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Test of Mathematics Solution Subjective 59 - Number of squares

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 59 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Consider the set of point S = { (x,y) : x,y are non-negative integers {\le {n}} }.
Find the number of squares that can be formed with vertices belonging to S and sides parallel to the axes.

Solution

S = {(x,y) : x,y are non-negative integers {\le {n}} }
We calculate number of squares by calculating number of |x| squares ,& number of squares number of {{n}* {n}} squares.
Now number of |x| squares = number of choosing one pair of lines with difference 1 parallel to x axis & integer distance x number of choosing one pair of lines to y axis with distance 1 & integer distance from y axis = {{n}*{n}} = {n^2}
Similarly number of {{k}*{k}} squares
= {(n-k+1)^2}
So total number of squares
= {\sum_{k=1}^{n}}{{k}^{2}} = {\frac{n(n+1)(2n+1)}{6}}

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 59 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Consider the set of point S = { (x,y) : x,y are non-negative integers {\le {n}} }.
Find the number of squares that can be formed with vertices belonging to S and sides parallel to the axes.

Solution

S = {(x,y) : x,y are non-negative integers {\le {n}} }
We calculate number of squares by calculating number of |x| squares ,& number of squares number of {{n}* {n}} squares.
Now number of |x| squares = number of choosing one pair of lines with difference 1 parallel to x axis & integer distance x number of choosing one pair of lines to y axis with distance 1 & integer distance from y axis = {{n}*{n}} = {n^2}
Similarly number of {{k}*{k}} squares
= {(n-k+1)^2}
So total number of squares
= {\sum_{k=1}^{n}}{{k}^{2}} = {\frac{n(n+1)(2n+1)}{6}}

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