multiplication. By the time we're done explaining, we'll have a 2 hour video.
It's true that it is better to know why something works the way it does, but there are already so many sources that you can find that info that repeating it in every math video would be counterproductive. If you were a teacher and introduced the subject to your class then that would be important that you explain why it works. But if you do a 1000000th video about the same topic it seems redundant. It's the same with Pythagorean theorem or difference of squares. But it is always ok to ask, so if you want explaination I would gladly provide 😊
Hello, thanks for your videos! I solved this the lazy way: the first number is twice the second number, so the subtraction leaves us with 2^99... May I also propose the word maths instead of math? regards
- Pourquoi ne pas écrire directement que mettre 2^99 en facteur, reviens à soustraire 99 de chaque exposant: 2^100 - 2^99 = 2^99 ( 2^1- 2^0) = 2^99 (2- 1) = 2^99. C'est en transmettant ces petites techniques de calcul que l'on fait progresser nos élèves. - On peut aussi remarquer que 2^100 est le double de 2^99, et que "DEUX PATATES " moins "UNE PATATE " égale "UNE PATATE ".
The way I solved it was 2^3 = 8 & 2^2 = 4...8-4 = 4...2^3 - 2^2 = 2^2. I did one or two more simple examples and it was the same pattern so.... 2^100 - 2^99 will = 2^99
@@toby9999 I didn't remember the rule, but it is easy enough to work it out if you try a couple of simple examples like 2^3-2^2, 2^8-2^7, etc. It didn't occur to me to even try to work 2^99 out. I have trouble after 2^16, after 2^18 I usually need paper. For an order of magnitude, you usually add 3 digits in base 10 for every 2^10. 2^10~=1K, 2^20~=1M, 2^30~=1G, 2^40~=1T, etc.
I watch you solve several of these puzzles and questions and it makes me feel really dumb. I'm sure that isn't the intent of the videos but I used to think I was smart but the longer I live the more I realize how dumb I am, which is a little saddening. Ah well, thanks for solving the the problem for me.
Just by looking at it I knew the answer was 2^99 but I thought the question was ultimately asking you to work out what 2^99 actually was. I'm sitting here thinking there's no way I can work that out in my head.
Yes, I expected that too! But I estimated it to somewhere over 500 octillion! 2¹⁰ = 1,024 so every 10th time just add 3 zeros: 2¹⁰⁰ is around 1 nonillion; 2⁹⁹ is half that! Gives you a rough answer... :)
This doesn't make sense. When subtracting two powers you cannot simply subtract the exponents. Otherwise 2^3 - 2^2 should be 2^(3-2)=2^1 = 2, while it should be 4. Moreover in this case A = 2, x = 100. Which means you claim: 2^100 - 2^99 = 2^(100 - 99) = 2^1 = 2, and it just isn't.
I did it the 2nd way (I knew 2^99 was half 2^100), and I offer the following formula: xⁿ+¹ - xⁿ = xⁿ•(x-1) I believe this works for any positive integer, at least!
Approximately 2^99... Disclaimer: They wouldn't let me within 1000 miles of Harvard Business School, but I did graduate with honors from the Janitorial Services Business School, which I think does qualify me to take on high level mathematics problems like this one with some degree confidence.
I got 2^99 in two or three seconds, and spent the other 7 to wonder how we were going to calculate 2^99 and didn't come up with a solution... Unfortunately the video also stops at 2^99, not at 6.3e+29 (rounded)...
This was unusually easy.. took about 3 secs... each power of 2 is double the power below it. So 2^100 = 2(2^99).. So it becomes ... 2^99 +2^99 - 2^99.. .
![Why you didn't learn tetration in school[Tetration]](http://i.ytimg.com/vi/Ea1_1aVfwl4/mqdefault.jpg)
![Why you didn't learn tetration in school[Tetration]](/img/tr.png)
Hey math friends! If you’re enjoying this video, could you double-check that you’ve liked it and subscribed to the channel? It’s a simple equation: your support + my passion = more great content! Thanks for helping me keep this going - you’re the best!
2¹⁰⁰ - 2⁹⁹ = (2 - 1)*2⁹⁹ = 2⁹⁹ .
