tag:blogger.com,1999:blog-2400513859305780710.post7337808365289210175..comments2023-04-04T07:53:53.789-07:00Comments on Mathematical Food For Thought: Egyptian. Topic: Number Theory. Level: AIME.Jeffrey Wanghttp://www.blogger.com/profile/11114458640271201663noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-2400513859305780710.post-7905175882702356462006-06-17T13:13:19.000-07:002006-06-17T13:13:19.000-07:00n^2+3n+2 factors as (n+1)(n+2)For all odd values o...n^2+3n+2 factors as (n+1)(n+2)<br><br>For all odd values of n, (n+1) is even and so contains 2 as a factor. So we want either odd values of n that are 2 less than a multiple of 3, or one less than a multiple of 6. There are 663 multiples of 3 less than 1991, 332 of which are odd - each one has a corresponding n value. There are 331 multiples of 6 - each one also has a corresponding n value, so a total of 663 odd values of n that work.<br><br>For even numbers, there are 331 values of n that result in (n+2) containing a factor of 6, and 332 values that make (n+1) contain a factor of 3.<br><br>Total number of possible values for n is then 1326 ... I thinkAdunakhornoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-51801855312030569432006-06-19T15:41:46.000-07:002006-06-19T15:41:46.000-07:00n^2 + 3n + 2 is always even, so we only want to co...n^2 + 3n + 2 is always even, so we only want to concern ourselves with divisibility b 3. We have it for n \equiv 1, 2 \bmod 3, and for each case we have 663 possibilities, for a total of 1326.QCnoreply@blogger.com