## 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…

### 4 Responses to “Le Monde puzzle”

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