But I'm not expanding that. I only know up to 2³² . I suppose I can figure out 2³³ , and cube it with a calculator:
(2³³)³ = 8,589,934,592³ = 633,825,300,114,114,700,748,351,602,688 .
i solved it just after watching the thumbnail within 5 secs, thanks to your videos now my brain is functioning at pace like it was 20 years ago.
I also did it in my head. I agree that this lady is an excellent teacher, who so meticulously explains things.
yeah thanks Poindexter
Just remember, with great power comes great responsibility
Nice one:)
@@igormiovski1803 autant rester stupide, c'est tellement plus simple. Votre phrase sonne bien, mais elle est juste idiote.
Easy to understand and simple as always. Thanks, Susanne.
I appreciate your confidence in humanity when you said ‘let me know which method you used to solve this’ 😅
I am trying to get another brain cell working. That way they can both get some company.
And your videos are helping with just the right amount of explanation.
Thank you your majesty. 🇬🇧
What I did is replace 2⁹⁹ by a variable. Let's say "a", on the second line. Then you can see 2a-a which is clearly a.
First time I saw this type of question, I solved it in 3 seconds. Now I can solve it in 3 milliseconds. PURE GENIUS!!!
Multiplying something by 2, doubles the number and when you subtract a number from its double, you get back the same number. In this case 2^100 is simply double of 2^99, so if you subtract 2^99 from it you will end up with 2^99.
My solution was exactly like yours, though the outcome initially surprised me. Note that the problem can be generalized to
a^(n+1) - a^n = aa^n -a^n = a^n(a - 1). It's convenient that when a = 2, the (a - 1) term is just one, so it disappears.
Very good short math video to watch in the morning and jog the brain
I used the second method. 2⁹⁹ is the half of 2¹⁰⁰. Nice video!
This blew me away. I'm very good at math and totally got the wrong answer at first attempt. Wow love your videos
Another way is to use x = 2^100 - 2^99. Divide both sides of the equation by 2^99 to get x / 2^99 = 2^100 / 2^99 - 2^99 / 2^99 which becomes x / 2^99 = 2 -1 or x = 2^99 = 1 then x = 2^99.
I watch these videos mostly to get to the end where she smiles and wishes us a wonderful day. 😁Okay, I'm exaggerating a little, but it is a very welcome touch.
My solution was very short and easy. 2^100=2*2^99, 2^99 =1x2^99, 2x2^99-1x2^99 =1x2^99. Done.
substitution of 2 to the power of 99 by a gives: 2a-1a=a and the reverse substitution of a thus: 2 to the power of 99
A= 1,267,650,600,228,229,401,496,703,205,376 - 633,825,300,114,114,700,748,351,602,688 = 633,825,300,114,114,700,748,351,602,688
Easy peasy.
🤯🤯
Wow you must have a super calculator that goes out that many places. 😮
@@DonShoemaker-b6o No, in Python it's just one line of code:
print(f"{2**99:,}")
Result:
633,825,300,114,114,700,748,351,602,688
@@kenhaley4Thanks for letting me know.
Thanks that’s what I was looking for 🗣️🗣️🗣️
Probably something I learnt in school (1960-ies) but haven´t used since then. So nice for der (die oder das?) Erinnerung, Vielen Dank!
Each power of 2 is double the previous one so 2^100 is 2*2^99 or 2^99+2^99 so 2^100-2^99 is 2^99
As she said in the video 😅
Thanks
Fifty years ago I would have got this in an instant but time hasn’t been on my side…..applying the same principle 2cubed minus 2squared = 2squared….i can sleep easy now.
omg..i got it noooow
You give great lectures. I wish you success.
My way of solution ▶
2¹⁰⁰ - 2⁹⁹
= 2⁹⁹(2¹ - 1)
= 2⁹⁹(2 - 1)
= 2⁹⁹
smart!
Let n=2^99
then 2n=2*2^99=2^100
so given that 2n-n=n
therefore 2^100-2^99=2^99.
That’s literally what she did…
That is exactly what this excellent teacher did.
Hey, how did you get those superscripts to appear in a RUclips comment?
No, not within 10 seconds, but then I thought about it and it took about 30 seconds while I was watching! My brain took your second approach!
Calculate or simplify?
