Olympiad Maths Problem : Can you solve for Exponents X and Y

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  • Опубліковано 14 жов 2024
  • Join me in this new video on how to solve a challenging olympiad type maths problem. Can you do it?
    The problem is find x and y from 2^x-2^y=496
    #exponentialequation #exponential #olympiadmathematics
    #mathsplus
    If you are new to the channel don't forget to give a like , subscribe and share.

КОМЕНТАРІ • 60

  • @SuperYoonHo
    @SuperYoonHo 2 роки тому +4

    Oh yes! Awesome question!!! Thank you so much.
    P.S. TOO EASY BRO

    • @mathsplus01
      @mathsplus01  2 роки тому +1

      Thank you. Have a great day.

    • @SuperYoonHo
      @SuperYoonHo 2 роки тому +1

      @@mathsplus01 You are most welcome and thank you for your kind words and have a good day

  • @mathsplus01
    @mathsplus01  2 роки тому +5

    Welcome everyone to this Olympiad Maths Question. If anyone has any questions about it I will be delighted to answer you below🤔

  • @Grim_Reaper_from_Hell
    @Grim_Reaper_from_Hell 2 роки тому +2

    In old days 1st you try to guess and then you solve. This one is easy to guess.
    Take the closest power of 2 that is higher than 496 and it is 512 and the difference is power of 2 as well. Problem solved in 2 seconds.

  • @onenote3769
    @onenote3769 2 роки тому +5

    Very interesting problem. Thank you for your videos!

    • @mathsplus01
      @mathsplus01  2 роки тому

      Cheers onenote. Thanks for the support and bless you

  • @RexxSchneider
    @RexxSchneider Рік тому +1

    Motivation: when you have a difference on one side and a number on the other, you always try to convert the difference into a product and then you can compare the product with the factors of the number. We know that 2^n > 0 so it follows that 2^x > 2^y, i.e. x > y. So we take the smaller factor 2^y out of both of the terms on the LHS and factorise the RHS:
    2^y * (2^(x-y) - 1) = 2^4 * 31. The only way that can be true (for integer x and y) is if 2^y = 2^4, and 2^(x-y) - 1 = 31, meaning 2^(x-y) = 32, so 2^(x-y) = 2^5. That gives y = 4 and x = 4+5 = 9.

  • @peterbyrne6394
    @peterbyrne6394 2 роки тому +2

    Thank you, grand solution.

  • @richardl6751
    @richardl6751 Місяць тому +1

    2^x-2^y can be used to find perfect numbers where x=1, 3, 5, 7... and y=0, 1, 2, 3... Not all numbers from this formula are perfect but all perfect numbers fit this formula.

    • @mathsplus01
      @mathsplus01  Місяць тому

      Thank you Richard, very interesting

  • @jerrypaquette5470
    @jerrypaquette5470 Рік тому +2

    They make these things more complicated than they have to. Yes 2^x > 496 and 2^x > 2^y. Remember the powers of 2: 2,4,8,16, etc. The first one that is greater than 496 is 512 which equal to 2^9. 512 - 496 = 16 = 2^4. Therefore X = 9, and Y =4

  • @tatayyanaidugaraga801
    @tatayyanaidugaraga801 2 роки тому +3

    I will support you bro
    Now it is 397 subs it will raise in few days wait for that

    • @mathsplus01
      @mathsplus01  2 роки тому +1

      Thank you. Blessings to you and your family.❤

    • @tatayyanaidugaraga801
      @tatayyanaidugaraga801 2 роки тому +2

      @@mathsplus01 i did it
      I had raisen your subscribers
      Keep that confident and keep growing
      Best of luck my dear friend

    • @mathsplus01
      @mathsplus01  2 роки тому +1

      @@tatayyanaidugaraga801 I am very grateful for all your help. You have great spirit of kindness within you. Bless you. A million Thanks for you

    • @tatayyanaidugaraga801
      @tatayyanaidugaraga801 2 роки тому +1

      @@mathsplus01 Now iam quiting
      Good bye my dear friend

  • @mathanalylitics2375
    @mathanalylitics2375 2 роки тому +3

    Good solution. Impressive as it appears simple in the end.

    • @mathsplus01
      @mathsplus01  2 роки тому

      Always a big thanks for your support. Bless you from Ireland!

  • @satrajitghosh8162
    @satrajitghosh8162 2 роки тому +2

    496 = 16*31 = 16* ( 16+8+4+2+1)
    = 16*(32-1) = 2^9 - 2^4
    Hereby x = 9 and y = 4

  • @SamaySanyasi
    @SamaySanyasi 2 роки тому +2

    Let's take
    x = log2(p) and y = log2(q)
    The equation will becomes:
    p-q = 496 ........ (Eq1)
    Because of log2 we also have following constraints:
    p>0 ...... (Eq2)
    q>0 ...... Eq3)
    Any tuple (p,q) satisfies Eq 1 2 3 is the solution. There are infinite solution.

    • @kohwenxu
      @kohwenxu Рік тому

      Video should really state that we are finding integer values of x and y.

  • @owlsschoolofmath9732
    @owlsschoolofmath9732 Рік тому +1

    I clicked on this thumbnail without knowing what it was and then I was like “hey that sounds kind of like Maths Plus!” 😂😂😂 Nice video!

    • @mathsplus01
      @mathsplus01  Рік тому +1

      Haha Thank you Owls School of Maths Sir , glad to know the videos still circulate like that also.

