catmg

f^1 (1) = 1/sqrt(2)

f^2 (1) = 1/sqrt(3)

f^3 (1) = 1/sqrt(4)

Do you see a pattern?

I hope this helped. :))

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1 + 2 + 3 + 4  + ... + 10 = 55

Let the number she didn't include be called x. 

55 - x = a square number. 

x has to be in the interval [1, 10]

The largest square number less than 55 is 49. 

55 - x = 49

x = 6

I hope this helped. :))

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Nice, but I think there is something wrong with the colours. 

60 will result in 3 terms. 

120 will result in 2 terms.

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For a polynomial with rational coefficients, the the irrational roots come in pairs. 

If one root is 1 + sqrt(2), the other one must be 1 - sqrt(2). 

Similarly, if one root is 2 + sqrt(7), the other one must be 2 - sqrt(7). 

Now that the roots are found, you can find the factored form of the quartic. :))

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Hello :))

To start, the x intercept is found when y = 0, and the x intercept is found when x = 0. 

x intercept

0 = -3/2x + 6

y intercept

y = -3/4(0) + 6

Any idea how to go from here?

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Hi Melody,

I love your diagrams, they are always so helpful for visualizing the problem. 

One small note, since one length is in cm and the other is in mm, I think it would be helpful to convert one of them so that the units are the same. 

550mm = 55cm

:))

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As GA pointed out, this problem is not just asking for integer solutions. 

This solution would be great if x was an integer. :))

Be careful with these small details.

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I suggest trying to connect the center of the circles and then drawing lines from those centers onto the belt. :))

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Yep, when you're completely stuck on a problem, it's good to sometimes just bash it out.

Try 2, 3, 5, 7 then 3, 5, 7, 11 then 5, 7, 11, 13. 

This works well on this problem becuase it asks for the smallest value, meaning we can test from small to big sums to guarentee there isn't a smaller solution that we didn't find. 

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