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Old problem bugging out

I tried to solve this problem (from a while ago):

https://brilliant.org/mathematics-problem/evaluating-a-function/?group=tUgjRZ0bXcg5

I, because of silly mistakes, used all three tries, and then wanted to view the solution. It said I had to spend points, so I did, but then the screen went blank and there was no solution. Even reloading now doesn't help.

Could you please help me either 1) view the solution on the actual page, or 2) help me get the solution?

Thanks, Michael

Note by Michael Tang
3 years, 11 months ago

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I can't figure out what I did wrong. Please help me.

(Spoilers!)

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So I got \(g(x+y) = g(x) + g(y) + xy\) as in the official solution (where \(g(t) = \dfrac{f(t)}{t+1}\).) Then I did \(x = y\) to get \(g(2x) = 2g(x) + x^2.\) After that I did \(y = 2x\) to get \[g(3x) = g(x) + g(2x) + 2x^2 = g(x) + [2g(x) + x^2] + 2x^2 \\ = 3g(x) + 3x^2.\] Sticking \(x = 1/3\) in we get \[g(1) = 3g(1/3) + 1/3\] or \[2 = 3g(1/3) + 1/3\] so \(g(1/3) = 5/9.\)

Then using the double formula we get

\[g(2/3) = 2g(1/3) + (1/3)^2 = 10/9 + 1/9 = 11/9.\]

Finally, plugging in \(y = 1\) and \(x = 2/3\) into the very first equation, we get

\[g(5/3) = g(2/3) + g(1) + 2/3 \\ = 11/9 + 2 + 2/3 = 35/9,\]

so \[f(5/3) = g(5/3) \cdot 3/8 = (35/9)(3/8) = 35/24.\]

Michael Tang - 3 years, 11 months ago

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It is all okay until the final step. Note that \[g(t)(t+1)=f(t)\] not \[f(t)(t+1)=g(t)\] Then, \[f(5/3)=g(5/3)\cdot 8/3=(35/9)(8/3)=280/27\]

Daniel Chiu - 3 years, 11 months ago

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Ohhhh.... of course I mess up the algebra at the very end. Thanks a lot!!

Michael Tang - 3 years, 11 months ago

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Hi Michael,

I think we fixed it, at least it is no longer bugging out for me. For the last few days we've had some finicky solution displaying behavior, but it should be all sorted out now.

Peter Taylor Staff - 3 years, 11 months ago

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Yes, it's fixed. Thanks so much!

Michael Tang - 3 years, 11 months ago

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