Put 1, 2, 3, 4, 8 and 12 in the circle so that the product along the sides of triangle are equal.
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Put 1, 2, 3, 4, 8 and 12 in the circle so that the product along the sides of triangle are equal.
So... what did you do?
I'm thinking that perhaps the correct set of numbers is this: 1, 2, 3, 4, 6, and 12.
Well, the question sure is sort of ambiguous, but if one has to put three numbers on one side of the triangle, two on the vertex and one in the middle so that the two on the vertex are used in the next two multiplications, then the 8 fits perfectly in. Granted, you can get a similar result with 6 instead of 8.
I'm guessing that the triangle has one number on each vertex and one number in the middle of each edge. In the attached picture you want A x B x D = A x C x F = D x E x F. But like I said, it's just a guess.
Ah, yup, I got 8 to fit in there well :)
Just like 6 instead of 8.
Ok Jerry, you made me rethink it, and yes indeed the numbers 1,2,3,4,8,and 12 can be made to work. Thanks for making me take another shot at it.
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