<PRE><B>Question 1045002</B>

How can the problem: Find 3 consecutive odd integers such that 4 times the first is 44 more than the sum of the second and third?
&lt;pre&gt;Let &lt;b&gt;&lt;u&gt;1&lt;sup&gt;st&lt;/b&gt;&lt;/u&gt;&lt;/sup&gt; integer be F
Then &lt;b&gt;&lt;u&gt;2&lt;sup&gt;nd&lt;/b&gt;&lt;/u&gt;&lt;/sup&gt; and &lt;b&gt;&lt;u&gt;3&lt;sup&gt;rd&lt;/b&gt;&lt;/u&gt;&lt;/sup&gt; are: F + 2, and F + 4, respectively
We then get: 4F = F + 2 + F + 4 + 44
4F = 2F + 50
4F - 2F = 50
2F = 50
F, or &lt;b&gt;&lt;u&gt;1&lt;sup&gt;st&lt;/b&gt;&lt;/u&gt;&lt;/sup&gt; integer = {{{50/2}}}, or {{{highlight_green(25)}}}
The &lt;b&gt;&lt;u&gt;2&lt;sup&gt;nd&lt;/b&gt;&lt;/u&gt;&lt;/sup&gt; and &lt;b&gt;&lt;u&gt;3&lt;sup&gt;rd&lt;/b&gt;&lt;/u&gt;&lt;/sup&gt; integers are: {{{highlight_green(matrix(1,9, 25 + 2, or, 27, and, 25 + 4, or, 29, &quot;,&quot;, respectively))}}}
Yes, it is that simple...not complex at all!</PRE>