A cute problem form India's UPSC exam

 UPSC is India's civil services exam, and surprisingly, the following cute problem appeared. Admittedly, it was a multiple choice question, and could be guessed easily.

Anyway, here is the question:

$a,b,c$ are positive integers such that 

$$\dfrac{bc + 1}{abc + a + c} = \frac{11}{43}$$

Then what are the possible values of $abc$?

Solution.

We can find all solutions. In fact there is a unique solution.

Take the reciprocal and do some minor algebra to get

$$a + \dfrac{c}{bc + 1} = \dfrac{43}{11} = 3 + \dfrac{10}{11}$$ 

Since $\frac{c}{bc + 1} \lt 1$, we must have that $a=3$ and $\frac{c}{bc+1} = \frac{10}{11}$.

Taking reciprocal again gives

$$b + \dfrac{1}{c} = \dfrac{11}{10} = 1 + \dfrac{1}{10}$$

$c = 1$ is not possible (LHS is an integer, while RHS is not). Thus $\frac{1}{c} \lt 1$ and so

$b = 1$ and $c=10$

Thus $(a,b,c) = (3,1,10)$ is the unique solution and $abc = 30$.

Powers of 2 series convergence from Peter Winkler

 Here is a cute problem from Peter Winkler.

The series below converges in the interval $[0,1)$

$$ x - x^2 + x^4 - x^8 + \dots + (-1)^n x^{2^n} + \dots $$

For $x \in [0, 1)$, call the value $f(x)$.

The question is, does the limit 

$$ \lim_{x \to 1-} f(x) $$

exist?

Solution.

A proof by Noam Elkies goes as follows (rephrased, not verbatim):

We can show that for $x \in [0,1)$, 

$$f(x) + f(x^2) = x $$

and so (by a repeated application of the above)

$$ f(x) = x - x^2 + f(x^4)$$

By the first identity, if the limit exists, it must be $\frac{1}{2}$.

Now for $0 \lt x \lt 1$ we must have that (using the second identity above)

$$ f(x) \gt f(x^4) \gt f(x^{16}) \gt \dots \gt f(x^{4^n}) \gt \dots$$

Thus if there was some $c$ such that $f(c) \gt \frac{1}{2}$, we must have that

$$f(c^{4^{-n}}) \gt f(c) \gt \frac{1}{2}$$

And since $c^{4^{-n}} \to 1$ as $n \to \infty$, the limit cannot be $\frac{1}{2}$ if $f(c) \gt \frac{1}{2}$.

A quick computation (wrote python code) upto $15$ terms shows that $f(0.9999) \gt 0.5006$.