2^100 = 2^(99+1)=2*2^99. So 2^99 is 1/2 of 2^100. Something minus one half of something will be one half of something. So answer is 2^99 or 0b1(0….0) with 99 zeros
I didn't utilize the a^m * a^n rule and just recognized that because the bases were the same I could just factor it to 2^99(2 - 1). (2-1) is (1) then x * 1 is just x. (x is the portion factored out. 2^99 in this case)
I solved it using the intuitive method outlined at the end but appreciated seeing the factoring method.
Susanne....you ROCK!!!
I did it like you did at the end, but wasn't happy. I thought, that there had to come out a real value fast and easy.
Slow learner like me need reasoning and explain. For me this is very good 👍👍👍.
Very good explanation
Three cheers for sequences and series.
I solved that much your way and then, knowing that log(2)=0.301 (I'm an audio engineer, this is one of those things we just know, 'cos decibels), easily worked out that 2exp99=10exp29.8 which is roughly 6.5x10exp29. Anything much more accurate than that I'm clearly going to need a calculator or log tables, or some tricky longhand.
Very neat.
Easier way to explain it is to write out an example of 2^4 - 2^3, as 2.2.2.2 - 2.2.2, leaving 2, then solve by example.
Well explained, thank you!
2^100 =1024^10 or slightly more than 1,24 x 10^30
2 ^99 =2^100 /2 or slightly more than 6,2 x 10^29
I checked 2^4 - 2^3 = 16 - 8 = 8 = 2^3 and did it again for 2^5 - 2^4 = 2^4 and realized that the patten is that the answer would always be the 2^smaller power, meaning 2^99. Thanks for reminding me of the x^(a+b) = x^a * x^b.
It's easy if you visualize the numbers in binary.
2^99 is 1 followed by 99 zeroes, 2^100 is 1 followed by 100 zeroes.
Adding 2^99 + 2^99 carries the leading 1 bit to the next position, equaling 2^100...
Agree, that's what I did - much easier to visualise that way- my algebra is almost non-existent.
Some of your lovely’s are old enough to be your dad. lol. I enjoy so much your detail in solving the equations. Thanks
Well explianed and solved
Thank you, Susanne❤❤❤
Easy One! I understood everything Thanks!
I looked at 2^3 - 2^2 and noted that this was 8 - 4, giving 4, ie 2^2, the second term.
Did the same with 2^4 - 2^3, making 16 - 8, ie 8, which is 2^3, again the second term.
So, in that basis, the answer had to be 2^99.
I did basically the same thing.
Yep that is how I attacked it as well. 2^2 - 2^1 =2 or 2^1, 2^3 - 2^2=4 or 2^2, 2^4 - 2^3 =8 or 2^3. That’s enough of a pattern for me.
Merci bien Professore
You are the one that does not scribble
The one that rightly puts = under =
But no need for the second line
Just take out the 2 to power 99 from the first line to get
2 ^99[2 -1
= 2^99
Hello, new sub here. Love the channel and the material you present. All the best to you!
3:20 I solved it using this alternative method, but the factoring method works very well too.
in my head I came up with 2^99 = (2^10)^9*(2^9) = 1024^9*512. 1024 is approx 1000+2%, 512 is approx 500+2%, so 2^99 = 500 followed by 27 zeros +a little over 20% so a little over 600,000,000,000,000,000,000,000,000,000 , which is not far off.
It is not a matter of solving but of calculating, and what is the value of 2^99 ?
Ha ha, Susanne ist die Meth Queen 😂🤣
Viel Erfolg mit deinem "international channel"!
You can factor 2^99 immediately, this implies 2^99(2 - 1) implies 2^99 * 1 = 2^99
Faster:
/2"99
or multiply by 2^-99
(and multiply it later)
I think that it would be really helpful for maths students if, instead of just presenting it as a rule, you’d explain WHY x^(a+b) = x^a . x^b. Teaching maths should be about ensuring that students understand why things are as they are, otherwise it’ll go in one ear and out the other.
But then she should explain that 2^99 = 2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2 and then explain the commutative property of multiplication. By the time we're done explaining, we'll have a 2 hour video.
