Ok. So 2^22 is 2^11 squared. 2^11 = 2048. So, picture a square grid, 2048 by 2048. First take care of the 2000 by 2000 part - that's 4,000,000. Then add 2*2000*48 = 192,000, so we're up to 4,192,000. That leaves a 48 by 48 grid in the top right corner to cover. Let's add a 50 by 50 grid instead, which adds 2,500, so we're up to 4,194,500. Now subtract the two extra rows and the two extra columns - 100 each. So 4,194,300. Finally, add back the 2 by 2 grid we subtracted twice, so 4,194,304. Finally, subtract the 1. 4,194,303. Q.E.D. You can do this in your head if you can hold a couple of numbers in memory, and you can ABSOLUTELY do it on scratch paper in under a minute.
Anyone who has done a lot of programming and worked in binary and hexadecimal can go through powers of two in seconds. A few common powers known for memory are 2^10 = 1024 or 1K and 2^20 = 1049576 or 1M, So multiply 2^20 by 4 and subtract 1 you get 4194303. You would think by now the Math Olympiad wouldn't use this type of question.
The well known of "2 power N", for large N in infornation system is 256 (2^8,, 8-bit), 1024 (2^10, kilobyte), 65.536 (2^16, 16-bits), and 16.777.216 (2^24, 24-bits, RGB color).. So, 2^22 - 1 = 2^16 x 2^6 - 1 = 65.536 x 64 - 1
Before watching: 2^22 -1 = (2^11 +1)(2^11 -1) 2^11 = 2048, so this is (2049)(2047) = 14343 + 81960 + 000000 + 4098000 = 96303+4098000 = 94303+4100000 = 4194303 Alternatively: Know your powers of 2. (Admittedly that's a bit difficult to know them up to the 22nd power)
I did it another way, but it also works. Thank you teacher. 2^22 - 1 = 2^5 x 2^5 x 2^5 x 2^5 x 2^2 - 1 = 32 x 32 = 1024 x 32 = 32768 x 32 = 1048576 x 4 = 4194304 - 1 = 4194303
Greetings. Thanks always for sharing. That was a very good exercise. Mentally, we could determine the value of 49 times 47 by considering base 40 times 56, the sum of 49 and 7 or 47 and 9, plus 9 times 7. That is (40×56)+(9×7)= 2240+63=2303.
2 to the nth power being 22. 2x10 = 20 (2x2) = 4 (2x2x2) = 8 now we have 2048 2048x2048 = 4194304 You cannot end it with a 3 in anyones mathamatical language. Your method is far too complcated. Imagine having to memorise all those steps. Oh boy !!!!! CRAZY is Right 😂😂😂😂😂
this is an example of how to make a complicated solution for simple question.
Ok. So 2^22 is 2^11 squared. 2^11 = 2048. So, picture a square grid, 2048 by 2048. First take care of the 2000 by 2000 part - that's 4,000,000. Then add 2*2000*48 = 192,000, so we're up to 4,192,000. That leaves a 48 by 48 grid in the top right corner to cover. Let's add a 50 by 50 grid instead, which adds 2,500, so we're up to 4,194,500. Now subtract the two extra rows and the two extra columns - 100 each. So 4,194,300. Finally, add back the 2 by 2 grid we subtracted twice, so 4,194,304. Finally, subtract the 1. 4,194,303. Q.E.D.
You can do this in your head if you can hold a couple of numbers in memory, and you can ABSOLUTELY do it on scratch paper in under a minute.
2^22 is 2^10 x 2^12 = 1024 x 4096 which is quite easy by hand, then subsract 1
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Anyone who has done a lot of programming and worked in binary and hexadecimal can go through powers of two in seconds. A few common powers known for memory are 2^10 = 1024 or 1K and 2^20 = 1049576 or 1M, So multiply 2^20 by 4 and subtract 1 you get 4194303. You would think by now the Math Olympiad wouldn't use this type of question.
2^22 = (2^10)^2 x 2^2
You're always doing great...
And make mathematics simplest
Thank you
The well known of "2 power N", for large N in infornation system is 256 (2^8,, 8-bit), 1024 (2^10, kilobyte), 65.536 (2^16, 16-bits), and 16.777.216 (2^24, 24-bits, RGB color)..
So,
2^22 - 1
= 2^16 x 2^6 - 1
= 65.536 x 64 - 1
Nice but calculator is not allowed
No calculator needed. Just paper and pencil .@@JJONLINEMATHSCLASSchannel
Before watching:
2^22 -1 = (2^11 +1)(2^11 -1)
2^11 = 2048, so this is (2049)(2047)
= 14343
+ 81960
+ 000000
+ 4098000
= 96303+4098000 = 94303+4100000 = 4194303
Alternatively: Know your powers of 2. (Admittedly that's a bit difficult to know them up to the 22nd power)
I did it another way, but it also works. Thank you teacher.
2^22 - 1 =
2^5 x 2^5 x 2^5 x 2^5 x 2^2 - 1 =
32 x 32 = 1024 x 32 = 32768 x 32 = 1048576 x 4 = 4194304 - 1 = 4194303
✅👍
Greetings. Thanks always for sharing. That was a very good exercise. Mentally, we could determine the value of 49 times 47 by considering base 40 times 56, the sum of 49 and 7 or 47 and 9, plus 9 times 7. That is (40×56)+(9×7)=
2240+63=2303.
😲😲, nice one. Thanks for the contribution
2^22 = (2^11)^2
2^11 = 2048
It's not that difficult toto multiply 2048 x 2048 = 4194304
than subtract 1 you get 4194303
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I don’t have to think about it.
It is well
2^22=2^11*2
=2048^2
=2048*2048
=4194304
2^22-1=4194304-1
=4194303
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Thank you
normal calculation
2^10 =1024
you multiply that with 2 until 2°22 is arrived.
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Binary 2^22 = 1000000000000000000000 - 1 = 1111111111111111111111
2^11 ➖ 1 2^11^1 ➖ 1 2^1^1 ➖ 1 21 (x ➖ 2x+1).
Thank you, mam, I am stupid in mathematics. I would get better from now on.
👍👍👍👍🤣, but you are smart ok?
2 to the nth power being 22. 2x10 = 20 (2x2) = 4 (2x2x2) = 8 now we have 2048 2048x2048 = 4194304 You cannot end it with a 3 in anyones mathamatical language. Your method is far too complcated. Imagine having to memorise all those steps. Oh boy !!!!! CRAZY is Right 😂😂😂😂😂
483
Muito linda esta mulher
😅2^22-1=4(10^3+24)(10^3+24)-1=4(10^6+4.8x10^4+.0576x10^4)-1=4(1.048576x10^6)-1=
4194303
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2^22-1=4194303 Final Answer JJ
👏🎓👏👏
it is < 4 194 303 > and done in no time, what is not all too crazy, since up to 2^30, I got the numbers all stored in my mind.
Le p'tit Daniel
Alla fine è solo fare una moltiplicazione che stronzata