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(This article was first published on <strong><a href="https://xianblog.wordpress.com/2017/04/18/le-monde-puzzle-1003/"> R – Xi'an's Og</a></strong>, and kindly contributed to <a rel="nofollow" href="https://www.r-bloggers.com/">R-bloggers)</a> </div>
<p><strong><img height="185" alt="" width="273" src="https://encrypted-tbn3.gstatic.com/images?q=tbn:ANd9GcRsQd3v7efgSlmsNPKddzl-oQS9fZOF8a68V6KPmf5dnmI_EJ66rw">A</strong> purely arithmetic <a title="Le Monde puzzle [website]" target="_blank" rel="nofollow" href="https://xianblog.wordpress.com/2011/09/03/le-monde-puzzle-website/">Le Monde mathematical puzzle</a>:</p>
<blockquote>
<p><em>Find the four integers w, x, y, z such that the four smallest pairwise sums among the six pairwise sums are 59, 65, 66, and 69. Similarly, find the four smallest of the five integers v, x, y, z such that the five smallest pairwise sums among the ten pairwise sums are 56, 64 , 66, 69 and 70.<br>
</em></p>
</blockquote>
<p><strong>T</strong>he first question is rather straightforward since there are only two possible orderings when x≤y≤z≤w :</p>
<p>x+y≤x+z≤x+w≤y+z≤y+w≤z+w and x+y≤x+z≤y+z≤x+w≤y+w≤z+w</p>
<p>which means</p>
<p>x+y=59, x+z=65, x+w=66, y+z=69, and x+y=59, x+z=65, y+z=66, x+w=69</p>
<p>but since the first system does not allow for an integer solution, the only possibility is the second system, with solution (x,y,z,w)=(29,30,36,40). And the second question is of the same complexity, with, when x≤y≤z≤w≤v :</p>
<p>x+y=56, x+z=64, y+z=66, x+w=69, x+v=70 or x+y=56, x+z=64, x+w=66, x+v=69, y+z=70</p>
<p>with solutions (x,y,z,w,v)=(27,29,37,42,43) and (x,y,z,w,v)=(25,31,39,41,44).</p>
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