Number Grid

Second example

73 74 75

83 84 85

93 94 95

Top left hand number = 73 Bottom right hand number = 95

Top right hand number = 75 Bottom left hand number = 93

So, 73 x 95 = 6935 75 x 93 = 6975

So, difference is,

largest number - smallest number = 6975 - 6935 = 40 (same as above)

For the 3 x 3 grid the difference seems to be 40.

Following the same procedure as before,

n n+1 n+2

n+10 n+11 n+12

n+20 n+21 n+22

Top left hand number = n Bottom right hand number = n+22

Top right hand number = n+2 Bottom left hand number = n+20

Top left hand number x Bottom right hand number

n x (n+22) = n + 22n

Top right hand number x Bottom left hand number

(n+2) x (n+20) = nxn + nx20 + 2xn + 2x20 = n + 20n + 2n + 40 = n + 22n + 40

Difference (n + 22n + 40) - (n + 22n) = 40

Therefore, the difference will always be 40 for a 3x3 square on a number grid 10 long.

Extend investigation

Same number grid but now choose 4x4 grid, but now go to the general case.

n n+1 n+2 n+3

n+10 n+11 n+12 n+13

n+20 n+21 n+22 n+23

n+30 n+31 n+32 n+33

Top left hand number = n Bottom right hand number = n+33

Top right hand number = n+3 Bottom left hand number = n+30

Top left hand number x Bottom right hand number

n x (n+33) = n + 33n

Top right hand number x Bottom left hand number

(n+3) x (n+30) = nxn + nx30 + 3xn + 3x30 = n + 30n + 3n + 90 = n + 33n + 90

Difference (n + 33n + 90) - (n + 33n) = 90

Now look at square size and differences

2 x 2 10

3 x 3 40

4 x 4 90

predict for 5 x 5 the difference will be 160. Why?

This is because

2 x 2 10 = 1 x 10

3 x 3 40 = 2 x 10

4 x 4 90 = 3 x 10

so,

I predict 5 x 5 will be 4 x 10 = 16 x 10 = 160

Check

n n+1 n+2 n+3 n+4

n+10 n+11 n+12 n+13 n+14

n+20 n+21 n+22 n+23 n+24

n+30 n+31 n+32 n+33 n+34

n+40 n+41 n+42 n+43 n+44

Top left hand number = n Bottom right hand number = n+44

Top right hand number = n+4 Bottom left hand number = n+40

Top left hand number x Bottom right hand number

n x (n+44) = n + 44n

Top right hand number x Bottom left hand number

(n+4) x (n+40) = nxn + nx40 + 4xn + 4x40 = n + 40n + 4n +160 = n + 44n + 160

Difference (n + 44n + 160) - (n + 44n) = 160

General rule for squares of size w

Rules so far

2 x 2 10 = (2-1=1) 1 = 1 x 10 = 10

3 x 3 40 = (3-1=2) 2 = 4 x 10 = 40

4 x 4 90 = (4-1=3) 3 = 9 x 10 = 90

5 x 5 160 = (5-1=4) 4 = 16 x 10 =160

The size of the square subtract 1, square the answer, multiply it by 10

the pattern continues as

w x w = (w-1) x 10

The general rule for the difference using squares of size w on a number grid 10 long is

(w-1) x 10 or 10(w-1)

Proof for a square w x w

Starting position

Any number width = w

n n+1 n+2 n+3 . . n+(w-1) The amount added

n+(1x10) . . . . . . onto n is one less

height n+(2x10) . . . . . . than the square (w-1)

= w n+(3x10) . . . . . .

. . . . . . .

. . . . . . .

n+10(w-1) . . . . . n+10(w-1) +(w-1)

Each row goes up 10, there are one less Need to add a further (w-1)

lots of 10 than the height w, so we need to add on to the first square in this row

(w-1) lots of 10 (same as the first row)

Top left hand number = n

Top right hand number n+(w-1) = n+w-1

Bottom left hand number n+10(w-1) = n+10w+10

Bottom right hand number n+10(w-1)+(w-1) = n+10w-10+w-1 = n+11w-11

Top left hand number x Bottom right hand number

n x (n+11w-11) = n + 11nw-11n

Top right hand number x Bottom left hand number

(n+w-1)(n+10w-10)

which gives

nxn + nx10w +nx(-10)+ wxn+wx10w+wx(-10)+(-1)xn+(-1)x10w+(-1)x(-10)

which gives

n + 10nw - 10n +nw +10w -10w - n - 10w + 10 = n + 11nw-11n + 10w - 20w + 10

Difference equals (n + 11nw-11n + 10w - 20w + 10) - (n + 11nw-11n)

Which gives

10w - 20w + 10 = 10(w - 2w + 1) = 10(w-1)

Extend investigation

Instead of considering squares now look at rectangles.