It's true that it is better to know why something works the way it does, but there are already so many sources that you can find that info that repeating it in every math video would be counterproductive. If you were a teacher and introduced the subject to your class then that would be important that you explain why it works. But if you do a 1000000th video about the same topic it seems redundant. It's the same with Pythagorean theorem or difference of squares. But it is always ok to ask, so if you want explaination I would gladly provide 😊
2^3+4 = 2^7. Same as (2x 2x2x2x2x2x2) = (2×2×2) ×(2×2×2×2) = (2^3)x (2^4) or x^3+4 = (x^3) × (x^4)
Math teacher here. I DO explain why the rules are what they are. I tell my students "Let's look at what's going on behind the math curtain."
Dear Susanne,
Could you please share which electronic board you are using?
Best regards,
Przemek
i have just seen you and your channel so you are not queen that you are an angel escaped from paradise as my opinion
Hello, thanks for your videos! I solved this the lazy way: the first number is twice the second number, so the subtraction leaves us with 2^99... May I also propose the word maths instead of math? regards
2^(n+1)-2^(n)=2^n for all n, so 2^100-2^99=2^99
2^100-2^99=633,825,300,114,114,700,748,351,602,688=2^99 It’s in my head.
- Pourquoi ne pas écrire directement que mettre 2^99 en facteur, reviens à soustraire 99 de chaque exposant:
2^100 - 2^99 = 2^99 ( 2^1- 2^0) = 2^99 (2- 1) = 2^99. C'est en transmettant ces petites techniques de calcul que l'on fait progresser nos élèves.
- On peut aussi remarquer que 2^100 est le double de 2^99, et que "DEUX PATATES " moins "UNE PATATE " égale "UNE PATATE ".
I used the second way: 2^100 = 2 * 2^99 and went from there. Not sure if I finished in under 10 seconds but it wasn’t much more than
Very good!
I wonder if anybody who just started saw the solution 'intuitively'. I'm not ashamed to say I didn't.
how much is 2^99?
3^100-3^99=3*3^99-3^99=2*3^99.
The way I solved it was 2^3 = 8 & 2^2 = 4...8-4 = 4...2^3 - 2^2 = 2^2. I did one or two more simple examples and it was the same pattern so.... 2^100 - 2^99 will = 2^99
Forget all the big exponents and just consider 2 squared and 2 cubed e.g. 2^2 and 2^3 and it becomes obvious.
I recognised that 2^100 is twice 2^99 so therefore 2^99 is the remaining ‘half’ after subtraction.
How did you recognise that? My first thought was that the answer would be 2. Obviously, it's wrong in hindsight.
@@toby9999 I didn't remember the rule, but it is easy enough to work it out if you try a couple of simple examples like 2^3-2^2, 2^8-2^7, etc.
It didn't occur to me to even try to work 2^99 out. I have trouble after 2^16, after 2^18 I usually need paper.
For an order of magnitude, you usually add 3 digits in base 10 for every 2^10. 2^10~=1K, 2^20~=1M, 2^30~=1G, 2^40~=1T, etc.
@Math queen What program do you use to show the problems.
I watch you solve several of these puzzles and questions and it makes me feel really dumb. I'm sure that isn't the intent of the videos but I used to think I was smart but the longer I live the more I realize how dumb I am, which is a little saddening. Ah well, thanks for solving the the problem for me.
Just by looking at it I knew the answer was 2^99 but I thought the question was ultimately asking you to work out what 2^99 actually was. I'm sitting here thinking there's no way I can work that out in my head.
I agree. The title was a little bit misleading.
Yes, I expected that too! But I estimated it to somewhere over 500 octillion!
2¹⁰ = 1,024 so every 10th time just add 3 zeros:
2¹⁰⁰ is around 1 nonillion; 2⁹⁹ is half that!
Gives you a rough answer... :)
@@HarryC-Smith Yep. Not understanding the question/assignment has let me down so many times.
@@HarryC-SmithLittle bit more accuracy gives you this method:
2^100=(2^10)^10=(1,024e3)^10≈1,24*e30
2^99=(2^100)/2≈6,2e29
I think in general x^(n+1) - x^n = (x-1)*x^n
2^99(2-1)=2^99×1=2^99
x^(n+1)-x^n=(x-1)*x^n.
half of 2^100? for example, 2^7 = 128 and 2^8 = 256, so, 2^8 - 2^7 = 256 - 128 = 128 which is half of 256.
