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?

PeculiarVi Aug 18, 2018

#1**+2 **

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

s >0

So the smallest integer greater than 0 is 1.

Edited to correct mistake. Thanks to Guest below for noting this.

Alan Aug 18, 2018

#2**0 **

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.

Guest Aug 18, 2018