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Le Monde puzzle

The puzzle in Le Monde is quite straightforward (!) this weekend: it can be rewritten as to figure out the values of the sums

10! \displaystyle{\left.\left\{1 - \sum_{j_1=1}^{10} \frac{1}{j_1}\left\{1-\sum_{\stackrel{j_2=1}{j_2\ne j_1}}^{10}\frac{1}{j_2}\right\{1-\ldots\right.\right.}

\displaystyle{\qquad\ldots\left.\left.\left.\left\{1-\sum_{\stackrel{j_9=1}{j_9\ne j_1,\ldots,j_8}}^{10} \frac{1}{j_9}\right\}\ldots\right\}\right\}\right\}}

and

\displaystyle{\left.(49!)^2 \left\{1 - \sum_{j_1=1}^{49} \frac{1}{j_1^2}\left\{1-\sum_{\stackrel{j_2=1}{j_2\ne j_1}}^{49}\frac{1}{j_2^2}\right\{1-\ldots\qquad\right.\right.}

\displaystyle{\qquad\ldots\left.\left.\left.\left\{1-\sum_{\stackrel{j_{48}=1}{j_{48}\ne j_1,\ldots,j_{47}}}^{49} \frac{1}{j_{48}^2}\right\}\ldots\right\}\right\}\right\}}

which are easily displayed and as easily solved.

The first sum can indeed be written as

\displaystyle{\sum_{i=1}^k (-1)^{k-i} \sum_{\stackrel{j_1\ne\cdots\ne j_i}{\in\{1,\ldots,k\}}} \prod_{u=1}^i j_u}

for k=10. This is simply

\displaystyle{\sum_{i=0}^k (-1)^{k-i} \sum_{\stackrel{j_1\ne\cdots\ne j_i}{\in\{1,\ldots,k\}}} \prod_{u=1}^i j_u}- (-1)^k = \prod_{i=1}^k (i-1) - (-1)^k

and the solution is thus (-1)^{k+1}, equal to -1 for k=10. Once we realise the fact this is a product with one missing term, the second sum is similar: it can be written as

\displaystyle{\sum_{i=0}^k (-1)^{k-i} \sum_{\stackrel{j_1\ne\cdots\ne j_i}{\in\{1,\ldots,k\}}} \prod_{u=1}^i j_u^2}- (-1)^k

\qquad = \prod_{i=1}^k (i^2-1) - (-1)^k = - (-1)^k

This is equal to 1 for k=49. A bit disappointing because it amounts to reformulate the question with the proper algebraic formula…

This entry was posted on February 21, 2010 at 7:35 am and is filed under Statistics with tags , . You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

4 Responses to “Le Monde puzzle”

  1. Le Monde puzzle [52] « Xi'an's Og Says:
    December 31, 2010 at 1:14 pm

    […] Monde puzzle [52] The last puzzle of the year in Le Monde reads as follows (as far as I understand its wording!): Iter(n,x,y) is the […]

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  2. New Le Monde puzzle « Xi'an's Og Says:
    March 4, 2010 at 6:49 am

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  3. Welcome, Robin! « Xi'an's Og Says:
    February 26, 2010 at 12:07 am

    […] Robin Ryder started his new blog with his different solutions to Le Monde puzzle, solutions that are much more elegant than my pedestrian rendering. I particularly like the one […]

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  4. Le Monde puzzle on sums of products « Robin Ryder's Blog Says:
    February 25, 2010 at 6:53 pm

    […] Monde puzzle on sums of products By robinryder Christian Robert is disappointed by last week-end’s Le Monde’s mathematical problem : Take the integers 1 to 10. Group […]

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