Anyone know the answers and working out for these two questions to do with problem solving and equations?

Question

I need help with the following questions:


1) The sum of four consecutive ODD numbers is 216. Find the numbers.


2) In an arithmagon, the number in a square is the sum of the two numbers in the two circles either side of it. Find x in these arithmagons.

(a) http://www.picturehost.co.uk/links/arith...

(b) http://www.picturehost.co.uk/links/arith...

(c) http://www.picturehost.co.uk/links/arith...


Thanks,
p_q

Here are the fixed links for Q2:

2)

(a) http://www.imagehosting.com/show.php/181151_arithmagongif.GIF.html

(b) http://www.imagehosting.com/show.php/181151_arithmagongif.GIF.html

(c) http://www.imagehosting.com/show.php/181160_Arithmagon3gif.GIF.html

TAKE NOTE: YOU WILL HAVE TO VIEW THEM QUICKLY BEFORE THEY EXPIRE

Answer 1

1. X + X+2 + X+4 + x+6 = 216

Solve for X, pick the next 3 consecutive odd numbers.

2. a
x is the top circle
y is the left bottom circle
z is the right bottom circle

So for each square:
26=x+y
18=x+z
22=y +z

You can then find x by solving this system of equations.

Answer 2

Can only help you with #1 right now. Four consecutive odd numbers can be represented as: x + 1, x + 3, x + 5, and x + 7. Now, (x + 1) + (x + 3) + ( x + 5) + ( x + 7) = 216. Now simplify: 4x + 16 = 216 => 4x = 200 => x = 50. So, the four odd numbers are 51, 53, 55, and 57. I'll get back to you on #2 when I research the damn things. Never dealt with them before.

Answer 3

I'll help you out with #1 at least.
n+n+2+n+4+n+6=216
4n+12=216
4n=204
n=51
the 4 numbers are 51, 53, 55, and 57
sorry, don't know how to do the second one.