https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_69&feed=atom&action=history 2008 iTest Problems/Problem 69 - Revision history 2021-05-10T00:07:56Z Revision history for this page on the wiki MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_69&diff=97346&oldid=prev Rockmanex3: Solution to Problem 69 — pentagonal numbers 2018-08-22T14:19:57Z <p>Solution to Problem 69 — pentagonal numbers</p> <p><b>New page</b></p><div>==Problem==<br /> <br /> In the sequence in the previous problem, how many of &lt;math&gt;u_1,u_2,u_3,\ldots, u_{2008}&lt;/math&gt; are pentagonal numbers? <br /> <br /> '''Note: The sequence is &lt;math&gt;1,2,5,6,9,12 \cdots,&lt;/math&gt; where the first term is 1, the next 2 terms are congruent to 2 modulo 3, the next 3 terms are congruent to 3 modulo 3, and so on.'''<br /> <br /> ==Solution==<br /> <br /> The first few pentagonal numbers are &lt;math&gt;1,5,12,22 \cdots,&lt;/math&gt; which seems to match the last term of each grouping. Notice that the common differences are the same.<br /> <br /> &lt;br&gt;<br /> By checking with the formula &lt;math&gt;p_n = \tfrac{3n^2 - n}{2}&lt;/math&gt; (or using a visual representation of pentagonal numbers), we confirm that the first difference is always one more than a multiple of 3, and the second difference is 3. This confirms that the last number of each grouping are pentagonal numbers. In addition, the last number of each grouping is term number &lt;math&gt;\tfrac{n(n+1)}{2}.&lt;/math&gt;<br /> <br /> &lt;br&gt;<br /> With this in mind, the highest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;\tfrac{n(n+1)}{2} &lt; 2008&lt;/math&gt; is &lt;math&gt;62&lt;/math&gt;, so there are &lt;math&gt;\boxed{62}&lt;/math&gt; pentagonal numbers in the sequence.<br /> <br /> ==See Also==<br /> {{2008 iTest box|num-b=68|num-a=70}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Rockmanex3