In a set of five consecutive integers, the largest integer is less than twice the average of the five integers. What is the smallest integer that could be in the set?
Let s be the smallest integer. Then the 5 numbers are: s, s+1, s+2, s+3, s+4
Their average is (5s + 10)/5 or s + 2
So: s + 4 < 2(s + 2)
s + 4 < 2s + 4
So the smallest integer greater than 0 is 1.
Edited to correct mistake. Thanks to Guest below for noting this.
If s is the smallest of the five integers, the mean is \((5s+10)/5=s+2\) (not (s+5) as previously suggested). We therefore require \(s+4<2(s+2)\), which reduces to \(s>0\). Thus the smallest integer must be 1 or greater.
As a reality check note that if s=1 we have the set 1,2,3,4,5 with average 3 and \(2\times3\) is indeed greater than 5, but with s=0 we have 0,1,2,3,4 with mean 2 and since \(2\times2=4\) the inequality just fails.