Question 971109
what is the smallest of 3 consecutive positive integers if the product of the
smaller two integers is 11 more than 5 times the largest integer 
<pre>
{{{n(n+1)}}}{{{""=""}}}{{{5(n+2)+11}}}

{{{n^2+n}}}{{{""=""}}}{{{5n+10+11}}}

{{{n^2+n}}}{{{""=""}}}{{{5n+21}}}

Get 0 on the right side:

{{{n^2-4n-21}}}{{{""=""}}}{{{"0"}}}

Factor the left side:

{{{(n-7)(n+3)}}}{{{""=""}}}{{{"0"}}}

Set each factor = 0

n-7 = 0;  n+3 = 0
  n = 7;    n = -3

Since the problem specifically states "positive": 
we ignore the negative answer.

Answer: 7,8,9

Checking product of smaller two = 7*8 = 56

5 times the largest = 5*9 = 45

11 more than 45 is 56.

So the solution 7,8,9 is correct.
  
Edwin</pre>