But I still don't know what 2^99 is, at least in decimal notation. Because in binary notation it's 1 with 99 zeros
Math queen i challange you :)) which number is bigger: 2^55 or 3^36.
And 2⁹⁹ is how much ?
Reminds me of "If you are running a race and you passed the second person then what's your position?".
What if the sign was ADDITION? Please make a video for that. Thank you.
3*(2^99)
With a similar way as in the video, it is easy to show, that 2^100+2^99=3*2^99
2^100 = 2 * 2^99 = 2^99 + 2^99
Then subtract 2^99 and the answer is 2^99
No need to factorise! 2x - x = x where x= 2^99
autre méthode :
2^99 + 2^99 = 2 X 2^99 = 2^100
2^100 - 2^99 = 2^99 +2^99 - 2^99 = 2^99
Every assembly language developer knows the answer out of their head: 10, but in binary 🙂
2 ^ 100 = 2(2^99) = 2^99 + 2^99. Subtract 2^99 and only 2^99 remains.
Mental math. 2^100 = 2×2^99 so you're subtracting half of 2^100, so you're left w the number after the minus sign.
2^99?
It was common sense (to me) but I had no idea how to do the math on paper (a problem I always had).
I think it’s 2. A ^ x - A ^ (x - 1) should be A ^ (x - ((x - 1)) = A ^ ( x - x + 1) = A ^ 1 = A
This doesn't make sense. When subtracting two powers you cannot simply subtract the exponents. Otherwise 2^3 - 2^2 should be 2^(3-2)=2^1 = 2, while it should be 4. Moreover in this case A = 2, x = 100. Which means you claim: 2^100 - 2^99 = 2^(100 - 99) = 2^1 = 2, and it just isn't.
2^99(2^1---1)= 2^99×1= 2^99
I did it the 2nd way (I knew 2^99 was half 2^100), and I offer the following formula:
xⁿ+¹ - xⁿ = xⁿ•(x-1)
I believe this works for any positive integer, at least!
You can easily simplify the expression as you have shown. But you haven't really "solved" it. Unless you know a way to evaluate 2^99 in 10 seconds!
It took to you 3 minutes and 52 second. You got to shape it up
For ever natural number n, 2(n+1)-2n=2^n*2--2^n+1=2^n*(2-1)=2^n. So the excercise in the video is simply the special case for n=99 ....
2^100 - 2^99 = ???
2^99 = x
2x - x = x
2^100 - 2^99 = 2^99
Approximately 2^99... Disclaimer: They wouldn't let me within 1000 miles of Harvard Business School, but I did graduate with honors from the Janitorial Services Business School, which I think does qualify me to take on high level mathematics problems like this one with some degree confidence.
I got 2^99 in two or three seconds, and spent the other 7 to wonder how we were going to calculate 2^99 and didn't come up with a solution... Unfortunately the video also stops at 2^99, not at 6.3e+29 (rounded)...
2^99 is the solution.
@@AaronSwenson Only if 1 + 2 is also a valid solution (to another equation, obviously). There is still an arithmetic operator in the "solution"...
@paulnieuwkamp8067 guess what....it is.
@@AaronSwenson I don't think any math teacher would accept "1 + 2"; they all want to see that 3 there.
@paulnieuwkamp8067 yes, but none will count it wrong unless they asked for the solution to be simplified completely.
I worked it out without watching the show. My answer was 2^(99). Let me know if I got it right.
This was unusually easy.. took about 3 secs... each power of 2 is double the power below it. So 2^100 = 2(2^99)..
So it becomes ...
2^99 +2^99 - 2^99.. .
No, it was hard. Easy for you doesn't mean easy for everyone. Not everyone has a PhD
I haven't a PhD but mentally I did 2³ - 2² = 8 - 4 = 4 = 2² then answered after about 5 secs 2⁹⁹
Took me ~20 seconds but I'm old so my brain needed a bit of starter fluid first :D
You forgot to evaluate 2^99 in seconds....
2^99.(2)-2^99
=2^99(2-1)
=2^99