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What else can I help you with?
How do you solve an arithmagon?
okay, so obviously you already know what an arithmagon is, otherwise you wouldn't have asked this question, but because it really needs to be shown to you, a good definition of an arithmagon is on this website; http://www.nzmaths.co.nz/algebra/units/arithmagons.aspx hope this helps!
Can you find the circle numbers if you only have the box numbers in a triangular arithmagon?
yesThe way to solve an arithmagon is to take the 3 box numbers, add them up and divide by 2. That gives your "center" number. For each corner circle, take the opposite box and subtract from the center number.As long as you are allowed to have fractions and/or negative numbers, I think this method always works. Do you have an example of an impossible arithmagon?For example:a -- 7 -- b.\ ........ /. 8 ..... 9... \ .. /..... cAdd up 7+8+9 = 24Divide by 2 = 12a = 12 - 9 = 3b = 12 - 8 = 4c = 12 - 7 = 5
How do you solve an arithmagon?
okay, so obviously you already know what an arithmagon is, otherwise you wouldn't have asked this question, but because it really needs to be shown to you, a good definition of an arithmagon is on this website; http://www.nzmaths.co.nz/algebra/units/arithmagons.aspx hope this helps!
What is a arithmagon?
It is a triangle with 3 rectangles and 3 circles!
What is an square Arithmagon?
A square Arithmagon is a specific type of polygon, particularly a quadrilateral, where the vertices are labeled with numbers such that the numbers on the edges equal the sums of the numbers at the opposite vertices. In the case of a square Arithmagon, it has four vertices and four edges, with each edge representing the sum of the two adjacent vertices. This structure often leads to interesting mathematical properties and relationships among the vertex values and the edge sums.
How do you solve the arithmagon with products 30.38 24.18 and 19.11?
6.2 4.9 3.9 by mj yap g6 cbc-clc
Can you find the circle numbers if you only have the box numbers in a triangular arithmagon?
yesThe way to solve an arithmagon is to take the 3 box numbers, add them up and divide by 2. That gives your "center" number. For each corner circle, take the opposite box and subtract from the center number.As long as you are allowed to have fractions and/or negative numbers, I think this method always works. Do you have an example of an impossible arithmagon?For example:a -- 7 -- b.\ ........ /. 8 ..... 9... \ .. /..... cAdd up 7+8+9 = 24Divide by 2 = 12a = 12 - 9 = 3b = 12 - 8 = 4c = 12 - 7 = 5
