Guest Posted April 7, 2010 Report Share Posted April 7, 2010 (edited) Susan was investigating patterns in square grids of numbers. The diagram show the pattern that she found Edited April 7, 2010 by Eragon Quote Link to post Share on other sites

0 Guest Posted April 7, 2010 Report Share Posted April 7, 2010 (edited) x + (x+1) * (x-1)*1/2 = 5335 x- is the length of the side ->solution x = 22 so number in the right bottom corner is 22^2 = 484 Edited April 7, 2010 by det Quote Link to post Share on other sites

0 Guest Posted April 7, 2010 Report Share Posted April 7, 2010 the answer is 484 which would be a 22x22 grid The math behind this, and I may have made it a bit complicated is as follows: The sum of the diagonal is the corner number x the hight of the grid, divided by 2 (two) and then added to half the number of the hight of the grid. Or: X = (FNxGH)/2 + (GH/2) (where X is the sum of the diagonal, FN is the final number/corner number and GH is the grid hight) Sorry if I have expressed this wrong, but it has been many years since high school maths!!! Quote Link to post Share on other sites

0 Guest Posted April 7, 2010 Report Share Posted April 7, 2010 Sorry Det, but I can't make your equation work, I have the same answer as you, but when I feed in the numbers to your equation, I don't get 5335!! what am I doing wrong? Quote Link to post Share on other sites

0 Guest Posted April 7, 2010 Report Share Posted April 7, 2010 Sorry Det, but I can't make your equation work, I have the same answer as you, but when I feed in the numbers to your equation, I don't get 5335!! what am I doing wrong? sorry, my mistake x + (x+1) * (x-1)*x/2 = 5335 Quote Link to post Share on other sites

0 Templeton 0 Posted July 4, 2020 Report Share Posted July 4, 2020 For a given NxN square there are N diagonals, starting with N in the top right corner. Each number in the sequence is below and one to the left of the previous number. So it is N - 1 more than the previous number. For a given N this forms an arithmetic progression with d = N -1. So S = (n/2) (2a + (n-1)d) where S = 5335, n = N, a = N, d = N - 1 Substitute in and arrange to get: N^{3} + N = 10670 Solve for N and find N^{2} for the bottom right entry. This equation is a depressed cubic (no N^{2} term) and can be solved by the Cardano method ( http://www.sosmath.com/algebra/factor/fac11/fac11.html ) to give N = 22. Alternatively, note that, for N much greater than 10, N^{ }is less than 1% of N^{3} and can be ignored. This gives N^{3} ~ 10670 N ~ 22 So the bottom right entry is 22^{2} = 484 Quote Link to post Share on other sites

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Susan was investigating patterns in square grids of numbers. The diagram show the pattern that she found

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