19 thoughts on “Modulus Functions in Two Variables – 2

  1. If I apply pick’s theorem,
    A = I + b/2 -1
    A + 1 – b/2 = I
    Here, A =128
    b = (7*4) + 4 = 32
    we get a different I. Can you please tell where am I going wrong.
    Thanks in advance.

    • Hi Arpita, I am not actually well-versed with Pick’s theorem, but as I recall the “I” over there is the interior points (excluding the boundary points). Here the question was effectively asking for “I + b” as it asks “within or on the boundary of…”

      Consider A = I + b/2 – 1 => 128 = I + 16 – 1 => I = 113….now if we want point in the interior and on the boundary, that is I + b = 113 + 32 = 145 as above.

      This is why I am a little wary of direct formulae – you have to be doubly careful to ensure you are applying them correctly and if you go wrong there is no indication…but in a logical (though slightly slower) approach, accuracy is more likely for someone like me.

      regards
      J

      • Hi J,
        yes. I missed that. true about that formulae and approach thing. You can never be sure with the formulae. I just wanted to see its application. Thanks a lot.

  2. Jaspreet, if you like you can make it a formula in that manner. But it almost certainly quicker to just count for small numbers 🙂 Also, this will not work if the coefficients of x and y are different (something future posts will dwell on)

    regards
    J

  3. Sir, first of all, your blogs are very insightul. I get to learn a lot. Thank you!

    I ‘m not understanding the solution to number of integer points on or within |x|+|y|=8. Please help! Thank you!

    • Parveen, I am just saying that x cannot go beyond the range of [-8, +8]. Plugging in the values of x, and finding what are the possible values of y corresponding to each x, we can quickly establish a pattern and count. Try listing them out manually for a smaller case first (say |x| + |y| <= 3, which should give you around 25 points I guess) and you will get a feel for the pattern of the problem.

      regards
      J

  4. Is there any shorter way to find integer solutions without plotting those points . We will not be given a graph sheet in the exam.IT looks like its time consuming

    • I didn’t need to draw the graph to that level of accuracy 🙂 I drew a precise graph only to illustrate the pattern; once you are comfortable with the behaviour of such a mod function you don’t even need to draw the graph in many cases! See the next few posts and try and create a few questions of your own; you will get the hang of it soon enough.

      regards
      J

  5. You calculated the integer points by pattern. Will there be a case where there is no pattern and it that case how should we proceed?

    • Not impossible, but unlikely. In that case, you will have to consider all the values of x individually, which will require about twice as much effort, but is still doable.

      regards
      J

  6. hi .. I tried calculating the number of points in the following manner :
    it took all the points on the vertices first
    (4 + 4(7)) for the outermost square + (4+4(6)) for the second outermost square … and so on + 1 in the last as the central point
    solving I get 28+4(1+2+4+… +7) +1 =29+112=141 can you tell me where am I going wrong ?

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