19 thoughts on “Modulus Functions in Two Variables – 2”
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.
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.
for |x| + |y| = n,this implies no. of pts will be given by 2*sum of AP – (n+1),AP will be 1,3,…upto n+1 terms..k?
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)
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.
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.
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.
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 ?
CAT-holics 
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.
for |x| + |y| = n,this implies no. of pts will be given by 2*sum of AP – (n+1),AP will be 1,3,…upto n+1 terms..k?
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
Can we use the formula for |a| + |b| <= r as (n+r-1)C(r-1) using the same logic you mentioned in this post.
If it is possible can i also use the logic mentioned here in this post
https://cat100percentile.com/2013/05/16/distribution-of-some-or-all-objects-identical-groups/
yes mr.scrabbler.. will wait for complex stuff..hit us with whatever u got :P.. all ready 😀
sir how will be the curve of |X| |Y|=8
2 rectangular hyperbolas given by xy=8 and xy=-8 matlab each quadrant mein u’ll have 1/x kinda curve..right sir?
Yes Jaspreet, that seems accurate 🙂
regards
J
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
I get it sir..Thank you!
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
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
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 ?
oops .. got it .. it will go till 4 + 4(0) .. so 4 points will be added
thanks 😛