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Explore the Back-StoryThis 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

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.

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}} $

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

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.

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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