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

post-28691-12706397335696.png

post-28691-12706398955699.png

Edited by Eragon

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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 by det

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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!!!

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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?

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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

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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:

N3 + N = 10670

Solve for N and find N2 for the bottom right entry.

This equation is a depressed cubic (no N2 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 N3 and can be ignored.

This gives N3 ~ 10670

N ~ 22 

So the bottom right entry is 222 = 484

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