    • @owlsschoolofmath9732
      @owlsschoolofmath9732 Рік тому +1

      @@mathsplus01 right!!! The algorithm is still serving it up! 🤣

  • @owlsmath
    @owlsmath 2 роки тому +2

    Nice! Love these types of problems.

    • @mathsplus01
      @mathsplus01  2 роки тому

      They are very popular also. I keep them simple and keep to real solutions so most people can understand.

    • @mathsplus01
      @mathsplus01  2 роки тому

      did you try using screen record and writing on a digital whiteboard? All you need a cheap writing tablet. I would be able to give you a few tips if you would to do your videos like i do. But the way you do them on video works just fine also

    • @owlsmath
      @owlsmath 2 роки тому

      @@mathsplus01 yes! So if you happen to scroll back in my history all my old videos were on the digital whiteboard and I actually kind of like it! But currently my digital setup is broken and I haven't gotten around to fixing it. I am not sure if i would go 100% digital again but would like to actually try and compare. So yes, when I give that a try again I could probably use some tips because I was actually struggling quite a bit with the setup! 🤣

    • @owlsmath
      @owlsmath 2 роки тому +1

      It's kind of ironic because I created my website obviously but don't know how to work the whiteboard 🤣

    • @mathsplus01
      @mathsplus01  2 роки тому +1

      @@owlsmath ok i understand, i think your maths is very good, the main way i improved in recent weeks was to do the video in sections, then merge together in adobe merge video for free. There is less pressure do all in one go also.

  • @maydin34
    @maydin34 2 роки тому +3

    496 = 31 * 16 = 31 * 2^4 , then 2^(x-4) - 2^(y-4) = 31 , 31 = 32 - 1 --> 31 = 2^5 - 2^0 , then x-4 = 5 -> x=9, and similarly, y - 4 = 0 --> y =4

    • @mathsplus01
      @mathsplus01  2 роки тому

      Thank you so much. Really good solution there👍

  • @هشامحلمىالسيد
    @هشامحلمىالسيد 2 роки тому +3

    In one step 496=512_16 =2^9_2^4 then x=9, y=4

  • @ATeima-kk5ps
    @ATeima-kk5ps 2 роки тому +2

    Interesting problem. I had a really different approach.
    let’s represent 496 in binary
    2^x-2^y=2^8+2^7+2^6+2^5+2^4
    2^x=2^8+2^7+2^6+2^5+2^4+2^y
    It’s easy to see that this is only possible when y = 4, because each term will get added to the one on its left eventually leading to one term remaining.
    2^x=2^9
    x=9
    Therefore x=9, y=4

  • @ellyjauharaha8190
    @ellyjauharaha8190 2 роки тому +2

    Thank you so much

    • @mathsplus01
      @mathsplus01  2 роки тому

      You are most welcome❤ Have a nice day!

  • @adammohamed5256
    @adammohamed5256 2 роки тому +2

    Yes enjoyed the whole video. It's class A work

    • @mathsplus01
      @mathsplus01  2 роки тому +1

      Thank you Adam Mohamed, I am glad you liked it. Blessings from Ireland

  • @zavian999
    @zavian999 2 роки тому +1

    I have been doing a chapter for the last few days... I thought I'll complete it today... 2 questions were left and I was just finding no interest in it for now... I came through your video... I found it really interesting, so I clicked and loved the problem... Great explanation.... Thank you so much 🙏

    • @mathsplus01
      @mathsplus01  2 роки тому +1

      Always do the maths in small steps. This is how you make progress. I know you will succeed!

  • @barryjackson405
    @barryjackson405 2 роки тому +3

    X=9,y=4

  • @vishwambharmahant9485
    @vishwambharmahant9485 2 роки тому +2

    X =9&y =4

    • @mathsplus01
      @mathsplus01  2 роки тому

      Fantastic work, correct. Keep going and well done to you!

  • @gnyandevkshirsagar
    @gnyandevkshirsagar 2 роки тому +2

    X=9; Y= 4

  • @irwandasaputra9315
    @irwandasaputra9315 2 роки тому +2

    x=9 dan y=4

    • @mathsplus01
      @mathsplus01  2 роки тому

      Yes, correct. You really get them right. Well done!

  • @SirNobleIZH
    @SirNobleIZH 2 роки тому +2

    Just memorizing my powers of 2 was enough

    • @mathsplus01
      @mathsplus01  2 роки тому

      Still solved it so you did it. Well done!

  • @himo3485
    @himo3485 2 роки тому

    x=k+y 2^k+y - 2^y = 496 2^y(2^k-1) = 496
    2^y=16=2^4 y=4 2^k-1=31 2^k=32=2^5 k=5 x=5+4=9
    ( x=9 , y=4 )

  • @e1woqf
    @e1woqf 2 роки тому +2

    512 - 16 = 496
    512 = 2⁹ and 16 = 2⁴
    => x = 9 and y = 4
    takes less than 30 seconds

    • @mathsplus01
      @mathsplus01  2 роки тому

      Aren't you a genius

    • @e1woqf
      @e1woqf 2 роки тому +1

      @@mathsplus01 Sorry, but to solve this problem in 30 seconds doesn't take a genius, if you have seen a similar problem on another channel. 😃

  • @irwandasaputra9315
    @irwandasaputra9315 2 роки тому +1

    x=9 dan y=4