2016 Putnam Problem B1

2016 Putnam B1. Let x_0, x_1, x_2, \dots be the sequence such that x_0 = 1 and for n \geq 0, x_{n+1} = \ln(e^{x_n} - x_n) (as usual, the function  \ln is the natural logarithm.

Show that the infinite series x_0 + x_1 + x_2 + \dots converges and find its sum.

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Data on IMO results

Following the recent IMO 2016, I have been meaning to do some analysis on IMO results. Unfortunately I have not had time to do so…

In the meantime, I thought I’d share the data I’ve scraped so far so that others who have the time and interest might have a go at analysing the data. Data is available at my Github repo.

All data was scraped from imo-official.org; the scripts I used to scrape them are in the ETL folder of the same repo. The data generally looks clean except some minor issues for “Contestant” (i.e. names of contestants). For example, my name is written in one order for 2003 and 2005 but in a different way for 2004. I have no idea how widespread this issue is, although a cursory glance at contestants in my country suggest the issue is a minor one.

Posted in Grade 12, Intl/Regional | Tagged | 1 Comment

IMO 2016

This year’s International Mathematical Olympiad (IMO) took place in Hong Kong from 6-16 July. The problems can be downloaded from this page or viewed at the Art of Problem Solving (AoPS) forum page for IMO 2016 (here).

Congratulations to my country, Singapore’s team, for coming in 4th overall! And for winning the country’s:

  • Best medal haul (4G 2S, previous best was 4G 1S 1B in 2011)
  • Highest absolute score (196 out of a possible 252, previous best was 179 in 2011)
  • 2nd best overall team ranking (4th, best was 3rd in 2011)

Some factoids on the results:

  • The Gold, Silver and Bronze medal score cutoffs were 29, 22 and 16 respectively. Nothing too out of the ordinary. (Interestingly, IMO 2014 had the exact same cutoffs.)
  • The top 3 countries were the United States, South Korea and China. (These 3 countries were the top 3 in 2015 as well, just a different order: USA, China, South Korea.)
    • Team USA won a best possible 6 Gold medals. (Next best was 4G 2S by South Korea, China and Singapore.)
  • There were 6 contestants with perfect scores: 3 from South Korea, 2 from USA and 1 from China.
    • 3 of them were first-time participants of the competition. 2 of them had won Gold in 2015, while the remaining contestant (Allen Liu, USA) had won Gold in 2014 and 2015.
  • Question 3 was the most difficult question, with a mean score of 0.251, 91% of students scoring 0 points and only 10 contestants scoring a perfect 7 points. The easiest was Question 1 with a mean score of 5.272. (The questions from easiest to hardest: 1, 4, 2, 5, 6, 3.)

And finally, some useful links:

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[Soln] 2013 AIME I Problem 15

2013 AIME I 15. Let N be the number of ordered triples (A, B, C) of integers satisfying the conditions

(a) 0 \leq A < B < C \leq 99,
(b) there exist integers a, b, and c, and prime p where 0 \leq b < a < c < p,
(c) p divides A - a, B - b, and C - c, and
(d) each ordered triple (A, B, C) and each ordered triple (b, a, c) form arithmetic sequences.

Find N.

Click for solution

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2013 AIME I Problem 15

2013 AIME I 15. Let N be the number of ordered triples (A, B, C) of integers satisfying the conditions

(a) 0 \leq A < B < C \leq 99,
(b) there exist integers a, b, and c, and prime p where 0 \leq b < a < c < p,
(c) p divides A - a, B - b, and C - c, and
(d) each ordered triple (A, B, C) and each ordered triple (b, a, c) form arithmetic sequences.

Find N.

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[Soln] 2013 AIME I Problem 14

2013 AIME I 14. For \pi \leq \theta < 2\pi, let

\begin{aligned} P &= \displaystyle\frac{1}{2}\cos \theta - \frac{1}{4} \sin 2\theta - \frac{1}{8} \cos 3 \theta + \frac{1}{16} \sin 4\theta + \frac{1}{32}\cos 5\theta \\  &\hspace{2em} - \frac{1}{64}\sin 6\theta - \frac{1}{128} \cos 7\theta + \dots \end{aligned}

and

\begin{aligned} Q &= 1 - \displaystyle\frac{1}{2}\sin \theta - \frac{1}{4} \cos 2\theta + \frac{1}{8} \sin 3 \theta + \frac{1}{16} \cos 4\theta - \frac{1}{32}\sin 5\theta \\  &\hspace{2em} - \frac{1}{64}\cos 6\theta + \frac{1}{128} \sin 7\theta + \dots \end{aligned}

so that \frac{P}{Q} = \frac{2\sqrt{2}}{7}. Then \sin \theta = -\frac{m}{n} where m and n are relatively prime positive integers. Find m + n.

Click for solution

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2013 AIME I Problem 14

2013 AIME I 14. For \pi \leq \theta < 2\pi, let

\begin{aligned} P &= \displaystyle\frac{1}{2}\cos \theta - \frac{1}{4} \sin 2\theta - \frac{1}{8} \cos 3 \theta + \frac{1}{16} \sin 4\theta + \frac{1}{32}\cos 5\theta \\  &\hspace{2em} - \frac{1}{64}\sin 6\theta - \frac{1}{128} \cos 7\theta + \dots \end{aligned}

and

\begin{aligned} Q &= 1 - \displaystyle\frac{1}{2}\sin \theta - \frac{1}{4} \cos 2\theta + \frac{1}{8} \sin 3 \theta + \frac{1}{16} \cos 4\theta - \frac{1}{32}\sin 5\theta \\  &\hspace{2em} - \frac{1}{64}\cos 6\theta + \frac{1}{128} \sin 7\theta + \dots \end{aligned}

so that \frac{P}{Q} = \frac{2\sqrt{2}}{7}. Then \sin \theta = -\frac{m}{n} where m and n are relatively prime positive integers. Find m + n.

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