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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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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 $f(5/3) = g(5/3) \cdot 3/8 = (35/9)(3/8) = 35/24.$

- 3 years, 11 months ago

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

- 3 years, 11 months ago

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

- 3 years, 11 months ago

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.

Staff - 3 years, 11 months ago

Yes, it's fixed. Thanks so much!

- 3 years, 11 months ago