# Urgent Help Needed PLEASE

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• Apr 13th 2007, 12:05 AM
nbukhari
Urgent Help Needed PLEASE
Hi i am trying to solve a number grid problem by 4x4 block finding the product of the top left number and the bottom right number in the block doing the same with the top right and bottom left numbers. Calculating the difference so far this is what i have done

1 2 3 4
11 12 13 14
21 22 23 24
31 32 33 34

I have chosen 1 as the upper left
Multiply 1 x 12 x 23 x 34 = 9384
Multiply 4 x 13 x 22 x 31 = 35464
Difference = 26080

Next 4x4 block

2 3 4 5
12 13 14 15
22 23 24 25
32 33 34 35

Multiply 2 x 13 x 24 x 35 =21840
Multiply 5 x 14 x 23 x 32 = 51520
Difference = 29680

Notice that there is a difference of 3600 between the 1st and the 2nd block (29680 - 2608 = 3600)

I then did another 4x4 block where 3 was the upper left and another 4x4 block 4 as the upper left and got a difference of 3800 i am so confused can any kind person pls explain am i doing this right and how can i generate a formula of D (DIFFERENCE) involving N (NUMBER OF SQUARES). your help would be very much appreciated also if you can explain the 5x5 block
Thank you:confused: :confused: :confused: :confused:
• Apr 14th 2007, 04:53 AM
topsquark
Quote:

Originally Posted by nbukhari
Hi i am trying to solve a number grid problem by 4x4 block finding the product of the top left number and the bottom right number in the block doing the same with the top right and bottom left numbers. Calculating the difference so far this is what i have done

1 2 3 4
11 12 13 14
21 22 23 24
31 32 33 34

I have chosen 1 as the upper left
Multiply 1 x 12 x 23 x 34 = 9384
Multiply 4 x 13 x 22 x 31 = 35464
Difference = 26080

Next 4x4 block

2 3 4 5
12 13 14 15
22 23 24 25
32 33 34 35

Multiply 2 x 13 x 24 x 35 =21840
Multiply 5 x 14 x 23 x 32 = 51520
Difference = 29680

Notice that there is a difference of 3600 between the 1st and the 2nd block (29680 - 2608 = 3600)

I then did another 4x4 block where 3 was the upper left and another 4x4 block 4 as the upper left and got a difference of 3800 i am so confused can any kind person pls explain am i doing this right and how can i generate a formula of D (DIFFERENCE) involving N (NUMBER OF SQUARES). your help would be very much appreciated also if you can explain the 5x5 block
Thank you:confused: :confused: :confused: :confused:

I am assuming the number block is 10x(something)?

If you have a 4x4 block and a number n in the upper left corner, then the next number in the diagonal leading to the lower right corner will be n + 11, the number after that will be (n + 11) + 11 = n + 22, and the last number in the diagonal (the lower right corner) will be (n + 22) + 11. So the product along this diagonal will be:
n(n + 11)(n + 22)(n + 33)

To do the other diagonal, the number in the top right corner will be n + 3, the next number in the diagonal will be (n + 3) + 9 = n + 12, then (n + 12) + 9 = n + 21, then (n + 21) + 9. This product will be:
(n + 3)(n + 12)(n + 21)(n + 30)

So the difference formula will be:
(n + 3)(n + 12)(n + 21)(n + 30) -
n(n + 11)(n + 22)(n + 33)

Now for the ugly part. You really ought to multiply these out and find the difference. I get:
100n^2 + 3300n + 22680

-Dan
• Apr 14th 2007, 06:33 AM
Soroban
Hello, nbukhari!

I'm still working on the 5×5 problem,
. . but I can verify what Dan did for the 4×4.

At the same time, I'll dazzle you with some algebraic gymnastics (I hope).

Reading down the southeast diagonal, the product is:

. . n(n + 11)(n + 22)(n + 33)

. . . = .[n(n + 33)]·[(n + 11)(n + 22)]

. . . = .(n² + 33)(n² + 33n + 242)

. . . = .[(n² + 33n + 121) - 121]·[(n² + 33n + 121) + 121]

. . . = .(n² + 33n + 121)² - 121²

Reading down the southwest diagonal, the product is:

. . (n + 3)(n + 12)(n + 21)(n + 30)

. . . = .[(n + 3)(n + 30)]·[(n + 12)(n + 21)]

. . . = .(n² + 33n + 90)(n² + 33n + 252)

. . . = .[(n² + 33n + 171) - 81]·[(n² + 33n + 171) + 81]

. . . = .(n² + 33n + 171)² - 81²

The difference is: .[(n² + 33n + 171)² - 81²] - [(n² + 33n + 121)² - 121²]

. . . = .(n² + 33n + 171)² - (n² + 33n + 121)² + 121² - 81²
. . . . . \_______________________________/
. . . . . . . . . . . . .
difference of squares

. . . = .[(n² + 33n + 171) - (n² + 33n + 121)]·[(n² + 33n + 171) + (n² + 33n + 121)] + 8080

. . . = .50(2n² + 66n + 292) + 8080

. . . = .100n² + 3300n + 22680 . . . . There!