6 17 18

26 27 28

So, 16 x 28 = 448 18 x 26 = 468

Difference is largest - smallest = 468 - 448 = 20

We can consider the solution to the squares since the rectangles just alters the length of one of the sides. The following is a 3x5 rectangle, all that has happened is the length has increased, (or the height has decreased, doesn't matter, one of the dimensions has changed)

n n+1 n+2 n+3 n+4

n+10 n+11 n+12 n+13 n+14

n+20 n+21 n+22 n+23 n+24

The solution to the above rectangle is the same as before

Top left hand number = n Bottom right hand number = n+24

Top right hand number = n+4 Bottom left hand number = n+20

Top left hand number x Bottom right hand number

n x (n+24) which gives n + 24n

Top right hand number x Bottom left hand number

(n+4) x (n+20) = nxn + nx20 + 4xn + 4x20 = n + 20n + 4n +80 = n + 24n + 80

Difference equals (n + 24n + 80) - (n + 24n) = 80

So for all 3x5 rectangles the difference is 80

How do we get to the general rule for rectangles?

Answer, we consider the proof for squares and change it slightly.

For a square w x w, change it to w x h for the rectangles

Starting position width = w

Any number

n n+1 n+2 n+3 . . n+(w-1)The amount added

n+(1x10) . . . . . . onto n is one less

height n+(2x10) . . . . . . than the width.

= h n+(3x10) . . . . . .

. . . . . . .

. . . . . . .

n+10(h-1) . . . . . n+10(h-1) +(w-1)

Each row goes up 10, there are one less Need to add a further (w-1)

lots of 10 than the height w, so we need to add on to the first square in this row

(h-1) lots of 10 (same as the first row)

Top left hand number = n

Top right hand number n+(w-1) = n+w-1

Bottom left hand number n+10(h-1) = n+10h+10

Bottom right hand number n+10(h-1)+(w-1) = n+10h-10+w-1 = n+10h+w-11

Top left hand number x Bottom right hand number

n x (n+10h+w-11) = n + 10nh+nw-11n

Top right hand number x Bottom left hand number

(n+w-1)(n+10h-10)

which gives

nxn + nx10h +nx(-10)+ wxn+wx10h+wx(-10)+(-1)xn+(-1)x10h+(-1)x(-10)

which gives

n + 10hn - 10n +nw +10wh -10w - n - 10h + 10

collecting terms which gives

n + 10hn + nw - 11n + 10wh -10h - 10w + 10

Difference equals

( n + 10hn + nw - 11n + 10wh -10h - 10w + 10) - (n + 10nh+nw-11n)

Which gives

10wh - 10h - 10w + 10 = 10(wh - h - w + 1) = 10(w-1)(h-1)

The general rule for rectangles on a 10 long number grid is 10(w-1)(h-1)

(At this point this piece of work is worth about a grade B probably a mark of 6 in each of the three strands)

Extend investigation

Change the size of the number grid. E.g.

2 3 4 5 6 7

8 9 10 11 12 13 14

5 16 17 18 19 20 21

22 23 24 25 26 27 28

29 30 31 32 33 34 35

. . . . . . .

Let the length of the number grid be L.

In this example the length L = 7, each row going down adds on another 7 (same as the length of the number grid)

So in our proof for rectangles above, instead of adding 10 each time we must add on the length of the number grid (L) instead.

How do we get to the general rule for any size number grid?

Answer, we consider the proof for the 10 long number grid and change it slightly.

Proof : instead of 10 long change to L long

So, for any rectangle w x h

Starting position

Any number width = w

n n+1 n+2 n+3 . . n+(w-1) The amount added

n+(1xL) . . . . . . nto n is one less

height n+(2xL) . . . . . . than the square (w-1)

= h n+(3xL) . . . . . .

. . . . . . .

. . . . . . .

n+L(h-1) . . . . . n+10(h-1) +(w-1)

Each row goes up L, there are one less Need to add a further (w-1)

lots of L than the height h, so we need to add on to the first square in this row

(h-1) lots of L (same as the first row)

Now to find the difference for any size square.

Top left hand number = n

Bottom right hand number = n+L(h-1)+(w-1) = n+Lh-L+w-1

Top right hand number = n+(w-1) = n+w-1

Bottom left hand number = n+L(h-1) = n+Lh-L

Top left hand number x Bottom right hand number

n(n+Lh-L+w-1) = n + Lhn-Ln+wn-n

Top right hand number x Bottom left hand number

(n+w-1) x (n+Lh-L)

which gives

nxn + nxLh +nx(-L)+ wxn+wxLh+wx(-L)+(-1)xn+(-1)xLh+(-1)x(-L)

which gives

n + Lhn - Ln +nw +Lwh -Lw - n - Lh + L

Difference

( n + Lhn - Ln +nw +Lwh -Lw - n - Lh + L ) - (n + Lhn-Ln+wn-n)

Which gives

Lwh - Lh - Lw + L = L(wh - h - w + 1) = L(w-1)(h-1)

The general rule for rectangles on a L long number grid is L(w-1)(h-1)