Skip to main content
ttokoro
21-Topaz I
ttokoro21-Topaz I
21-Topaz I
November 5, 2021
Solved

How to find the answer?

  • November 5, 2021
  • 3 replies
  • 4267 views

image.pngimage.png

 

2^11 is 2048. How to find a and b? a and b are integers. 

Best answer by StuartBruff
@LucMeekes wrote:

Mathcad /Maple will not be fooled:

 

If you evaluate that numerically, it says 2048,

but that's due to poor (with respect to symbolic) numerical precision,

similar to the numerical results I presented above.

 

Luc

Yes, I'd expect that.  However, I do note that Tetsuro specifically mentioned Prime 7.

 

If, as I implied by my mention of infinities, you plug in -∞ to the expression in Maple you get 2048 symbolically (at least, in Maple Flow you do).

 

(Mathematica gives "Sum[2^n, {n, -Infinity, 10}] = 2048", as well.  I would hope Mathcad Prime 7 gives the same result?).

 

Stuart

 

(Of course, there's the thorny issue of whether -Infinity counts as an integer)

 

 

3 replies

S
StuartBruff23-Emerald V
23-Emerald V
November 5, 2021
@ttokoro wrote:

image.pngimage.png

 

2^11 is 2048. How to find a and b? a and b are integers. 

AFAIA, the binary expansion of any integer is unique.   

 

2021 11 05 A.png

 

Stuart

    ttokoro
    21-Topaz I
    ttokoro21-Topaz IAuthor
    21-Topaz I
    November 5, 2021

    Prime 7 shows another a and b. 

    t.t.
    S
    StuartBruff23-Emerald V
    23-Emerald V
    November 5, 2021
    @ttokoro wrote:

    Prime 7 shows another a and b. 

    Given the Puzzles and Games Label, I thought it might.  😈

     

    Stuart

    L
    LucMeekes23-Emerald IV
    23-Emerald IV
    November 5, 2021

    Hmmm,

     

    LucMeekes_0-1636106807423.png

    Never expected this would happen.

     

    But then, if b is a positive integer <10 then a is complex

    for b=10 I get an exception due to trying to take ln(0).

    for b=11 , a=11.

    for b>11 a is a positive real.

    Ah! b=22 gives a=23 (numerically), but:

    LucMeekes_1-1636107583494.png

    And with every higher multiple of 11 for b you get closer to a being b+1. for b=1012:

    LucMeekes_2-1636107803468.png

    and you don't even need multiples of 11 anymore. For b=1013 the evaluation gives:

    LucMeekes_3-1636107877950.png

    But at b=1023 the numerical evaluation fails.

    LucMeekes_4-1636108053736.png

     

    Luc

      S
      StuartBruff23-Emerald V
      23-Emerald V
      November 5, 2021

      Neat. 

       

      Something else that occurred to me in my sleep-deprived state early this morning, but which I promptly forgot about until now, was based on one of the many peculiarities of infinities.

       

      2021 11 05 D.png

       

      (a = -38 does the trick numerically, but I wanted to give the impression of a really large number)

       

      I don't have a symbolic processor to hand to check its result ...

       

      Stuart

      L
      LucMeekes23-Emerald IV
      23-Emerald IV
      November 5, 2021

      Mathcad /Maple will not be fooled:

      LucMeekes_0-1636118929777.png

      If you evaluate that numerically, it says 2048,

      but that's due to poor (with respect to symbolic) numerical precision,

      similar to the numerical results I presented above.

       

      Luc

      W
      Werner_E25-Diamond I
      25-Diamond I
      November 5, 2021

      > a and b are integers

      Mathcad never was able to solve diophantic equations and the symbolics ever so often ignore an "assume, xxx=integer"

      Werner_E_0-1636109505281.png

      You probably are aware thats its quite easy to manually arrive at the unique solution a=b=11.

      Given that the sum is

      Werner_E_1-1636109633320.png

      with a<=b<=11, you can do as follows

      Werner_E_0-1636110818942.png

      Somewhat shorter:

      Werner_E_0-1636120601654.png

       

      But the shortest sure is Stuarts remark about the binary representation of a number being unique.

       

       

       

        Terms & ConditionsAccessibility statement