[Guest]

Question 1

(x-a)(x-b)(x-c),,,(x-z)=

Solve for

Question 2

You have a chessboard, on which the two opposing diagonal corner squares have been removed. You have an infinite number of domino's, each of which cover exactly two squares on the chessboard. Without any domino overlapping another, and no domino hanging over an edge (So each domino must cover two squares) can you completely cover the chessboard with domino's?

Question 3

You have 10 trees in pots. Is it possible to plant the trees in five rows of four trees?

[Guest]

Question 3

You have 10 trees in pots. Is it possible to plant the trees in five rows of four trees?

If you plant all 10 trees in a single line, you'd effectively have 7 rows (linear groupings) of 4 trees.

Edited July 1, 2009 by LichWriter

[Guest]

Question 1

(x-a)(x-b)(x-c),,,(x-z)=

Solve for

[Guest]

For Questions 3

plant them so they make the shape of a five pointer star. With one tree at each of the points, and on tree where the two "lines" would cross

[Guest]

Question 1

(x-a)(x-b)(x-c),,,(x-z)=

Solve for

my apologies for empty post

= 0, since somewhere in the equation ...(x-x)... appears which is obviously 0, so is 0

[Guest]

This was done recently on the forum. There are 32 black and white square, getting rid of 2 diagonal squares will leave only 30 of one color. Since each domino covers 1 black and 1 white square it's impossible to cover 32 black and 30 white square with dominoes.

[Guest]

At some point you find the term (x-x), which equals 0. So you don't need to know any of the variables; one zero in the mix and the whole problem will always equal 0.

[Guest]

The third problem can be extended to planting n trees in n/2 rows of 4 trees, where n/2 is an odd number.

The answer always works in the same way, using end-points and crossing-points of lines which make an n/2 -pointed star

for example, 14 trees can be planted in 7 rows of 4 trees each, etc.

Donjar

[Guest]

where n/2 is an odd number.

Good observation, Donjar.

To add, I believe that "n" need only be greater than 4, not only odd.

[Guest]

Um...technically the expression in question one is already solved for

. "Find the value of " would be a more appropriate request, in which case the previous couple of posters are correct, =0

[Guest]

Zero. (x-x) = 0. Anything and everything times zero, is zero.

Yes. On the sides, have pieces going the opposite way than the inner pieces, tracing a square ended zig-zag from blank to blank.

Draw a circle of 5 points, evenly spaced. Draw straight lines to connect each of the points to eachother. At each intersection put a point. There will be 5 sets of 4, because each segment (5 segments) will have 2 endpoints and two intersections (2+2=4).

Yay. Time for me to sleep.

[Guest]

Zero. (x-x) = 0. Anything and everything times zero, is zero.

Nice one!

Yes. On the sides, have pieces going the opposite way than the inner pieces, tracing a square ended zig-zag from blank to blank.

Sorry, dude. Incorrect.

Draw a circle of 5 points, evenly spaced. Draw straight lines to connect each of the points to eachother. At each intersection put a point. There will be 5 sets of 4, because each segment (5 segments) will have 2 endpoints and two intersections (2+2=4).

Yay. Time for me to sleep.

Nice one!